message
stringlengths 2
23.4k
| message_type
stringclasses 2
values | message_id
int64 0
1
| conversation_id
int64 129
108k
| cluster
float64 6
6
| __index_level_0__
int64 258
216k
|
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Codehorses has just hosted the second Codehorses Cup. This year, the same as the previous one, organizers are giving T-shirts for the winners.
The valid sizes of T-shirts are either "M" or from 0 to 3 "X" followed by "S" or "L". For example, sizes "M", "XXS", "L", "XXXL" are valid and "XM", "Z", "XXXXL" are not.
There are n winners to the cup for both the previous year and the current year. Ksenia has a list with the T-shirt sizes printed for the last year cup and is yet to send the new list to the printing office.
Organizers want to distribute the prizes as soon as possible, so now Ksenia is required not to write the whole list from the scratch but just make some changes to the list of the previous year. In one second she can choose arbitrary position in any word and replace its character with some uppercase Latin letter. Ksenia can't remove or add letters in any of the words.
What is the minimal number of seconds Ksenia is required to spend to change the last year list to the current one?
The lists are unordered. That means, two lists are considered equal if and only if the number of occurrences of any string is the same in both lists.
Input
The first line contains one integer n (1 ≤ n ≤ 100) — the number of T-shirts.
The i-th of the next n lines contains a_i — the size of the i-th T-shirt of the list for the previous year.
The i-th of the next n lines contains b_i — the size of the i-th T-shirt of the list for the current year.
It is guaranteed that all the sizes in the input are valid. It is also guaranteed that Ksenia can produce list b from the list a.
Output
Print the minimal number of seconds Ksenia is required to spend to change the last year list to the current one. If the lists are already equal, print 0.
Examples
Input
3
XS
XS
M
XL
S
XS
Output
2
Input
2
XXXL
XXL
XXL
XXXS
Output
1
Input
2
M
XS
XS
M
Output
0
Note
In the first example Ksenia can replace "M" with "S" and "S" in one of the occurrences of "XS" with "L".
In the second example Ksenia should replace "L" in "XXXL" with "S".
In the third example lists are equal.
|
instruction
| 0
| 6,572
| 6
| 13,144
|
Tags: greedy, implementation
Correct Solution:
```
n = int(input())
a = []
b = []
for i in range(n):
a.append(input())
for i in range(n):
k = input()
if k in a:
a.remove(k)
else:
b.append(k)
c = [0,0,0,0]
for i in range(len(a)):
c[len(a[i])-1] += 1
print(sum(c))
```
|
output
| 1
| 6,572
| 6
| 13,145
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Codehorses has just hosted the second Codehorses Cup. This year, the same as the previous one, organizers are giving T-shirts for the winners.
The valid sizes of T-shirts are either "M" or from 0 to 3 "X" followed by "S" or "L". For example, sizes "M", "XXS", "L", "XXXL" are valid and "XM", "Z", "XXXXL" are not.
There are n winners to the cup for both the previous year and the current year. Ksenia has a list with the T-shirt sizes printed for the last year cup and is yet to send the new list to the printing office.
Organizers want to distribute the prizes as soon as possible, so now Ksenia is required not to write the whole list from the scratch but just make some changes to the list of the previous year. In one second she can choose arbitrary position in any word and replace its character with some uppercase Latin letter. Ksenia can't remove or add letters in any of the words.
What is the minimal number of seconds Ksenia is required to spend to change the last year list to the current one?
The lists are unordered. That means, two lists are considered equal if and only if the number of occurrences of any string is the same in both lists.
Input
The first line contains one integer n (1 ≤ n ≤ 100) — the number of T-shirts.
The i-th of the next n lines contains a_i — the size of the i-th T-shirt of the list for the previous year.
The i-th of the next n lines contains b_i — the size of the i-th T-shirt of the list for the current year.
It is guaranteed that all the sizes in the input are valid. It is also guaranteed that Ksenia can produce list b from the list a.
Output
Print the minimal number of seconds Ksenia is required to spend to change the last year list to the current one. If the lists are already equal, print 0.
Examples
Input
3
XS
XS
M
XL
S
XS
Output
2
Input
2
XXXL
XXL
XXL
XXXS
Output
1
Input
2
M
XS
XS
M
Output
0
Note
In the first example Ksenia can replace "M" with "S" and "S" in one of the occurrences of "XS" with "L".
In the second example Ksenia should replace "L" in "XXXL" with "S".
In the third example lists are equal.
|
instruction
| 0
| 6,573
| 6
| 13,146
|
Tags: greedy, implementation
Correct Solution:
```
n = int(input())
a , b = [] , []
for i in range(n):
x = input()
a.append(x)
for i in range(n):
x = input()
if x in a :
a.remove(x)
print(len(a))
```
|
output
| 1
| 6,573
| 6
| 13,147
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Codehorses has just hosted the second Codehorses Cup. This year, the same as the previous one, organizers are giving T-shirts for the winners.
The valid sizes of T-shirts are either "M" or from 0 to 3 "X" followed by "S" or "L". For example, sizes "M", "XXS", "L", "XXXL" are valid and "XM", "Z", "XXXXL" are not.
There are n winners to the cup for both the previous year and the current year. Ksenia has a list with the T-shirt sizes printed for the last year cup and is yet to send the new list to the printing office.
Organizers want to distribute the prizes as soon as possible, so now Ksenia is required not to write the whole list from the scratch but just make some changes to the list of the previous year. In one second she can choose arbitrary position in any word and replace its character with some uppercase Latin letter. Ksenia can't remove or add letters in any of the words.
What is the minimal number of seconds Ksenia is required to spend to change the last year list to the current one?
The lists are unordered. That means, two lists are considered equal if and only if the number of occurrences of any string is the same in both lists.
Input
The first line contains one integer n (1 ≤ n ≤ 100) — the number of T-shirts.
The i-th of the next n lines contains a_i — the size of the i-th T-shirt of the list for the previous year.
The i-th of the next n lines contains b_i — the size of the i-th T-shirt of the list for the current year.
It is guaranteed that all the sizes in the input are valid. It is also guaranteed that Ksenia can produce list b from the list a.
Output
Print the minimal number of seconds Ksenia is required to spend to change the last year list to the current one. If the lists are already equal, print 0.
Examples
Input
3
XS
XS
M
XL
S
XS
Output
2
Input
2
XXXL
XXL
XXL
XXXS
Output
1
Input
2
M
XS
XS
M
Output
0
Note
In the first example Ksenia can replace "M" with "S" and "S" in one of the occurrences of "XS" with "L".
In the second example Ksenia should replace "L" in "XXXL" with "S".
In the third example lists are equal.
|
instruction
| 0
| 6,574
| 6
| 13,148
|
Tags: greedy, implementation
Correct Solution:
```
#!/usr/bin/env python3
A = []
B = []
N = int(input())
for i in range(N):
A.append(input())
for i in range(N):
B.append(input())
ans = 0
for i in B:
if i in A:
x = A.index(i)
A = A[:x] + A[x + 1:]
else:
ans += 1
print(ans)
```
|
output
| 1
| 6,574
| 6
| 13,149
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Codehorses has just hosted the second Codehorses Cup. This year, the same as the previous one, organizers are giving T-shirts for the winners.
The valid sizes of T-shirts are either "M" or from 0 to 3 "X" followed by "S" or "L". For example, sizes "M", "XXS", "L", "XXXL" are valid and "XM", "Z", "XXXXL" are not.
There are n winners to the cup for both the previous year and the current year. Ksenia has a list with the T-shirt sizes printed for the last year cup and is yet to send the new list to the printing office.
Organizers want to distribute the prizes as soon as possible, so now Ksenia is required not to write the whole list from the scratch but just make some changes to the list of the previous year. In one second she can choose arbitrary position in any word and replace its character with some uppercase Latin letter. Ksenia can't remove or add letters in any of the words.
What is the minimal number of seconds Ksenia is required to spend to change the last year list to the current one?
The lists are unordered. That means, two lists are considered equal if and only if the number of occurrences of any string is the same in both lists.
Input
The first line contains one integer n (1 ≤ n ≤ 100) — the number of T-shirts.
The i-th of the next n lines contains a_i — the size of the i-th T-shirt of the list for the previous year.
The i-th of the next n lines contains b_i — the size of the i-th T-shirt of the list for the current year.
It is guaranteed that all the sizes in the input are valid. It is also guaranteed that Ksenia can produce list b from the list a.
Output
Print the minimal number of seconds Ksenia is required to spend to change the last year list to the current one. If the lists are already equal, print 0.
Examples
Input
3
XS
XS
M
XL
S
XS
Output
2
Input
2
XXXL
XXL
XXL
XXXS
Output
1
Input
2
M
XS
XS
M
Output
0
Note
In the first example Ksenia can replace "M" with "S" and "S" in one of the occurrences of "XS" with "L".
In the second example Ksenia should replace "L" in "XXXL" with "S".
In the third example lists are equal.
|
instruction
| 0
| 6,575
| 6
| 13,150
|
Tags: greedy, implementation
Correct Solution:
```
n = int(input())
a = []
for i in range(n):
a.append(input())
b = []
for i in range(n):
cur = input()
if cur in a:
a.remove(cur)
else:
b.append(cur)
print(len(b))
```
|
output
| 1
| 6,575
| 6
| 13,151
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Codehorses has just hosted the second Codehorses Cup. This year, the same as the previous one, organizers are giving T-shirts for the winners.
The valid sizes of T-shirts are either "M" or from 0 to 3 "X" followed by "S" or "L". For example, sizes "M", "XXS", "L", "XXXL" are valid and "XM", "Z", "XXXXL" are not.
There are n winners to the cup for both the previous year and the current year. Ksenia has a list with the T-shirt sizes printed for the last year cup and is yet to send the new list to the printing office.
Organizers want to distribute the prizes as soon as possible, so now Ksenia is required not to write the whole list from the scratch but just make some changes to the list of the previous year. In one second she can choose arbitrary position in any word and replace its character with some uppercase Latin letter. Ksenia can't remove or add letters in any of the words.
What is the minimal number of seconds Ksenia is required to spend to change the last year list to the current one?
The lists are unordered. That means, two lists are considered equal if and only if the number of occurrences of any string is the same in both lists.
Input
The first line contains one integer n (1 ≤ n ≤ 100) — the number of T-shirts.
The i-th of the next n lines contains a_i — the size of the i-th T-shirt of the list for the previous year.
The i-th of the next n lines contains b_i — the size of the i-th T-shirt of the list for the current year.
It is guaranteed that all the sizes in the input are valid. It is also guaranteed that Ksenia can produce list b from the list a.
Output
Print the minimal number of seconds Ksenia is required to spend to change the last year list to the current one. If the lists are already equal, print 0.
Examples
Input
3
XS
XS
M
XL
S
XS
Output
2
Input
2
XXXL
XXL
XXL
XXXS
Output
1
Input
2
M
XS
XS
M
Output
0
Note
In the first example Ksenia can replace "M" with "S" and "S" in one of the occurrences of "XS" with "L".
In the second example Ksenia should replace "L" in "XXXL" with "S".
In the third example lists are equal.
|
instruction
| 0
| 6,576
| 6
| 13,152
|
Tags: greedy, implementation
Correct Solution:
```
n = int(input())
a, b = [], []
for i in range(n):
a.append( input() )
for i in range(n):
b.append( input() )
for i in a:
try:
b.pop(b.index(i))
except Exception as e:
pass
print(len(b))
```
|
output
| 1
| 6,576
| 6
| 13,153
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Codehorses has just hosted the second Codehorses Cup. This year, the same as the previous one, organizers are giving T-shirts for the winners.
The valid sizes of T-shirts are either "M" or from 0 to 3 "X" followed by "S" or "L". For example, sizes "M", "XXS", "L", "XXXL" are valid and "XM", "Z", "XXXXL" are not.
There are n winners to the cup for both the previous year and the current year. Ksenia has a list with the T-shirt sizes printed for the last year cup and is yet to send the new list to the printing office.
Organizers want to distribute the prizes as soon as possible, so now Ksenia is required not to write the whole list from the scratch but just make some changes to the list of the previous year. In one second she can choose arbitrary position in any word and replace its character with some uppercase Latin letter. Ksenia can't remove or add letters in any of the words.
What is the minimal number of seconds Ksenia is required to spend to change the last year list to the current one?
The lists are unordered. That means, two lists are considered equal if and only if the number of occurrences of any string is the same in both lists.
Input
The first line contains one integer n (1 ≤ n ≤ 100) — the number of T-shirts.
The i-th of the next n lines contains a_i — the size of the i-th T-shirt of the list for the previous year.
The i-th of the next n lines contains b_i — the size of the i-th T-shirt of the list for the current year.
It is guaranteed that all the sizes in the input are valid. It is also guaranteed that Ksenia can produce list b from the list a.
Output
Print the minimal number of seconds Ksenia is required to spend to change the last year list to the current one. If the lists are already equal, print 0.
Examples
Input
3
XS
XS
M
XL
S
XS
Output
2
Input
2
XXXL
XXL
XXL
XXXS
Output
1
Input
2
M
XS
XS
M
Output
0
Note
In the first example Ksenia can replace "M" with "S" and "S" in one of the occurrences of "XS" with "L".
In the second example Ksenia should replace "L" in "XXXL" with "S".
In the third example lists are equal.
|
instruction
| 0
| 6,577
| 6
| 13,154
|
Tags: greedy, implementation
Correct Solution:
```
n = int(input())
L1 = [ [] for i in range(5) ]
L2 = [ [] for i in range(5) ]
ans = 0
for i in range(n):
a = input()
L1[len(a)].append(a)
for j in range(n):
b = input()
L2[len(b)].append(b)
s = 0
m = 0
l = 0
for i in L1[1]:
if i == 'S':
s += 1
elif i == 'M':
m += 1
else:
l += 1
s1 = 0
m1 = 0
l1 = 0
for i in L2[1]:
if i == 'S':
s1 += 1
elif i == 'M':
m1 += 1
else:
l1 += 1
ans += (abs(m - m1) + abs(s - s1) + abs(l - l1)) // 2
for i in range(2, 5):
s = 0
l = 0
s1 = 0
l1 = 0
for j in range(len(L1[i])):
if L1[i][j][i - 1] == 'S':
s += 1
else:
l += 1
if L2[i][j][i - 1] == 'S':
s1 += 1
else:
l1 += 1
ans += (abs(l - l1) + abs(s - s1)) // 2
print(ans)
```
|
output
| 1
| 6,577
| 6
| 13,155
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Codehorses has just hosted the second Codehorses Cup. This year, the same as the previous one, organizers are giving T-shirts for the winners.
The valid sizes of T-shirts are either "M" or from 0 to 3 "X" followed by "S" or "L". For example, sizes "M", "XXS", "L", "XXXL" are valid and "XM", "Z", "XXXXL" are not.
There are n winners to the cup for both the previous year and the current year. Ksenia has a list with the T-shirt sizes printed for the last year cup and is yet to send the new list to the printing office.
Organizers want to distribute the prizes as soon as possible, so now Ksenia is required not to write the whole list from the scratch but just make some changes to the list of the previous year. In one second she can choose arbitrary position in any word and replace its character with some uppercase Latin letter. Ksenia can't remove or add letters in any of the words.
What is the minimal number of seconds Ksenia is required to spend to change the last year list to the current one?
The lists are unordered. That means, two lists are considered equal if and only if the number of occurrences of any string is the same in both lists.
Input
The first line contains one integer n (1 ≤ n ≤ 100) — the number of T-shirts.
The i-th of the next n lines contains a_i — the size of the i-th T-shirt of the list for the previous year.
The i-th of the next n lines contains b_i — the size of the i-th T-shirt of the list for the current year.
It is guaranteed that all the sizes in the input are valid. It is also guaranteed that Ksenia can produce list b from the list a.
Output
Print the minimal number of seconds Ksenia is required to spend to change the last year list to the current one. If the lists are already equal, print 0.
Examples
Input
3
XS
XS
M
XL
S
XS
Output
2
Input
2
XXXL
XXL
XXL
XXXS
Output
1
Input
2
M
XS
XS
M
Output
0
Note
In the first example Ksenia can replace "M" with "S" and "S" in one of the occurrences of "XS" with "L".
In the second example Ksenia should replace "L" in "XXXL" with "S".
In the third example lists are equal.
|
instruction
| 0
| 6,578
| 6
| 13,156
|
Tags: greedy, implementation
Correct Solution:
```
n=int(input())
d=dict()
d1=dict()
for i in range(n):
l=input()
if l not in d:
d[l]=1
else:
d[l]+=1
x=0
for i in range(n):
l=input()
if l in d1:
d1[l]+=1
else:
d1[l]=1
for i in d1:
if i in d:
x+=max(d1[i]-d[i],0)
else:
x+=d1[i]
print(x)
```
|
output
| 1
| 6,578
| 6
| 13,157
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Codehorses has just hosted the second Codehorses Cup. This year, the same as the previous one, organizers are giving T-shirts for the winners.
The valid sizes of T-shirts are either "M" or from 0 to 3 "X" followed by "S" or "L". For example, sizes "M", "XXS", "L", "XXXL" are valid and "XM", "Z", "XXXXL" are not.
There are n winners to the cup for both the previous year and the current year. Ksenia has a list with the T-shirt sizes printed for the last year cup and is yet to send the new list to the printing office.
Organizers want to distribute the prizes as soon as possible, so now Ksenia is required not to write the whole list from the scratch but just make some changes to the list of the previous year. In one second she can choose arbitrary position in any word and replace its character with some uppercase Latin letter. Ksenia can't remove or add letters in any of the words.
What is the minimal number of seconds Ksenia is required to spend to change the last year list to the current one?
The lists are unordered. That means, two lists are considered equal if and only if the number of occurrences of any string is the same in both lists.
Input
The first line contains one integer n (1 ≤ n ≤ 100) — the number of T-shirts.
The i-th of the next n lines contains a_i — the size of the i-th T-shirt of the list for the previous year.
The i-th of the next n lines contains b_i — the size of the i-th T-shirt of the list for the current year.
It is guaranteed that all the sizes in the input are valid. It is also guaranteed that Ksenia can produce list b from the list a.
Output
Print the minimal number of seconds Ksenia is required to spend to change the last year list to the current one. If the lists are already equal, print 0.
Examples
Input
3
XS
XS
M
XL
S
XS
Output
2
Input
2
XXXL
XXL
XXL
XXXS
Output
1
Input
2
M
XS
XS
M
Output
0
Note
In the first example Ksenia can replace "M" with "S" and "S" in one of the occurrences of "XS" with "L".
In the second example Ksenia should replace "L" in "XXXL" with "S".
In the third example lists are equal.
|
instruction
| 0
| 6,579
| 6
| 13,158
|
Tags: greedy, implementation
Correct Solution:
```
def check(a,b):
if a==None or b==None:
return 0
count=0
if len(a)==len(b):
count+=1
elif len(b)>len(a):
if b[len(b)-1]!=a[len(a)-1]:
count+=1
count=count+len(b)-len(a)
else:
check(b,a)
return count
n=int(input())
a=[]
b=[]
n=2*n
for i in range(1,n+1):
if i <= n//2:
t=str(input())
a.append(t)
else:
z=str(input())
if a.count(z)==0:
b.append(z)
else:
a.remove(z)
a.sort()
b.sort()
#print(a,b)
sum=0
for i in range(len(a)):
sum=sum+check(a[i],b[i])
print(sum)
#print(a,b)`
```
|
output
| 1
| 6,579
| 6
| 13,159
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a matrix of size n × n filled with lowercase English letters. You can change no more than k letters in this matrix.
Consider all paths from the upper left corner to the lower right corner that move from a cell to its neighboring cell to the right or down. Each path is associated with the string that is formed by all the letters in the cells the path visits. Thus, the length of each string is 2n - 1.
Find the lexicographically smallest string that can be associated with a path after changing letters in at most k cells of the matrix.
A string a is lexicographically smaller than a string b, if the first different letter in a and b is smaller in a.
Input
The first line contains two integers n and k (1 ≤ n ≤ 2000, 0 ≤ k ≤ n^2) — the size of the matrix and the number of letters you can change.
Each of the next n lines contains a string of n lowercase English letters denoting one row of the matrix.
Output
Output the lexicographically smallest string that can be associated with some valid path after changing no more than k letters in the matrix.
Examples
Input
4 2
abcd
bcde
bcad
bcde
Output
aaabcde
Input
5 3
bwwwz
hrhdh
sepsp
sqfaf
ajbvw
Output
aaaepfafw
Input
7 6
ypnxnnp
pnxonpm
nxanpou
xnnpmud
nhtdudu
npmuduh
pmutsnz
Output
aaaaaaadudsnz
Note
In the first sample test case it is possible to change letters 'b' in cells (2, 1) and (3, 1) to 'a', then the minimum path contains cells (1, 1), (2, 1), (3, 1), (4, 1), (4, 2), (4, 3), (4, 4). The first coordinate corresponds to the row and the second coordinate corresponds to the column.
|
instruction
| 0
| 7,581
| 6
| 15,162
|
Tags: greedy
Correct Solution:
```
def solve(m, matrix, good, n):
c = 'z'
for x in range(n):
y = m - x
if y < 0 or y >= n: continue
if not good[x][y]: continue
if x < n - 1: c = min(c, matrix[x + 1][y])
if y < n - 1: c = min(c, matrix[x][y + 1])
for x in range(n):
y = m - x
if y < 0 or y >= n: continue
if not good[x][y]: continue
if x < n - 1 and matrix[x + 1][y] == c: good[x+1][y] = 1
if y < n - 1 and matrix[x][y + 1] == c: good[x][y + 1] = 1
return c
def main():
n, k = map(int, input().split())
matrix = []
for i in range(n):
s = list(input())
matrix.append(s)
dp = [[0 for i in range(n)] for j in range(n)]
good = [[0 for i in range(n)] for j in range(n)]
dp[0][0] = 0 if matrix[0][0] == 'a' else 1
for i in range(1, n):
dp[0][i] = dp[0][i - 1]
if matrix[0][i] != 'a':
dp[0][i] += 1
dp[i][0] = dp[i - 1][0]
if matrix[i][0] != 'a':
dp[i][0] += 1
for i in range(1, n):
for j in range(1, n):
dp[i][j] = min(dp[i-1][j], dp[i][j-1])
if matrix[i][j] != 'a':
dp[i][j] += 1
m = -1
for i in range(n):
for j in range(n):
if dp[i][j] <= k:
m = max(m, i + j)
if m == -1:
print(matrix[0][0], end = '')
m = 0
good[0][0] = 1
else:
for i in range(m + 1):
print('a', end = '')
for i in range(n):
y = m - i
if y < 0 or y >= n:
continue
if dp[i][y] <= k:
good[i][y] = 1
while m < 2*n - 2:
res = solve(m, matrix, good, n)
print(res, end = '')
m += 1
if __name__=="__main__":
main()
```
|
output
| 1
| 7,581
| 6
| 15,163
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Reading books is one of Sasha's passions. Once while he was reading one book, he became acquainted with an unusual character. The character told about himself like that: "Many are my names in many countries. Mithrandir among the Elves, Tharkûn to the Dwarves, Olórin I was in my youth in the West that is forgotten, in the South Incánus, in the North Gandalf; to the East I go not."
And at that moment Sasha thought, how would that character be called in the East? In the East all names are palindromes. A string is a palindrome if it reads the same backward as forward. For example, such strings as "kazak", "oo" and "r" are palindromes, but strings "abb" and "ij" are not.
Sasha believed that the hero would be named after one of the gods of the East. As long as there couldn't be two equal names, so in the East people did the following: they wrote the original name as a string on a piece of paper, then cut the paper minimum number of times k, so they got k+1 pieces of paper with substrings of the initial string, and then unite those pieces together to get a new string. Pieces couldn't be turned over, they could be shuffled.
In this way, it's possible to achive a string abcdefg from the string f|de|abc|g using 3 cuts (by swapping papers with substrings f and abc). The string cbadefg can't be received using the same cuts.
More formally, Sasha wants for the given palindrome s find such minimum k, that you can cut this string into k + 1 parts, and then unite them in such a way that the final string will be a palindrome and it won't be equal to the initial string s. It there is no answer, then print "Impossible" (without quotes).
Input
The first line contains one string s (1 ≤ |s| ≤ 5 000) — the initial name, which consists only of lowercase Latin letters. It is guaranteed that s is a palindrome.
Output
Print one integer k — the minimum number of cuts needed to get a new name, or "Impossible" (without quotes).
Examples
Input
nolon
Output
2
Input
otto
Output
1
Input
qqqq
Output
Impossible
Input
kinnikkinnik
Output
1
Note
In the first example, you can cut the string in those positions: no|l|on, and then unite them as follows on|l|no. It can be shown that there is no solution with one cut.
In the second example, you can cut the string right in the middle, and swap peaces, so you get toot.
In the third example, you can't make a string, that won't be equal to the initial one.
In the fourth example, you can cut the suffix nik and add it to the beginning, so you get nikkinnikkin.
|
instruction
| 0
| 7,598
| 6
| 15,196
|
Tags: constructive algorithms, hashing, strings
Correct Solution:
```
s = input()
n = len(s) // 2
k = len(s)
allsame = True
for i in range(n):
if s[i] != s[0]:
allsame = False
if allsame:
print("Impossible")
else:
ans = False
for i in range(1, n + 1):
str = s[i:] + s[:i]
if str[:n] == ''.join(reversed(str[k - n:])) and str != s:
print(1)
ans = True
break
if not ans:
print(2)
```
|
output
| 1
| 7,598
| 6
| 15,197
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Reading books is one of Sasha's passions. Once while he was reading one book, he became acquainted with an unusual character. The character told about himself like that: "Many are my names in many countries. Mithrandir among the Elves, Tharkûn to the Dwarves, Olórin I was in my youth in the West that is forgotten, in the South Incánus, in the North Gandalf; to the East I go not."
And at that moment Sasha thought, how would that character be called in the East? In the East all names are palindromes. A string is a palindrome if it reads the same backward as forward. For example, such strings as "kazak", "oo" and "r" are palindromes, but strings "abb" and "ij" are not.
Sasha believed that the hero would be named after one of the gods of the East. As long as there couldn't be two equal names, so in the East people did the following: they wrote the original name as a string on a piece of paper, then cut the paper minimum number of times k, so they got k+1 pieces of paper with substrings of the initial string, and then unite those pieces together to get a new string. Pieces couldn't be turned over, they could be shuffled.
In this way, it's possible to achive a string abcdefg from the string f|de|abc|g using 3 cuts (by swapping papers with substrings f and abc). The string cbadefg can't be received using the same cuts.
More formally, Sasha wants for the given palindrome s find such minimum k, that you can cut this string into k + 1 parts, and then unite them in such a way that the final string will be a palindrome and it won't be equal to the initial string s. It there is no answer, then print "Impossible" (without quotes).
Input
The first line contains one string s (1 ≤ |s| ≤ 5 000) — the initial name, which consists only of lowercase Latin letters. It is guaranteed that s is a palindrome.
Output
Print one integer k — the minimum number of cuts needed to get a new name, or "Impossible" (without quotes).
Examples
Input
nolon
Output
2
Input
otto
Output
1
Input
qqqq
Output
Impossible
Input
kinnikkinnik
Output
1
Note
In the first example, you can cut the string in those positions: no|l|on, and then unite them as follows on|l|no. It can be shown that there is no solution with one cut.
In the second example, you can cut the string right in the middle, and swap peaces, so you get toot.
In the third example, you can't make a string, that won't be equal to the initial one.
In the fourth example, you can cut the suffix nik and add it to the beginning, so you get nikkinnikkin.
|
instruction
| 0
| 7,599
| 6
| 15,198
|
Tags: constructive algorithms, hashing, strings
Correct Solution:
```
def isPalindrome( word ) :
return True if word == word[::-1] else False
word = input()
n = len( word )
possible = False
for i in range( n//2 + 1 ) :
new = word[i:] + word[:i]
if isPalindrome( new ) and new != word :
possible = True
print( 1 )
break
if not possible :
for i in range( 1 , n//2 if n%2 == 0 else (n+1)//2 ) :
if word[:i] != word[-i:] :
possible = True
print( 2 )
break
if not possible :
print('Impossible')
```
|
output
| 1
| 7,599
| 6
| 15,199
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Reading books is one of Sasha's passions. Once while he was reading one book, he became acquainted with an unusual character. The character told about himself like that: "Many are my names in many countries. Mithrandir among the Elves, Tharkûn to the Dwarves, Olórin I was in my youth in the West that is forgotten, in the South Incánus, in the North Gandalf; to the East I go not."
And at that moment Sasha thought, how would that character be called in the East? In the East all names are palindromes. A string is a palindrome if it reads the same backward as forward. For example, such strings as "kazak", "oo" and "r" are palindromes, but strings "abb" and "ij" are not.
Sasha believed that the hero would be named after one of the gods of the East. As long as there couldn't be two equal names, so in the East people did the following: they wrote the original name as a string on a piece of paper, then cut the paper minimum number of times k, so they got k+1 pieces of paper with substrings of the initial string, and then unite those pieces together to get a new string. Pieces couldn't be turned over, they could be shuffled.
In this way, it's possible to achive a string abcdefg from the string f|de|abc|g using 3 cuts (by swapping papers with substrings f and abc). The string cbadefg can't be received using the same cuts.
More formally, Sasha wants for the given palindrome s find such minimum k, that you can cut this string into k + 1 parts, and then unite them in such a way that the final string will be a palindrome and it won't be equal to the initial string s. It there is no answer, then print "Impossible" (without quotes).
Input
The first line contains one string s (1 ≤ |s| ≤ 5 000) — the initial name, which consists only of lowercase Latin letters. It is guaranteed that s is a palindrome.
Output
Print one integer k — the minimum number of cuts needed to get a new name, or "Impossible" (without quotes).
Examples
Input
nolon
Output
2
Input
otto
Output
1
Input
qqqq
Output
Impossible
Input
kinnikkinnik
Output
1
Note
In the first example, you can cut the string in those positions: no|l|on, and then unite them as follows on|l|no. It can be shown that there is no solution with one cut.
In the second example, you can cut the string right in the middle, and swap peaces, so you get toot.
In the third example, you can't make a string, that won't be equal to the initial one.
In the fourth example, you can cut the suffix nik and add it to the beginning, so you get nikkinnikkin.
|
instruction
| 0
| 7,600
| 6
| 15,200
|
Tags: constructive algorithms, hashing, strings
Correct Solution:
```
import sys
from collections import Counter
S = input()
N = len(S)
C = Counter(S)
if len(C) == 1:
print('Impossible')
sys.exit()
if len(C) == 2:
if min(list(C.values())) == 1:
print('Impossible')
sys.exit()
for i in range(1,N):
T = S[i:] + S[:i]
if T == T[::-1] and T != S:
print(1)
sys.exit()
print(2)
sys.exit()
```
|
output
| 1
| 7,600
| 6
| 15,201
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Reading books is one of Sasha's passions. Once while he was reading one book, he became acquainted with an unusual character. The character told about himself like that: "Many are my names in many countries. Mithrandir among the Elves, Tharkûn to the Dwarves, Olórin I was in my youth in the West that is forgotten, in the South Incánus, in the North Gandalf; to the East I go not."
And at that moment Sasha thought, how would that character be called in the East? In the East all names are palindromes. A string is a palindrome if it reads the same backward as forward. For example, such strings as "kazak", "oo" and "r" are palindromes, but strings "abb" and "ij" are not.
Sasha believed that the hero would be named after one of the gods of the East. As long as there couldn't be two equal names, so in the East people did the following: they wrote the original name as a string on a piece of paper, then cut the paper minimum number of times k, so they got k+1 pieces of paper with substrings of the initial string, and then unite those pieces together to get a new string. Pieces couldn't be turned over, they could be shuffled.
In this way, it's possible to achive a string abcdefg from the string f|de|abc|g using 3 cuts (by swapping papers with substrings f and abc). The string cbadefg can't be received using the same cuts.
More formally, Sasha wants for the given palindrome s find such minimum k, that you can cut this string into k + 1 parts, and then unite them in such a way that the final string will be a palindrome and it won't be equal to the initial string s. It there is no answer, then print "Impossible" (without quotes).
Input
The first line contains one string s (1 ≤ |s| ≤ 5 000) — the initial name, which consists only of lowercase Latin letters. It is guaranteed that s is a palindrome.
Output
Print one integer k — the minimum number of cuts needed to get a new name, or "Impossible" (without quotes).
Examples
Input
nolon
Output
2
Input
otto
Output
1
Input
qqqq
Output
Impossible
Input
kinnikkinnik
Output
1
Note
In the first example, you can cut the string in those positions: no|l|on, and then unite them as follows on|l|no. It can be shown that there is no solution with one cut.
In the second example, you can cut the string right in the middle, and swap peaces, so you get toot.
In the third example, you can't make a string, that won't be equal to the initial one.
In the fourth example, you can cut the suffix nik and add it to the beginning, so you get nikkinnikkin.
|
instruction
| 0
| 7,601
| 6
| 15,202
|
Tags: constructive algorithms, hashing, strings
Correct Solution:
```
def main():
s = input()
n = len(s)
for part in range(n - 1):
newstr = s[part + 1:] + s[:part + 1]
if newstr != s and newstr == newstr[::-1]:
print(1)
return
left = s[:n // 2]
right = s[(n + 1) // 2:]
if n == 1 or left == right and len({c for c in left}) == 1:
print("Impossible")
else:
print(2)
if __name__ == "__main__":
main()
```
|
output
| 1
| 7,601
| 6
| 15,203
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Reading books is one of Sasha's passions. Once while he was reading one book, he became acquainted with an unusual character. The character told about himself like that: "Many are my names in many countries. Mithrandir among the Elves, Tharkûn to the Dwarves, Olórin I was in my youth in the West that is forgotten, in the South Incánus, in the North Gandalf; to the East I go not."
And at that moment Sasha thought, how would that character be called in the East? In the East all names are palindromes. A string is a palindrome if it reads the same backward as forward. For example, such strings as "kazak", "oo" and "r" are palindromes, but strings "abb" and "ij" are not.
Sasha believed that the hero would be named after one of the gods of the East. As long as there couldn't be two equal names, so in the East people did the following: they wrote the original name as a string on a piece of paper, then cut the paper minimum number of times k, so they got k+1 pieces of paper with substrings of the initial string, and then unite those pieces together to get a new string. Pieces couldn't be turned over, they could be shuffled.
In this way, it's possible to achive a string abcdefg from the string f|de|abc|g using 3 cuts (by swapping papers with substrings f and abc). The string cbadefg can't be received using the same cuts.
More formally, Sasha wants for the given palindrome s find such minimum k, that you can cut this string into k + 1 parts, and then unite them in such a way that the final string will be a palindrome and it won't be equal to the initial string s. It there is no answer, then print "Impossible" (without quotes).
Input
The first line contains one string s (1 ≤ |s| ≤ 5 000) — the initial name, which consists only of lowercase Latin letters. It is guaranteed that s is a palindrome.
Output
Print one integer k — the minimum number of cuts needed to get a new name, or "Impossible" (without quotes).
Examples
Input
nolon
Output
2
Input
otto
Output
1
Input
qqqq
Output
Impossible
Input
kinnikkinnik
Output
1
Note
In the first example, you can cut the string in those positions: no|l|on, and then unite them as follows on|l|no. It can be shown that there is no solution with one cut.
In the second example, you can cut the string right in the middle, and swap peaces, so you get toot.
In the third example, you can't make a string, that won't be equal to the initial one.
In the fourth example, you can cut the suffix nik and add it to the beginning, so you get nikkinnikkin.
|
instruction
| 0
| 7,602
| 6
| 15,204
|
Tags: constructive algorithms, hashing, strings
Correct Solution:
```
from sys import *
s = input();
def check(t):
return (t == t[::-1]) and (t != s)
for i in range(1, len(s)):
t = s[i:] + s[:i]
if check(t):
print("1")
exit()
for i in range(1, len(s)//2 + (len(s)%2)):
t = s[-i:] + s[i:-i] + s[:i]
if check(t):
print("2")
exit()
print("Impossible")
```
|
output
| 1
| 7,602
| 6
| 15,205
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Reading books is one of Sasha's passions. Once while he was reading one book, he became acquainted with an unusual character. The character told about himself like that: "Many are my names in many countries. Mithrandir among the Elves, Tharkûn to the Dwarves, Olórin I was in my youth in the West that is forgotten, in the South Incánus, in the North Gandalf; to the East I go not."
And at that moment Sasha thought, how would that character be called in the East? In the East all names are palindromes. A string is a palindrome if it reads the same backward as forward. For example, such strings as "kazak", "oo" and "r" are palindromes, but strings "abb" and "ij" are not.
Sasha believed that the hero would be named after one of the gods of the East. As long as there couldn't be two equal names, so in the East people did the following: they wrote the original name as a string on a piece of paper, then cut the paper minimum number of times k, so they got k+1 pieces of paper with substrings of the initial string, and then unite those pieces together to get a new string. Pieces couldn't be turned over, they could be shuffled.
In this way, it's possible to achive a string abcdefg from the string f|de|abc|g using 3 cuts (by swapping papers with substrings f and abc). The string cbadefg can't be received using the same cuts.
More formally, Sasha wants for the given palindrome s find such minimum k, that you can cut this string into k + 1 parts, and then unite them in such a way that the final string will be a palindrome and it won't be equal to the initial string s. It there is no answer, then print "Impossible" (without quotes).
Input
The first line contains one string s (1 ≤ |s| ≤ 5 000) — the initial name, which consists only of lowercase Latin letters. It is guaranteed that s is a palindrome.
Output
Print one integer k — the minimum number of cuts needed to get a new name, or "Impossible" (without quotes).
Examples
Input
nolon
Output
2
Input
otto
Output
1
Input
qqqq
Output
Impossible
Input
kinnikkinnik
Output
1
Note
In the first example, you can cut the string in those positions: no|l|on, and then unite them as follows on|l|no. It can be shown that there is no solution with one cut.
In the second example, you can cut the string right in the middle, and swap peaces, so you get toot.
In the third example, you can't make a string, that won't be equal to the initial one.
In the fourth example, you can cut the suffix nik and add it to the beginning, so you get nikkinnikkin.
|
instruction
| 0
| 7,603
| 6
| 15,206
|
Tags: constructive algorithms, hashing, strings
Correct Solution:
```
s = input()
l = len(s)
c = s[0]
diff = False
for i in range(0,int(l/2)):
if s[i] != c:
diff = True
if not diff:
print('Impossible')
exit()
s_2 = s + s
for i in range(1,l):
is_palendrome = True
for j in range(int(l/2)):
if s_2[j + i] != s_2[i + l - j-1]:
is_palendrome = False
if is_palendrome and s_2[i:i+l] != s:
print(1)
exit()
print(2)
```
|
output
| 1
| 7,603
| 6
| 15,207
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Reading books is one of Sasha's passions. Once while he was reading one book, he became acquainted with an unusual character. The character told about himself like that: "Many are my names in many countries. Mithrandir among the Elves, Tharkûn to the Dwarves, Olórin I was in my youth in the West that is forgotten, in the South Incánus, in the North Gandalf; to the East I go not."
And at that moment Sasha thought, how would that character be called in the East? In the East all names are palindromes. A string is a palindrome if it reads the same backward as forward. For example, such strings as "kazak", "oo" and "r" are palindromes, but strings "abb" and "ij" are not.
Sasha believed that the hero would be named after one of the gods of the East. As long as there couldn't be two equal names, so in the East people did the following: they wrote the original name as a string on a piece of paper, then cut the paper minimum number of times k, so they got k+1 pieces of paper with substrings of the initial string, and then unite those pieces together to get a new string. Pieces couldn't be turned over, they could be shuffled.
In this way, it's possible to achive a string abcdefg from the string f|de|abc|g using 3 cuts (by swapping papers with substrings f and abc). The string cbadefg can't be received using the same cuts.
More formally, Sasha wants for the given palindrome s find such minimum k, that you can cut this string into k + 1 parts, and then unite them in such a way that the final string will be a palindrome and it won't be equal to the initial string s. It there is no answer, then print "Impossible" (without quotes).
Input
The first line contains one string s (1 ≤ |s| ≤ 5 000) — the initial name, which consists only of lowercase Latin letters. It is guaranteed that s is a palindrome.
Output
Print one integer k — the minimum number of cuts needed to get a new name, or "Impossible" (without quotes).
Examples
Input
nolon
Output
2
Input
otto
Output
1
Input
qqqq
Output
Impossible
Input
kinnikkinnik
Output
1
Note
In the first example, you can cut the string in those positions: no|l|on, and then unite them as follows on|l|no. It can be shown that there is no solution with one cut.
In the second example, you can cut the string right in the middle, and swap peaces, so you get toot.
In the third example, you can't make a string, that won't be equal to the initial one.
In the fourth example, you can cut the suffix nik and add it to the beginning, so you get nikkinnikkin.
|
instruction
| 0
| 7,604
| 6
| 15,208
|
Tags: constructive algorithms, hashing, strings
Correct Solution:
```
def pallin(s):
n= len(s)
for i in range(0,n//2):
if s[i]!=s[n-i-1]:
return False
return True
if __name__ == '__main__':
s= input()
#print(s)
#print(pallin(s))
n= len(s)
for i in range(n-1,0,-1):
s1= s[0:i]
s2= s[i:]
t= s2+s1
#print(s1)
#print(s2)
#print(t)
if s!=t and pallin(t):
print("1")
exit()
for i in range(1,n//2):
if s[i]!=s[i-1]:
print("2")
exit()
print("Impossible")
```
|
output
| 1
| 7,604
| 6
| 15,209
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Reading books is one of Sasha's passions. Once while he was reading one book, he became acquainted with an unusual character. The character told about himself like that: "Many are my names in many countries. Mithrandir among the Elves, Tharkûn to the Dwarves, Olórin I was in my youth in the West that is forgotten, in the South Incánus, in the North Gandalf; to the East I go not."
And at that moment Sasha thought, how would that character be called in the East? In the East all names are palindromes. A string is a palindrome if it reads the same backward as forward. For example, such strings as "kazak", "oo" and "r" are palindromes, but strings "abb" and "ij" are not.
Sasha believed that the hero would be named after one of the gods of the East. As long as there couldn't be two equal names, so in the East people did the following: they wrote the original name as a string on a piece of paper, then cut the paper minimum number of times k, so they got k+1 pieces of paper with substrings of the initial string, and then unite those pieces together to get a new string. Pieces couldn't be turned over, they could be shuffled.
In this way, it's possible to achive a string abcdefg from the string f|de|abc|g using 3 cuts (by swapping papers with substrings f and abc). The string cbadefg can't be received using the same cuts.
More formally, Sasha wants for the given palindrome s find such minimum k, that you can cut this string into k + 1 parts, and then unite them in such a way that the final string will be a palindrome and it won't be equal to the initial string s. It there is no answer, then print "Impossible" (without quotes).
Input
The first line contains one string s (1 ≤ |s| ≤ 5 000) — the initial name, which consists only of lowercase Latin letters. It is guaranteed that s is a palindrome.
Output
Print one integer k — the minimum number of cuts needed to get a new name, or "Impossible" (without quotes).
Examples
Input
nolon
Output
2
Input
otto
Output
1
Input
qqqq
Output
Impossible
Input
kinnikkinnik
Output
1
Note
In the first example, you can cut the string in those positions: no|l|on, and then unite them as follows on|l|no. It can be shown that there is no solution with one cut.
In the second example, you can cut the string right in the middle, and swap peaces, so you get toot.
In the third example, you can't make a string, that won't be equal to the initial one.
In the fourth example, you can cut the suffix nik and add it to the beginning, so you get nikkinnikkin.
|
instruction
| 0
| 7,605
| 6
| 15,210
|
Tags: constructive algorithms, hashing, strings
Correct Solution:
```
from collections import Counter
import sys
S=input()
L=len(S)
def pand(x):
if x==x[::-1]:
return True
else:
return False
for i in range(1,len(S)):
if S[i:]+S[:i]!=S and pand(S[i:]+S[:i])==True:
print(1)
sys.exit()
if len(Counter(S).keys())==1:
print("Impossible")
sys.exit()
if L%2==0:
print(2)
else:
for i in range(1,L//2+1):
if S[:i]!=S[-i:]:
print(2)
sys.exit()
else:
print("Impossible")
```
|
output
| 1
| 7,605
| 6
| 15,211
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Reading books is one of Sasha's passions. Once while he was reading one book, he became acquainted with an unusual character. The character told about himself like that: "Many are my names in many countries. Mithrandir among the Elves, Tharkûn to the Dwarves, Olórin I was in my youth in the West that is forgotten, in the South Incánus, in the North Gandalf; to the East I go not."
And at that moment Sasha thought, how would that character be called in the East? In the East all names are palindromes. A string is a palindrome if it reads the same backward as forward. For example, such strings as "kazak", "oo" and "r" are palindromes, but strings "abb" and "ij" are not.
Sasha believed that the hero would be named after one of the gods of the East. As long as there couldn't be two equal names, so in the East people did the following: they wrote the original name as a string on a piece of paper, then cut the paper minimum number of times k, so they got k+1 pieces of paper with substrings of the initial string, and then unite those pieces together to get a new string. Pieces couldn't be turned over, they could be shuffled.
In this way, it's possible to achive a string abcdefg from the string f|de|abc|g using 3 cuts (by swapping papers with substrings f and abc). The string cbadefg can't be received using the same cuts.
More formally, Sasha wants for the given palindrome s find such minimum k, that you can cut this string into k + 1 parts, and then unite them in such a way that the final string will be a palindrome and it won't be equal to the initial string s. It there is no answer, then print "Impossible" (without quotes).
Input
The first line contains one string s (1 ≤ |s| ≤ 5 000) — the initial name, which consists only of lowercase Latin letters. It is guaranteed that s is a palindrome.
Output
Print one integer k — the minimum number of cuts needed to get a new name, or "Impossible" (without quotes).
Examples
Input
nolon
Output
2
Input
otto
Output
1
Input
qqqq
Output
Impossible
Input
kinnikkinnik
Output
1
Note
In the first example, you can cut the string in those positions: no|l|on, and then unite them as follows on|l|no. It can be shown that there is no solution with one cut.
In the second example, you can cut the string right in the middle, and swap peaces, so you get toot.
In the third example, you can't make a string, that won't be equal to the initial one.
In the fourth example, you can cut the suffix nik and add it to the beginning, so you get nikkinnikkin.
Submitted Solution:
```
import collections
#import random
import heapq
import bisect
import math
import time
class Solution2:
def solve(self, A1, A2):
pass
def gcd(a, b):
if not b: return a
return gcd(b, a%b)
def lcm(a, b):
return b*a//gcd(b,a)
class Solution:
def solve(self, s):
count = collections.Counter(s)
if len(count) == 1: return "Impossible"
if len(count) == 2:
if any(v == 1 for v in count.values()): return "Impossible"
for i in range(len(s)):
k1, k2 = s[i:], s[:i]
new_s = k1 + k2
if new_s[::-1] == new_s and new_s != s: return '1'
return '2'
sol = Solution()
sol2 = Solution2()
#TT = int(input())
for test_case in range(1):
N = input()
#a = []
#for _ in range(int(N)-1):
#a.append([int(c) for c in input().split()])
#b = [int(c) for c in input().split()]
out = sol.solve(N)
#print(' '.join([str(o) for o in out]))
print(out)
# out2 = sol2.solve(s)
# for _ in range(100000):
# rand = [random.randrange(60) for _ in range(10)]
# out1 = sol.solve(rand)
# out2 = sol2.solve(rand)
# if out1 != out2:
# print(rand, out1, out2)
# break
```
|
instruction
| 0
| 7,606
| 6
| 15,212
|
Yes
|
output
| 1
| 7,606
| 6
| 15,213
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Reading books is one of Sasha's passions. Once while he was reading one book, he became acquainted with an unusual character. The character told about himself like that: "Many are my names in many countries. Mithrandir among the Elves, Tharkûn to the Dwarves, Olórin I was in my youth in the West that is forgotten, in the South Incánus, in the North Gandalf; to the East I go not."
And at that moment Sasha thought, how would that character be called in the East? In the East all names are palindromes. A string is a palindrome if it reads the same backward as forward. For example, such strings as "kazak", "oo" and "r" are palindromes, but strings "abb" and "ij" are not.
Sasha believed that the hero would be named after one of the gods of the East. As long as there couldn't be two equal names, so in the East people did the following: they wrote the original name as a string on a piece of paper, then cut the paper minimum number of times k, so they got k+1 pieces of paper with substrings of the initial string, and then unite those pieces together to get a new string. Pieces couldn't be turned over, they could be shuffled.
In this way, it's possible to achive a string abcdefg from the string f|de|abc|g using 3 cuts (by swapping papers with substrings f and abc). The string cbadefg can't be received using the same cuts.
More formally, Sasha wants for the given palindrome s find such minimum k, that you can cut this string into k + 1 parts, and then unite them in such a way that the final string will be a palindrome and it won't be equal to the initial string s. It there is no answer, then print "Impossible" (without quotes).
Input
The first line contains one string s (1 ≤ |s| ≤ 5 000) — the initial name, which consists only of lowercase Latin letters. It is guaranteed that s is a palindrome.
Output
Print one integer k — the minimum number of cuts needed to get a new name, or "Impossible" (without quotes).
Examples
Input
nolon
Output
2
Input
otto
Output
1
Input
qqqq
Output
Impossible
Input
kinnikkinnik
Output
1
Note
In the first example, you can cut the string in those positions: no|l|on, and then unite them as follows on|l|no. It can be shown that there is no solution with one cut.
In the second example, you can cut the string right in the middle, and swap peaces, so you get toot.
In the third example, you can't make a string, that won't be equal to the initial one.
In the fourth example, you can cut the suffix nik and add it to the beginning, so you get nikkinnikkin.
Submitted Solution:
```
s=input()
if len(set(s[:len(s)//2]))<=1:
print("Impossible");exit()
for i in range(1,len(s)):
n=s[i:]+s[:i]
if(n==n[::-1])and(n!=s):
print(1);exit()
print(2)
```
|
instruction
| 0
| 7,607
| 6
| 15,214
|
Yes
|
output
| 1
| 7,607
| 6
| 15,215
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Reading books is one of Sasha's passions. Once while he was reading one book, he became acquainted with an unusual character. The character told about himself like that: "Many are my names in many countries. Mithrandir among the Elves, Tharkûn to the Dwarves, Olórin I was in my youth in the West that is forgotten, in the South Incánus, in the North Gandalf; to the East I go not."
And at that moment Sasha thought, how would that character be called in the East? In the East all names are palindromes. A string is a palindrome if it reads the same backward as forward. For example, such strings as "kazak", "oo" and "r" are palindromes, but strings "abb" and "ij" are not.
Sasha believed that the hero would be named after one of the gods of the East. As long as there couldn't be two equal names, so in the East people did the following: they wrote the original name as a string on a piece of paper, then cut the paper minimum number of times k, so they got k+1 pieces of paper with substrings of the initial string, and then unite those pieces together to get a new string. Pieces couldn't be turned over, they could be shuffled.
In this way, it's possible to achive a string abcdefg from the string f|de|abc|g using 3 cuts (by swapping papers with substrings f and abc). The string cbadefg can't be received using the same cuts.
More formally, Sasha wants for the given palindrome s find such minimum k, that you can cut this string into k + 1 parts, and then unite them in such a way that the final string will be a palindrome and it won't be equal to the initial string s. It there is no answer, then print "Impossible" (without quotes).
Input
The first line contains one string s (1 ≤ |s| ≤ 5 000) — the initial name, which consists only of lowercase Latin letters. It is guaranteed that s is a palindrome.
Output
Print one integer k — the minimum number of cuts needed to get a new name, or "Impossible" (without quotes).
Examples
Input
nolon
Output
2
Input
otto
Output
1
Input
qqqq
Output
Impossible
Input
kinnikkinnik
Output
1
Note
In the first example, you can cut the string in those positions: no|l|on, and then unite them as follows on|l|no. It can be shown that there is no solution with one cut.
In the second example, you can cut the string right in the middle, and swap peaces, so you get toot.
In the third example, you can't make a string, that won't be equal to the initial one.
In the fourth example, you can cut the suffix nik and add it to the beginning, so you get nikkinnikkin.
Submitted Solution:
```
from collections import Counter
def solve(s):
n = len(s)
c = Counter(s)
if max(c.values()) >= n - 1:
return 'Impossible'
for i in range(1, n):
new_s = s[i:] + s[:i]
if new_s == s:
continue
for j in range(n // 2 + 1):
if new_s[j] != new_s[-j - 1]:
break
else:
return 1
return 2
if __name__ == '__main__':
s = input()
print(solve(s))
```
|
instruction
| 0
| 7,608
| 6
| 15,216
|
Yes
|
output
| 1
| 7,608
| 6
| 15,217
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Reading books is one of Sasha's passions. Once while he was reading one book, he became acquainted with an unusual character. The character told about himself like that: "Many are my names in many countries. Mithrandir among the Elves, Tharkûn to the Dwarves, Olórin I was in my youth in the West that is forgotten, in the South Incánus, in the North Gandalf; to the East I go not."
And at that moment Sasha thought, how would that character be called in the East? In the East all names are palindromes. A string is a palindrome if it reads the same backward as forward. For example, such strings as "kazak", "oo" and "r" are palindromes, but strings "abb" and "ij" are not.
Sasha believed that the hero would be named after one of the gods of the East. As long as there couldn't be two equal names, so in the East people did the following: they wrote the original name as a string on a piece of paper, then cut the paper minimum number of times k, so they got k+1 pieces of paper with substrings of the initial string, and then unite those pieces together to get a new string. Pieces couldn't be turned over, they could be shuffled.
In this way, it's possible to achive a string abcdefg from the string f|de|abc|g using 3 cuts (by swapping papers with substrings f and abc). The string cbadefg can't be received using the same cuts.
More formally, Sasha wants for the given palindrome s find such minimum k, that you can cut this string into k + 1 parts, and then unite them in such a way that the final string will be a palindrome and it won't be equal to the initial string s. It there is no answer, then print "Impossible" (without quotes).
Input
The first line contains one string s (1 ≤ |s| ≤ 5 000) — the initial name, which consists only of lowercase Latin letters. It is guaranteed that s is a palindrome.
Output
Print one integer k — the minimum number of cuts needed to get a new name, or "Impossible" (without quotes).
Examples
Input
nolon
Output
2
Input
otto
Output
1
Input
qqqq
Output
Impossible
Input
kinnikkinnik
Output
1
Note
In the first example, you can cut the string in those positions: no|l|on, and then unite them as follows on|l|no. It can be shown that there is no solution with one cut.
In the second example, you can cut the string right in the middle, and swap peaces, so you get toot.
In the third example, you can't make a string, that won't be equal to the initial one.
In the fourth example, you can cut the suffix nik and add it to the beginning, so you get nikkinnikkin.
Submitted Solution:
```
import sys
from collections import deque
import math
input_ = lambda: sys.stdin.readline().strip("\r\n")
ii = lambda : int(input_())
il = lambda : list(map(int, input_().split()))
ilf = lambda : list(map(float, input_().split()))
ip = lambda : input_()
fi = lambda : float(input_())
li = lambda : list(input_())
pr = lambda x : print(x)
prinT = lambda x : print(x)
f = lambda : sys.stdout.flush()
s = ip()
l = len(s)
for i in range (1,l-1) :
x = s[i:] + s[:i]
if (x == x[::-1] and s != x) :
print(1)
exit(0)
for j in range(1,l//2 + l%2) :
x = s[-j:] + s[j:-j] + s[:j]
#print(x)
if (x == x[::-1] and x!=s) :
print(2)
exit(0)
print("Impossible")
```
|
instruction
| 0
| 7,609
| 6
| 15,218
|
Yes
|
output
| 1
| 7,609
| 6
| 15,219
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Reading books is one of Sasha's passions. Once while he was reading one book, he became acquainted with an unusual character. The character told about himself like that: "Many are my names in many countries. Mithrandir among the Elves, Tharkûn to the Dwarves, Olórin I was in my youth in the West that is forgotten, in the South Incánus, in the North Gandalf; to the East I go not."
And at that moment Sasha thought, how would that character be called in the East? In the East all names are palindromes. A string is a palindrome if it reads the same backward as forward. For example, such strings as "kazak", "oo" and "r" are palindromes, but strings "abb" and "ij" are not.
Sasha believed that the hero would be named after one of the gods of the East. As long as there couldn't be two equal names, so in the East people did the following: they wrote the original name as a string on a piece of paper, then cut the paper minimum number of times k, so they got k+1 pieces of paper with substrings of the initial string, and then unite those pieces together to get a new string. Pieces couldn't be turned over, they could be shuffled.
In this way, it's possible to achive a string abcdefg from the string f|de|abc|g using 3 cuts (by swapping papers with substrings f and abc). The string cbadefg can't be received using the same cuts.
More formally, Sasha wants for the given palindrome s find such minimum k, that you can cut this string into k + 1 parts, and then unite them in such a way that the final string will be a palindrome and it won't be equal to the initial string s. It there is no answer, then print "Impossible" (without quotes).
Input
The first line contains one string s (1 ≤ |s| ≤ 5 000) — the initial name, which consists only of lowercase Latin letters. It is guaranteed that s is a palindrome.
Output
Print one integer k — the minimum number of cuts needed to get a new name, or "Impossible" (without quotes).
Examples
Input
nolon
Output
2
Input
otto
Output
1
Input
qqqq
Output
Impossible
Input
kinnikkinnik
Output
1
Note
In the first example, you can cut the string in those positions: no|l|on, and then unite them as follows on|l|no. It can be shown that there is no solution with one cut.
In the second example, you can cut the string right in the middle, and swap peaces, so you get toot.
In the third example, you can't make a string, that won't be equal to the initial one.
In the fourth example, you can cut the suffix nik and add it to the beginning, so you get nikkinnikkin.
Submitted Solution:
```
# |
# _` | __ \ _` | __| _ \ __ \ _` | _` |
# ( | | | ( | ( ( | | | ( | ( |
# \__,_| _| _| \__,_| \___| \___/ _| _| \__,_| \__,_|
import sys
import math
def read_line():
return sys.stdin.readline()[:-1]
def read_int():
return int(sys.stdin.readline())
def read_int_line():
return [int(v) for v in sys.stdin.readline().split()]
def reverse(s):
return s[::-1]
def isPalindrome(s):
# Calling reverse function
rev = reverse(s)
# Checking if both string are equal or not
if (s == rev):
return True
return False
s = read_line()
n = len(s)
d = {}
for i in s:
if i not in d:
d[i] = 1
else:
d[i] +=1
f = True
if n&1==1 and len(list(d.keys())) <= 2 :
f = False
elif n&1 != 1 and len(list(d.keys())) == 1:
f = False
ans = 2
for i in range(1,n):
pre = s[:i]
post = s[i:]
if isPalindrome(post+pre):
if post+pre != s:
ans = 1
break
if f:
print(ans)
else:
print("Impossible")
```
|
instruction
| 0
| 7,610
| 6
| 15,220
|
No
|
output
| 1
| 7,610
| 6
| 15,221
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Reading books is one of Sasha's passions. Once while he was reading one book, he became acquainted with an unusual character. The character told about himself like that: "Many are my names in many countries. Mithrandir among the Elves, Tharkûn to the Dwarves, Olórin I was in my youth in the West that is forgotten, in the South Incánus, in the North Gandalf; to the East I go not."
And at that moment Sasha thought, how would that character be called in the East? In the East all names are palindromes. A string is a palindrome if it reads the same backward as forward. For example, such strings as "kazak", "oo" and "r" are palindromes, but strings "abb" and "ij" are not.
Sasha believed that the hero would be named after one of the gods of the East. As long as there couldn't be two equal names, so in the East people did the following: they wrote the original name as a string on a piece of paper, then cut the paper minimum number of times k, so they got k+1 pieces of paper with substrings of the initial string, and then unite those pieces together to get a new string. Pieces couldn't be turned over, they could be shuffled.
In this way, it's possible to achive a string abcdefg from the string f|de|abc|g using 3 cuts (by swapping papers with substrings f and abc). The string cbadefg can't be received using the same cuts.
More formally, Sasha wants for the given palindrome s find such minimum k, that you can cut this string into k + 1 parts, and then unite them in such a way that the final string will be a palindrome and it won't be equal to the initial string s. It there is no answer, then print "Impossible" (without quotes).
Input
The first line contains one string s (1 ≤ |s| ≤ 5 000) — the initial name, which consists only of lowercase Latin letters. It is guaranteed that s is a palindrome.
Output
Print one integer k — the minimum number of cuts needed to get a new name, or "Impossible" (without quotes).
Examples
Input
nolon
Output
2
Input
otto
Output
1
Input
qqqq
Output
Impossible
Input
kinnikkinnik
Output
1
Note
In the first example, you can cut the string in those positions: no|l|on, and then unite them as follows on|l|no. It can be shown that there is no solution with one cut.
In the second example, you can cut the string right in the middle, and swap peaces, so you get toot.
In the third example, you can't make a string, that won't be equal to the initial one.
In the fourth example, you can cut the suffix nik and add it to the beginning, so you get nikkinnikkin.
Submitted Solution:
```
def reverse(s):
return s[::-1]
def isPalindrome(s):
# Calling reverse function
rev = reverse(s)
# Checking if both string are equal or not
if (s == rev):
return True
return False
if __name__ == '__main__':
s = input()
n = len(s)
a = s[0]
if n%2 == 0:
same = True
for i in range(n):
if not a == s[i]:
same = False
else:
same = True
for i in range(n):
if (not a == s[i]) and (not i == n//2):
same = False
if not same:
if n % 2 == 0:
print(1)
else:
possible = False
for i in range(1,n):
next = s[i:]+s[:i]
print(next)
if not possible:
possible = isPalindrome(next)
if possible:
print(1)
else:
print(2)
else:
print("Impossible")
```
|
instruction
| 0
| 7,611
| 6
| 15,222
|
No
|
output
| 1
| 7,611
| 6
| 15,223
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Reading books is one of Sasha's passions. Once while he was reading one book, he became acquainted with an unusual character. The character told about himself like that: "Many are my names in many countries. Mithrandir among the Elves, Tharkûn to the Dwarves, Olórin I was in my youth in the West that is forgotten, in the South Incánus, in the North Gandalf; to the East I go not."
And at that moment Sasha thought, how would that character be called in the East? In the East all names are palindromes. A string is a palindrome if it reads the same backward as forward. For example, such strings as "kazak", "oo" and "r" are palindromes, but strings "abb" and "ij" are not.
Sasha believed that the hero would be named after one of the gods of the East. As long as there couldn't be two equal names, so in the East people did the following: they wrote the original name as a string on a piece of paper, then cut the paper minimum number of times k, so they got k+1 pieces of paper with substrings of the initial string, and then unite those pieces together to get a new string. Pieces couldn't be turned over, they could be shuffled.
In this way, it's possible to achive a string abcdefg from the string f|de|abc|g using 3 cuts (by swapping papers with substrings f and abc). The string cbadefg can't be received using the same cuts.
More formally, Sasha wants for the given palindrome s find such minimum k, that you can cut this string into k + 1 parts, and then unite them in such a way that the final string will be a palindrome and it won't be equal to the initial string s. It there is no answer, then print "Impossible" (without quotes).
Input
The first line contains one string s (1 ≤ |s| ≤ 5 000) — the initial name, which consists only of lowercase Latin letters. It is guaranteed that s is a palindrome.
Output
Print one integer k — the minimum number of cuts needed to get a new name, or "Impossible" (without quotes).
Examples
Input
nolon
Output
2
Input
otto
Output
1
Input
qqqq
Output
Impossible
Input
kinnikkinnik
Output
1
Note
In the first example, you can cut the string in those positions: no|l|on, and then unite them as follows on|l|no. It can be shown that there is no solution with one cut.
In the second example, you can cut the string right in the middle, and swap peaces, so you get toot.
In the third example, you can't make a string, that won't be equal to the initial one.
In the fourth example, you can cut the suffix nik and add it to the beginning, so you get nikkinnikkin.
Submitted Solution:
```
s = input()
if(len(set(s)) == 1 or len(s) == 3):
print("Impossible")
else:
n = len(s)
if(n%2 != 0):
print(2)
elif(s[:n//2] == s[n//2:]):
if(n%4 == 0):
print(1)
else:
print(2)
else:
print(2)
```
|
instruction
| 0
| 7,612
| 6
| 15,224
|
No
|
output
| 1
| 7,612
| 6
| 15,225
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Reading books is one of Sasha's passions. Once while he was reading one book, he became acquainted with an unusual character. The character told about himself like that: "Many are my names in many countries. Mithrandir among the Elves, Tharkûn to the Dwarves, Olórin I was in my youth in the West that is forgotten, in the South Incánus, in the North Gandalf; to the East I go not."
And at that moment Sasha thought, how would that character be called in the East? In the East all names are palindromes. A string is a palindrome if it reads the same backward as forward. For example, such strings as "kazak", "oo" and "r" are palindromes, but strings "abb" and "ij" are not.
Sasha believed that the hero would be named after one of the gods of the East. As long as there couldn't be two equal names, so in the East people did the following: they wrote the original name as a string on a piece of paper, then cut the paper minimum number of times k, so they got k+1 pieces of paper with substrings of the initial string, and then unite those pieces together to get a new string. Pieces couldn't be turned over, they could be shuffled.
In this way, it's possible to achive a string abcdefg from the string f|de|abc|g using 3 cuts (by swapping papers with substrings f and abc). The string cbadefg can't be received using the same cuts.
More formally, Sasha wants for the given palindrome s find such minimum k, that you can cut this string into k + 1 parts, and then unite them in such a way that the final string will be a palindrome and it won't be equal to the initial string s. It there is no answer, then print "Impossible" (without quotes).
Input
The first line contains one string s (1 ≤ |s| ≤ 5 000) — the initial name, which consists only of lowercase Latin letters. It is guaranteed that s is a palindrome.
Output
Print one integer k — the minimum number of cuts needed to get a new name, or "Impossible" (without quotes).
Examples
Input
nolon
Output
2
Input
otto
Output
1
Input
qqqq
Output
Impossible
Input
kinnikkinnik
Output
1
Note
In the first example, you can cut the string in those positions: no|l|on, and then unite them as follows on|l|no. It can be shown that there is no solution with one cut.
In the second example, you can cut the string right in the middle, and swap peaces, so you get toot.
In the third example, you can't make a string, that won't be equal to the initial one.
In the fourth example, you can cut the suffix nik and add it to the beginning, so you get nikkinnikkin.
Submitted Solution:
```
def pal(s):
s=list(s)
s1=s[:len(s)//2]
if len(s)%2==0:
s2=s[len(s)//2:]
else:
s2=s[len(s)//2+1:]
s2.reverse()
if s1==s2:
return True
return False
s=input()
for i in range(len(s)//2+1):
if pal(s[:i+1])==False:
if len(s)%2==1:
print(2)
else:
print(1)
break
else:
print('Impossible')
```
|
instruction
| 0
| 7,613
| 6
| 15,226
|
No
|
output
| 1
| 7,613
| 6
| 15,227
|
Provide tags and a correct Python 3 solution for this coding contest problem.
One popular website developed an unusual username editing procedure. One can change the username only by deleting some characters from it: to change the current name s, a user can pick number p and character c and delete the p-th occurrence of character c from the name. After the user changed his name, he can't undo the change.
For example, one can change name "arca" by removing the second occurrence of character "a" to get "arc".
Polycarpus learned that some user initially registered under nickname t, where t is a concatenation of k copies of string s. Also, Polycarpus knows the sequence of this user's name changes. Help Polycarpus figure out the user's final name.
Input
The first line contains an integer k (1 ≤ k ≤ 2000). The second line contains a non-empty string s, consisting of lowercase Latin letters, at most 100 characters long. The third line contains an integer n (0 ≤ n ≤ 20000) — the number of username changes. Each of the next n lines contains the actual changes, one per line. The changes are written as "pi ci" (without the quotes), where pi (1 ≤ pi ≤ 200000) is the number of occurrences of letter ci, ci is a lowercase Latin letter. It is guaranteed that the operations are correct, that is, the letter to be deleted always exists, and after all operations not all letters are deleted from the name. The letters' occurrences are numbered starting from 1.
Output
Print a single string — the user's final name after all changes are applied to it.
Examples
Input
2
bac
3
2 a
1 b
2 c
Output
acb
Input
1
abacaba
4
1 a
1 a
1 c
2 b
Output
baa
Note
Let's consider the first sample. Initially we have name "bacbac"; the first operation transforms it into "bacbc", the second one — to "acbc", and finally, the third one transforms it into "acb".
|
instruction
| 0
| 7,816
| 6
| 15,632
|
Tags: *special, binary search, brute force, data structures, strings
Correct Solution:
```
# cook your dish here
n=int(input())
s=list(input())*n
d={}
for i in range(len(s)):
if s[i] not in d:
d[s[i]]=[i]
else:
d[s[i]].append(i)
m=int(input())
for i in range(m):
o,c=input().split()
s[d[c].pop(int(o)-1)]=''
print(''.join(s))
```
|
output
| 1
| 7,816
| 6
| 15,633
|
Provide tags and a correct Python 3 solution for this coding contest problem.
One popular website developed an unusual username editing procedure. One can change the username only by deleting some characters from it: to change the current name s, a user can pick number p and character c and delete the p-th occurrence of character c from the name. After the user changed his name, he can't undo the change.
For example, one can change name "arca" by removing the second occurrence of character "a" to get "arc".
Polycarpus learned that some user initially registered under nickname t, where t is a concatenation of k copies of string s. Also, Polycarpus knows the sequence of this user's name changes. Help Polycarpus figure out the user's final name.
Input
The first line contains an integer k (1 ≤ k ≤ 2000). The second line contains a non-empty string s, consisting of lowercase Latin letters, at most 100 characters long. The third line contains an integer n (0 ≤ n ≤ 20000) — the number of username changes. Each of the next n lines contains the actual changes, one per line. The changes are written as "pi ci" (without the quotes), where pi (1 ≤ pi ≤ 200000) is the number of occurrences of letter ci, ci is a lowercase Latin letter. It is guaranteed that the operations are correct, that is, the letter to be deleted always exists, and after all operations not all letters are deleted from the name. The letters' occurrences are numbered starting from 1.
Output
Print a single string — the user's final name after all changes are applied to it.
Examples
Input
2
bac
3
2 a
1 b
2 c
Output
acb
Input
1
abacaba
4
1 a
1 a
1 c
2 b
Output
baa
Note
Let's consider the first sample. Initially we have name "bacbac"; the first operation transforms it into "bacbc", the second one — to "acbc", and finally, the third one transforms it into "acb".
|
instruction
| 0
| 7,817
| 6
| 15,634
|
Tags: *special, binary search, brute force, data structures, strings
Correct Solution:
```
k = int(input())
string = list(input()) * k
d = {}
for i, letter in enumerate(string):
if letter not in d:
d[letter] = [i]
else:
d[letter].append(i)
n = int(input())
for i in range(n):
inp = input().split()
p, c = int(inp[0]), inp[1]
index = d[c][p-1]
string[index] = '$'
d[c].pop(p-1)
print(*filter(lambda x: x != '$', string), sep='')
```
|
output
| 1
| 7,817
| 6
| 15,635
|
Provide tags and a correct Python 3 solution for this coding contest problem.
One popular website developed an unusual username editing procedure. One can change the username only by deleting some characters from it: to change the current name s, a user can pick number p and character c and delete the p-th occurrence of character c from the name. After the user changed his name, he can't undo the change.
For example, one can change name "arca" by removing the second occurrence of character "a" to get "arc".
Polycarpus learned that some user initially registered under nickname t, where t is a concatenation of k copies of string s. Also, Polycarpus knows the sequence of this user's name changes. Help Polycarpus figure out the user's final name.
Input
The first line contains an integer k (1 ≤ k ≤ 2000). The second line contains a non-empty string s, consisting of lowercase Latin letters, at most 100 characters long. The third line contains an integer n (0 ≤ n ≤ 20000) — the number of username changes. Each of the next n lines contains the actual changes, one per line. The changes are written as "pi ci" (without the quotes), where pi (1 ≤ pi ≤ 200000) is the number of occurrences of letter ci, ci is a lowercase Latin letter. It is guaranteed that the operations are correct, that is, the letter to be deleted always exists, and after all operations not all letters are deleted from the name. The letters' occurrences are numbered starting from 1.
Output
Print a single string — the user's final name after all changes are applied to it.
Examples
Input
2
bac
3
2 a
1 b
2 c
Output
acb
Input
1
abacaba
4
1 a
1 a
1 c
2 b
Output
baa
Note
Let's consider the first sample. Initially we have name "bacbac"; the first operation transforms it into "bacbc", the second one — to "acbc", and finally, the third one transforms it into "acb".
|
instruction
| 0
| 7,818
| 6
| 15,636
|
Tags: *special, binary search, brute force, data structures, strings
Correct Solution:
```
n = int(input())
t = input()
l = len(t)
m = int(input())
q = {c: [] for c in 'abcdefghijklmnopqrstuvwxyz'}
for j, c in enumerate(t):
q[c].append(j)
q = {c: [i + j for i in range(0, n * l, l) for j in q[c]] for c in q}
t = n * list(t)
for i in range(m):
j, c = input().split()
t[q[c].pop(int(j) - 1)] = ''
print(''.join(t))
# Made By Mostafa_Khaled
```
|
output
| 1
| 7,818
| 6
| 15,637
|
Provide tags and a correct Python 3 solution for this coding contest problem.
One popular website developed an unusual username editing procedure. One can change the username only by deleting some characters from it: to change the current name s, a user can pick number p and character c and delete the p-th occurrence of character c from the name. After the user changed his name, he can't undo the change.
For example, one can change name "arca" by removing the second occurrence of character "a" to get "arc".
Polycarpus learned that some user initially registered under nickname t, where t is a concatenation of k copies of string s. Also, Polycarpus knows the sequence of this user's name changes. Help Polycarpus figure out the user's final name.
Input
The first line contains an integer k (1 ≤ k ≤ 2000). The second line contains a non-empty string s, consisting of lowercase Latin letters, at most 100 characters long. The third line contains an integer n (0 ≤ n ≤ 20000) — the number of username changes. Each of the next n lines contains the actual changes, one per line. The changes are written as "pi ci" (without the quotes), where pi (1 ≤ pi ≤ 200000) is the number of occurrences of letter ci, ci is a lowercase Latin letter. It is guaranteed that the operations are correct, that is, the letter to be deleted always exists, and after all operations not all letters are deleted from the name. The letters' occurrences are numbered starting from 1.
Output
Print a single string — the user's final name after all changes are applied to it.
Examples
Input
2
bac
3
2 a
1 b
2 c
Output
acb
Input
1
abacaba
4
1 a
1 a
1 c
2 b
Output
baa
Note
Let's consider the first sample. Initially we have name "bacbac"; the first operation transforms it into "bacbc", the second one — to "acbc", and finally, the third one transforms it into "acb".
|
instruction
| 0
| 7,819
| 6
| 15,638
|
Tags: *special, binary search, brute force, data structures, strings
Correct Solution:
```
from collections import defaultdict
k = int(input())
s = list(input() * k)
p = defaultdict(list)
for i in range(len(s)):
p[(s[i])].append(i)
for _ in range(int(input())):
x, c = input().split()
x = int(x)
s[p[(c)].pop(int(x) - 1)] = ""
print("".join(s))
```
|
output
| 1
| 7,819
| 6
| 15,639
|
Provide tags and a correct Python 3 solution for this coding contest problem.
One popular website developed an unusual username editing procedure. One can change the username only by deleting some characters from it: to change the current name s, a user can pick number p and character c and delete the p-th occurrence of character c from the name. After the user changed his name, he can't undo the change.
For example, one can change name "arca" by removing the second occurrence of character "a" to get "arc".
Polycarpus learned that some user initially registered under nickname t, where t is a concatenation of k copies of string s. Also, Polycarpus knows the sequence of this user's name changes. Help Polycarpus figure out the user's final name.
Input
The first line contains an integer k (1 ≤ k ≤ 2000). The second line contains a non-empty string s, consisting of lowercase Latin letters, at most 100 characters long. The third line contains an integer n (0 ≤ n ≤ 20000) — the number of username changes. Each of the next n lines contains the actual changes, one per line. The changes are written as "pi ci" (without the quotes), where pi (1 ≤ pi ≤ 200000) is the number of occurrences of letter ci, ci is a lowercase Latin letter. It is guaranteed that the operations are correct, that is, the letter to be deleted always exists, and after all operations not all letters are deleted from the name. The letters' occurrences are numbered starting from 1.
Output
Print a single string — the user's final name after all changes are applied to it.
Examples
Input
2
bac
3
2 a
1 b
2 c
Output
acb
Input
1
abacaba
4
1 a
1 a
1 c
2 b
Output
baa
Note
Let's consider the first sample. Initially we have name "bacbac"; the first operation transforms it into "bacbc", the second one — to "acbc", and finally, the third one transforms it into "acb".
|
instruction
| 0
| 7,820
| 6
| 15,640
|
Tags: *special, binary search, brute force, data structures, strings
Correct Solution:
```
k=int(input())
e={}
s=input()
t=list(s*k)
for i in range(len(t)):
if t[i] in e:e[t[i]]+=[i]
else:e[t[i]]=[i]
for i in range(int(input())):
q,w=input().split()
q=int(q)-1
t[e[w][q]]=""
e[w].pop(q)
print("".join(t))
```
|
output
| 1
| 7,820
| 6
| 15,641
|
Provide tags and a correct Python 3 solution for this coding contest problem.
One popular website developed an unusual username editing procedure. One can change the username only by deleting some characters from it: to change the current name s, a user can pick number p and character c and delete the p-th occurrence of character c from the name. After the user changed his name, he can't undo the change.
For example, one can change name "arca" by removing the second occurrence of character "a" to get "arc".
Polycarpus learned that some user initially registered under nickname t, where t is a concatenation of k copies of string s. Also, Polycarpus knows the sequence of this user's name changes. Help Polycarpus figure out the user's final name.
Input
The first line contains an integer k (1 ≤ k ≤ 2000). The second line contains a non-empty string s, consisting of lowercase Latin letters, at most 100 characters long. The third line contains an integer n (0 ≤ n ≤ 20000) — the number of username changes. Each of the next n lines contains the actual changes, one per line. The changes are written as "pi ci" (without the quotes), where pi (1 ≤ pi ≤ 200000) is the number of occurrences of letter ci, ci is a lowercase Latin letter. It is guaranteed that the operations are correct, that is, the letter to be deleted always exists, and after all operations not all letters are deleted from the name. The letters' occurrences are numbered starting from 1.
Output
Print a single string — the user's final name after all changes are applied to it.
Examples
Input
2
bac
3
2 a
1 b
2 c
Output
acb
Input
1
abacaba
4
1 a
1 a
1 c
2 b
Output
baa
Note
Let's consider the first sample. Initially we have name "bacbac"; the first operation transforms it into "bacbc", the second one — to "acbc", and finally, the third one transforms it into "acb".
|
instruction
| 0
| 7,821
| 6
| 15,642
|
Tags: *special, binary search, brute force, data structures, strings
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
##########################################################
from collections import Counter
# c=sorted((i,int(val))for i,val in enumerate(input().split()))
import heapq
# c=sorted((i,int(val))for i,val in enumerate(input().split()))
# n = int(input())
# ls = list(map(int, input().split()))
# n, k = map(int, input().split())
# n =int(input())
# e=list(map(int, input().split()))
from collections import Counter
#for _ in range(int(input())):
k=int(input())
ans=list(input()*k)
d={}
#print(ans)
for i,ch in enumerate(ans):
if ch not in d:
d[ch]=[i]
else:
d[ch].append(i)
#print(d)
for _ in range(int(input())):
var=input().split()
ch=var[1]
pos=int(var[0])
ans[d[ch][pos-1]]=""
d[ch].pop(pos-1)
print("".join(ans))
```
|
output
| 1
| 7,821
| 6
| 15,643
|
Provide tags and a correct Python 3 solution for this coding contest problem.
One popular website developed an unusual username editing procedure. One can change the username only by deleting some characters from it: to change the current name s, a user can pick number p and character c and delete the p-th occurrence of character c from the name. After the user changed his name, he can't undo the change.
For example, one can change name "arca" by removing the second occurrence of character "a" to get "arc".
Polycarpus learned that some user initially registered under nickname t, where t is a concatenation of k copies of string s. Also, Polycarpus knows the sequence of this user's name changes. Help Polycarpus figure out the user's final name.
Input
The first line contains an integer k (1 ≤ k ≤ 2000). The second line contains a non-empty string s, consisting of lowercase Latin letters, at most 100 characters long. The third line contains an integer n (0 ≤ n ≤ 20000) — the number of username changes. Each of the next n lines contains the actual changes, one per line. The changes are written as "pi ci" (without the quotes), where pi (1 ≤ pi ≤ 200000) is the number of occurrences of letter ci, ci is a lowercase Latin letter. It is guaranteed that the operations are correct, that is, the letter to be deleted always exists, and after all operations not all letters are deleted from the name. The letters' occurrences are numbered starting from 1.
Output
Print a single string — the user's final name after all changes are applied to it.
Examples
Input
2
bac
3
2 a
1 b
2 c
Output
acb
Input
1
abacaba
4
1 a
1 a
1 c
2 b
Output
baa
Note
Let's consider the first sample. Initially we have name "bacbac"; the first operation transforms it into "bacbc", the second one — to "acbc", and finally, the third one transforms it into "acb".
|
instruction
| 0
| 7,822
| 6
| 15,644
|
Tags: *special, binary search, brute force, data structures, strings
Correct Solution:
```
n = int(input())
t = input()
l = len(t)
m = int(input())
q = {c: [] for c in 'abcdefghijklmnopqrstuvwxyz'}
for j, c in enumerate(t):
q[c].append(j)
q = {c: [i + j for i in range(0, n * l, l) for j in q[c]] for c in q}
t = n * list(t)
for i in range(m):
j, c = input().split()
t[q[c].pop(int(j) - 1)] = ''
print(''.join(t))
```
|
output
| 1
| 7,822
| 6
| 15,645
|
Provide tags and a correct Python 3 solution for this coding contest problem.
One popular website developed an unusual username editing procedure. One can change the username only by deleting some characters from it: to change the current name s, a user can pick number p and character c and delete the p-th occurrence of character c from the name. After the user changed his name, he can't undo the change.
For example, one can change name "arca" by removing the second occurrence of character "a" to get "arc".
Polycarpus learned that some user initially registered under nickname t, where t is a concatenation of k copies of string s. Also, Polycarpus knows the sequence of this user's name changes. Help Polycarpus figure out the user's final name.
Input
The first line contains an integer k (1 ≤ k ≤ 2000). The second line contains a non-empty string s, consisting of lowercase Latin letters, at most 100 characters long. The third line contains an integer n (0 ≤ n ≤ 20000) — the number of username changes. Each of the next n lines contains the actual changes, one per line. The changes are written as "pi ci" (without the quotes), where pi (1 ≤ pi ≤ 200000) is the number of occurrences of letter ci, ci is a lowercase Latin letter. It is guaranteed that the operations are correct, that is, the letter to be deleted always exists, and after all operations not all letters are deleted from the name. The letters' occurrences are numbered starting from 1.
Output
Print a single string — the user's final name after all changes are applied to it.
Examples
Input
2
bac
3
2 a
1 b
2 c
Output
acb
Input
1
abacaba
4
1 a
1 a
1 c
2 b
Output
baa
Note
Let's consider the first sample. Initially we have name "bacbac"; the first operation transforms it into "bacbc", the second one — to "acbc", and finally, the third one transforms it into "acb".
|
instruction
| 0
| 7,823
| 6
| 15,646
|
Tags: *special, binary search, brute force, data structures, strings
Correct Solution:
```
#copied... idea
def main():
mode="filee"
if mode=="file":f=open("test.txt","r")
get = lambda :[int(x) for x in (f.readline() if mode=="file" else input()).split()]
gets = lambda :[str(x) for x in (f.readline() if mode=="file" else input()).split()]
[k]=get()
[g]=gets()
h = g*k
h = list(h)
p = []
for i in range(128):
p.append([0])
for i in range(len(h)):
p[ord(h[i])].append(i)
[n]=get()
for i in range(n):
[x,y]=gets()
x = int(x)
h[p[ord(y)][x]]=''
p[ord(y)].pop(x)
print("".join(h))
if mode=="file":f.close()
if __name__=="__main__":
main()
```
|
output
| 1
| 7,823
| 6
| 15,647
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One popular website developed an unusual username editing procedure. One can change the username only by deleting some characters from it: to change the current name s, a user can pick number p and character c and delete the p-th occurrence of character c from the name. After the user changed his name, he can't undo the change.
For example, one can change name "arca" by removing the second occurrence of character "a" to get "arc".
Polycarpus learned that some user initially registered under nickname t, where t is a concatenation of k copies of string s. Also, Polycarpus knows the sequence of this user's name changes. Help Polycarpus figure out the user's final name.
Input
The first line contains an integer k (1 ≤ k ≤ 2000). The second line contains a non-empty string s, consisting of lowercase Latin letters, at most 100 characters long. The third line contains an integer n (0 ≤ n ≤ 20000) — the number of username changes. Each of the next n lines contains the actual changes, one per line. The changes are written as "pi ci" (without the quotes), where pi (1 ≤ pi ≤ 200000) is the number of occurrences of letter ci, ci is a lowercase Latin letter. It is guaranteed that the operations are correct, that is, the letter to be deleted always exists, and after all operations not all letters are deleted from the name. The letters' occurrences are numbered starting from 1.
Output
Print a single string — the user's final name after all changes are applied to it.
Examples
Input
2
bac
3
2 a
1 b
2 c
Output
acb
Input
1
abacaba
4
1 a
1 a
1 c
2 b
Output
baa
Note
Let's consider the first sample. Initially we have name "bacbac"; the first operation transforms it into "bacbc", the second one — to "acbc", and finally, the third one transforms it into "acb".
Submitted Solution:
```
n = int(input())
s = input()
sens = [[]]
for i in s:
sens[-1].append(i)
if i in ['.', '!', '?']:
sens.append([])
for i in range(len(sens)):
if sens[i]:
sens[i] = ''.join(sens[i])
sens[i] = sens[i].strip()
if len(sens[i]) > n:
print('Impossible')
exit(0)
sens.pop()
i = 0
ans = 0
while i < len(sens):
l = len(sens[i])
while i + 1 < len(sens) and l + 1 + len(sens[i + 1]) <= n:
i += 1
l += len(sens[i + 1]) + 1
i += 1
ans += 1
print(ans)
```
|
instruction
| 0
| 7,824
| 6
| 15,648
|
No
|
output
| 1
| 7,824
| 6
| 15,649
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One popular website developed an unusual username editing procedure. One can change the username only by deleting some characters from it: to change the current name s, a user can pick number p and character c and delete the p-th occurrence of character c from the name. After the user changed his name, he can't undo the change.
For example, one can change name "arca" by removing the second occurrence of character "a" to get "arc".
Polycarpus learned that some user initially registered under nickname t, where t is a concatenation of k copies of string s. Also, Polycarpus knows the sequence of this user's name changes. Help Polycarpus figure out the user's final name.
Input
The first line contains an integer k (1 ≤ k ≤ 2000). The second line contains a non-empty string s, consisting of lowercase Latin letters, at most 100 characters long. The third line contains an integer n (0 ≤ n ≤ 20000) — the number of username changes. Each of the next n lines contains the actual changes, one per line. The changes are written as "pi ci" (without the quotes), where pi (1 ≤ pi ≤ 200000) is the number of occurrences of letter ci, ci is a lowercase Latin letter. It is guaranteed that the operations are correct, that is, the letter to be deleted always exists, and after all operations not all letters are deleted from the name. The letters' occurrences are numbered starting from 1.
Output
Print a single string — the user's final name after all changes are applied to it.
Examples
Input
2
bac
3
2 a
1 b
2 c
Output
acb
Input
1
abacaba
4
1 a
1 a
1 c
2 b
Output
baa
Note
Let's consider the first sample. Initially we have name "bacbac"; the first operation transforms it into "bacbc", the second one — to "acbc", and finally, the third one transforms it into "acb".
Submitted Solution:
```
n = int(input())
t = input()
l = len(t[0])
m = int(input())
q = {c: [] for c in 'abcdefghijklmnopqrstuvwxyz'}
for j, c in enumerate(t):
q[c].append(j)
q = {c: [i + j for i in range(0, n * l, l) for j in q[c]] for c in q}
t = n * list(t)
for i in range(m):
j, c = input().split()
t[q[c].pop(int(j) - 1)] = ''
print(''.join(t))
```
|
instruction
| 0
| 7,825
| 6
| 15,650
|
No
|
output
| 1
| 7,825
| 6
| 15,651
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One popular website developed an unusual username editing procedure. One can change the username only by deleting some characters from it: to change the current name s, a user can pick number p and character c and delete the p-th occurrence of character c from the name. After the user changed his name, he can't undo the change.
For example, one can change name "arca" by removing the second occurrence of character "a" to get "arc".
Polycarpus learned that some user initially registered under nickname t, where t is a concatenation of k copies of string s. Also, Polycarpus knows the sequence of this user's name changes. Help Polycarpus figure out the user's final name.
Input
The first line contains an integer k (1 ≤ k ≤ 2000). The second line contains a non-empty string s, consisting of lowercase Latin letters, at most 100 characters long. The third line contains an integer n (0 ≤ n ≤ 20000) — the number of username changes. Each of the next n lines contains the actual changes, one per line. The changes are written as "pi ci" (without the quotes), where pi (1 ≤ pi ≤ 200000) is the number of occurrences of letter ci, ci is a lowercase Latin letter. It is guaranteed that the operations are correct, that is, the letter to be deleted always exists, and after all operations not all letters are deleted from the name. The letters' occurrences are numbered starting from 1.
Output
Print a single string — the user's final name after all changes are applied to it.
Examples
Input
2
bac
3
2 a
1 b
2 c
Output
acb
Input
1
abacaba
4
1 a
1 a
1 c
2 b
Output
baa
Note
Let's consider the first sample. Initially we have name "bacbac"; the first operation transforms it into "bacbc", the second one — to "acbc", and finally, the third one transforms it into "acb".
Submitted Solution:
```
n = int(input())
t = input()
r = [''] + [t] * n
m = int(input())
q = {c: [] for c in set(t)}
for j, c in enumerate(t):
q[c].append(j)
p = {c: [] for c in q}
for i in range(m):
j, c = input().split()
p[c].append(int(j))
for c in p:
for i in range(len(p[c])):
k = p[c][i]
for j in range(i + 1, len(p[c])):
if p[c][j] >= k: p[c][j] += 1
for c in p:
l = len(q[c])
for i in range(len(p[c])):
x, y = p[c][i] // l, p[c][i] % l
r[x] = r[x][: q[c][y]] + ' ' + r[x][q[c][y] + 1: ]
t = ''.join(r)
print(t.replace(' ', ''))
```
|
instruction
| 0
| 7,826
| 6
| 15,652
|
No
|
output
| 1
| 7,826
| 6
| 15,653
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Limak is a little polar bear. He likes nice strings — strings of length n, consisting of lowercase English letters only.
The distance between two letters is defined as the difference between their positions in the alphabet. For example, <image>, and <image>.
Also, the distance between two nice strings is defined as the sum of distances of corresponding letters. For example, <image>, and <image>.
Limak gives you a nice string s and an integer k. He challenges you to find any nice string s' that <image>. Find any s' satisfying the given conditions, or print "-1" if it's impossible to do so.
As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use gets/scanf/printf instead of getline/cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.
Input
The first line contains two integers n and k (1 ≤ n ≤ 105, 0 ≤ k ≤ 106).
The second line contains a string s of length n, consisting of lowercase English letters.
Output
If there is no string satisfying the given conditions then print "-1" (without the quotes).
Otherwise, print any nice string s' that <image>.
Examples
Input
4 26
bear
Output
roar
Input
2 7
af
Output
db
Input
3 1000
hey
Output
-1
|
instruction
| 0
| 7,977
| 6
| 15,954
|
Tags: greedy, strings
Correct Solution:
```
s, k = map(int, input().split())
strin = list(input())
newStr = strin
d = 0
m = ord("m")
a = ord("a")
z = ord("z")
i = 0
while i < s and d < k:
diff = 0
c = ord(strin[i])
if c <= m:
diff = z - c
else:
diff = a - c
if (abs(diff) + d) <= k:
d += abs(diff)
newStr[i] = chr(c + diff)
else:
if c <= m:
c += k - d
else:
c -= k - d
d += abs(k - d)
newStr[i] = chr(c)
i += 1
if d == k:
print("".join(newStr))
else:
print("-1")
```
|
output
| 1
| 7,977
| 6
| 15,955
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Limak is a little polar bear. He likes nice strings — strings of length n, consisting of lowercase English letters only.
The distance between two letters is defined as the difference between their positions in the alphabet. For example, <image>, and <image>.
Also, the distance between two nice strings is defined as the sum of distances of corresponding letters. For example, <image>, and <image>.
Limak gives you a nice string s and an integer k. He challenges you to find any nice string s' that <image>. Find any s' satisfying the given conditions, or print "-1" if it's impossible to do so.
As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use gets/scanf/printf instead of getline/cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.
Input
The first line contains two integers n and k (1 ≤ n ≤ 105, 0 ≤ k ≤ 106).
The second line contains a string s of length n, consisting of lowercase English letters.
Output
If there is no string satisfying the given conditions then print "-1" (without the quotes).
Otherwise, print any nice string s' that <image>.
Examples
Input
4 26
bear
Output
roar
Input
2 7
af
Output
db
Input
3 1000
hey
Output
-1
|
instruction
| 0
| 7,978
| 6
| 15,956
|
Tags: greedy, strings
Correct Solution:
```
n, k = map(int, input().split())
b = list(input())
for i in range(n):
x = ord('z') - ord(b[i])
y = ord(b[i]) - ord('a')
if x > y:
if k < x:
x = k
b[i] = chr(ord(b[i]) + x)
k -= x
else:
if k < y:
y = k
b[i] = chr(ord(b[i]) - y)
k -= y
if k==0:
break
if k > 0:
print (-1)
else:
s = ''.join(b)
print (s)
```
|
output
| 1
| 7,978
| 6
| 15,957
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Limak is a little polar bear. He likes nice strings — strings of length n, consisting of lowercase English letters only.
The distance between two letters is defined as the difference between their positions in the alphabet. For example, <image>, and <image>.
Also, the distance between two nice strings is defined as the sum of distances of corresponding letters. For example, <image>, and <image>.
Limak gives you a nice string s and an integer k. He challenges you to find any nice string s' that <image>. Find any s' satisfying the given conditions, or print "-1" if it's impossible to do so.
As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use gets/scanf/printf instead of getline/cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.
Input
The first line contains two integers n and k (1 ≤ n ≤ 105, 0 ≤ k ≤ 106).
The second line contains a string s of length n, consisting of lowercase English letters.
Output
If there is no string satisfying the given conditions then print "-1" (without the quotes).
Otherwise, print any nice string s' that <image>.
Examples
Input
4 26
bear
Output
roar
Input
2 7
af
Output
db
Input
3 1000
hey
Output
-1
|
instruction
| 0
| 7,979
| 6
| 15,958
|
Tags: greedy, strings
Correct Solution:
```
n,k = map(int,input().split())
s = input()
s = list(s)
for i in range(n):
if ord(s[i])>=110:
m = min(ord(s[i])-97,k)
k-=m
s[i]=chr(ord(s[i])-m)
else:
m = min(122-ord(s[i]),k)
k-=m
s[i]=chr(ord(s[i])+m)
if k>0:
print(-1)
else:
print(''.join(s))
```
|
output
| 1
| 7,979
| 6
| 15,959
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Limak is a little polar bear. He likes nice strings — strings of length n, consisting of lowercase English letters only.
The distance between two letters is defined as the difference between their positions in the alphabet. For example, <image>, and <image>.
Also, the distance between two nice strings is defined as the sum of distances of corresponding letters. For example, <image>, and <image>.
Limak gives you a nice string s and an integer k. He challenges you to find any nice string s' that <image>. Find any s' satisfying the given conditions, or print "-1" if it's impossible to do so.
As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use gets/scanf/printf instead of getline/cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.
Input
The first line contains two integers n and k (1 ≤ n ≤ 105, 0 ≤ k ≤ 106).
The second line contains a string s of length n, consisting of lowercase English letters.
Output
If there is no string satisfying the given conditions then print "-1" (without the quotes).
Otherwise, print any nice string s' that <image>.
Examples
Input
4 26
bear
Output
roar
Input
2 7
af
Output
db
Input
3 1000
hey
Output
-1
|
instruction
| 0
| 7,980
| 6
| 15,960
|
Tags: greedy, strings
Correct Solution:
```
import collections
import math
n, k = map(int, input().split())
s = input()
ss = []
t = 0
for i in range(len(s)):
t += max(abs(ord('a') - ord(s[i])), abs(ord('z') - ord(s[i])))
if k > t:
print(-1)
exit(0)
else:
for i in range(len(s)):
if k == 0:
ss.append(s[i])
continue
x, y = abs(ord('a') - ord(s[i])), abs(ord('z') - ord(s[i]))
if x >= y:
if k >= x:
ss.append('a')
k -= x
else:
ss.append(chr(ord(s[i]) - k))
k = 0
else:
if k >= y:
ss.append('z')
k -= y
else:
ss.append(chr(ord(s[i]) + k))
k = 0
print(''.join(ss))
```
|
output
| 1
| 7,980
| 6
| 15,961
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Limak is a little polar bear. He likes nice strings — strings of length n, consisting of lowercase English letters only.
The distance between two letters is defined as the difference between their positions in the alphabet. For example, <image>, and <image>.
Also, the distance between two nice strings is defined as the sum of distances of corresponding letters. For example, <image>, and <image>.
Limak gives you a nice string s and an integer k. He challenges you to find any nice string s' that <image>. Find any s' satisfying the given conditions, or print "-1" if it's impossible to do so.
As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use gets/scanf/printf instead of getline/cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.
Input
The first line contains two integers n and k (1 ≤ n ≤ 105, 0 ≤ k ≤ 106).
The second line contains a string s of length n, consisting of lowercase English letters.
Output
If there is no string satisfying the given conditions then print "-1" (without the quotes).
Otherwise, print any nice string s' that <image>.
Examples
Input
4 26
bear
Output
roar
Input
2 7
af
Output
db
Input
3 1000
hey
Output
-1
|
instruction
| 0
| 7,981
| 6
| 15,962
|
Tags: greedy, strings
Correct Solution:
```
n, k = map(int, input().split())
st = input()
ans = ''
i = 0
while i < n and k > 0:
if abs(ord(st[i]) - 97) > abs(ord(st[i]) - 122):
ans += chr(max(97, ord(st[i]) - k))
k -= ord(st[i]) - ord(ans[-1])
else:
ans += chr(min(122, ord(st[i]) + k))
k -= ord(ans[-1]) - ord(st[i])
i += 1
if k == 0:
print(ans + st[i:])
else:
print(-1)
```
|
output
| 1
| 7,981
| 6
| 15,963
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Limak is a little polar bear. He likes nice strings — strings of length n, consisting of lowercase English letters only.
The distance between two letters is defined as the difference between their positions in the alphabet. For example, <image>, and <image>.
Also, the distance between two nice strings is defined as the sum of distances of corresponding letters. For example, <image>, and <image>.
Limak gives you a nice string s and an integer k. He challenges you to find any nice string s' that <image>. Find any s' satisfying the given conditions, or print "-1" if it's impossible to do so.
As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use gets/scanf/printf instead of getline/cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.
Input
The first line contains two integers n and k (1 ≤ n ≤ 105, 0 ≤ k ≤ 106).
The second line contains a string s of length n, consisting of lowercase English letters.
Output
If there is no string satisfying the given conditions then print "-1" (without the quotes).
Otherwise, print any nice string s' that <image>.
Examples
Input
4 26
bear
Output
roar
Input
2 7
af
Output
db
Input
3 1000
hey
Output
-1
|
instruction
| 0
| 7,982
| 6
| 15,964
|
Tags: greedy, strings
Correct Solution:
```
[n,k] = input().split(" ")
s = input()
ss = ''
n = int(n)
k = int(k)
MaxD = [[0,0]]*n
SumMaxD = 0
for i in list(range(0,n)):
xl = -97 + ord(s[i])
xh = 122 - ord(s[i])
if(xl > xh):
MaxD[i] = [xl,-1]
SumMaxD = SumMaxD + xl
else:
MaxD[i] = [xh,1]
SumMaxD = SumMaxD + xh
if(SumMaxD < k):
print(-1)
else:
for i in range(0,n):
if(MaxD[i][0]<k):
ss =ss + chr(ord(s[i]) + MaxD[i][0]*MaxD[i][1])
k = k - MaxD[i][0]
else:
ss =ss + chr(ord(s[i]) + k*MaxD[i][1]) + s[(i+1):len(s):1]
break
print(ss)
```
|
output
| 1
| 7,982
| 6
| 15,965
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Limak is a little polar bear. He likes nice strings — strings of length n, consisting of lowercase English letters only.
The distance between two letters is defined as the difference between their positions in the alphabet. For example, <image>, and <image>.
Also, the distance between two nice strings is defined as the sum of distances of corresponding letters. For example, <image>, and <image>.
Limak gives you a nice string s and an integer k. He challenges you to find any nice string s' that <image>. Find any s' satisfying the given conditions, or print "-1" if it's impossible to do so.
As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use gets/scanf/printf instead of getline/cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.
Input
The first line contains two integers n and k (1 ≤ n ≤ 105, 0 ≤ k ≤ 106).
The second line contains a string s of length n, consisting of lowercase English letters.
Output
If there is no string satisfying the given conditions then print "-1" (without the quotes).
Otherwise, print any nice string s' that <image>.
Examples
Input
4 26
bear
Output
roar
Input
2 7
af
Output
db
Input
3 1000
hey
Output
-1
|
instruction
| 0
| 7,983
| 6
| 15,966
|
Tags: greedy, strings
Correct Solution:
```
def check(s, k):
ans = 0
for i in range(len(s)):
ans += abs(ord(s[i]) - ord(k[i]))
return ans
n, k = map(int, input().split())
s = input()
cnt = 0
for i in s:
cnt += max(ord('z') - ord(i), ord(i) - ord('a'))
if k > cnt:
print(-1)
exit()
else:
ans = ''
cr = 0
while k != 0:
ps1 = ord(s[cr]) - ord('a')
ps2 = ord('z') - ord(s[cr])
if ps1 > k:
ans += chr(ord(s[cr]) - k)
k = 0
elif ps2 > k:
ans += chr(ord(s[cr]) + k)
k = 0
else:
if ps2 >= ps1:
ans += 'z'
k -= ps2
else:
ans += 'a'
k -= ps1
cr += 1
ans += s[len(ans):]
print(ans)
#print(check(ans, s))
```
|
output
| 1
| 7,983
| 6
| 15,967
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Limak is a little polar bear. He likes nice strings — strings of length n, consisting of lowercase English letters only.
The distance between two letters is defined as the difference between their positions in the alphabet. For example, <image>, and <image>.
Also, the distance between two nice strings is defined as the sum of distances of corresponding letters. For example, <image>, and <image>.
Limak gives you a nice string s and an integer k. He challenges you to find any nice string s' that <image>. Find any s' satisfying the given conditions, or print "-1" if it's impossible to do so.
As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use gets/scanf/printf instead of getline/cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.
Input
The first line contains two integers n and k (1 ≤ n ≤ 105, 0 ≤ k ≤ 106).
The second line contains a string s of length n, consisting of lowercase English letters.
Output
If there is no string satisfying the given conditions then print "-1" (without the quotes).
Otherwise, print any nice string s' that <image>.
Examples
Input
4 26
bear
Output
roar
Input
2 7
af
Output
db
Input
3 1000
hey
Output
-1
|
instruction
| 0
| 7,984
| 6
| 15,968
|
Tags: greedy, strings
Correct Solution:
```
a, b = map(int, input().split())
s = input()
if b / a > 25:
print(-1)
else:
ans = ""
c = 0
for i in s:
if b == 0:
ans += s[c:]
break
idx = ord(i) - 97
if idx >= 13:
if b > idx:
ans += "a"
b -= idx
else:
ans += chr(idx + 97 - b) # alepha[idx - b]
b = 0
else:
if b > 25-idx:
ans += "z"
b -= 25 - idx
else:
ans += chr(idx + 97 + b) # alepha[idx + b]
b = 0
c += 1
if b == 0:
print(ans.lower())
else:
print(-1)
```
|
output
| 1
| 7,984
| 6
| 15,969
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f
|
instruction
| 0
| 8,087
| 6
| 16,174
|
Tags: dp, greedy, implementation
Correct Solution:
```
def main():
s = input()
a = []
g = [elem for elem in "aeiou"]
p = []
for i in range(len(s)):
if (s[i] not in g):
a.append(s[i])
else:
a = []
if len(a) == 3:
if (a[0] != a[1]) or (a[1] != a[2]):
a = [ a[2] ]
p.append(i)
else:
a = [ a[1], a[2] ]
for i in range(len(s)):
if i in p:
print(" ", end='')
print(s[i], end='')
if __name__ == "__main__":
main()
```
|
output
| 1
| 8,087
| 6
| 16,175
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f
|
instruction
| 0
| 8,088
| 6
| 16,176
|
Tags: dp, greedy, implementation
Correct Solution:
```
s = input()
vowels = ('a', 'e', 'u', 'o', 'i')
res = ''
i = 0
while i < len(s) - 2:
if s[i] not in vowels and s[i + 1] not in vowels and s[i + 2] not in vowels and s[i: i + 3] != s[i] * 3:
res += s[i: i + 2] + ' '
i += 2
else:
res += s[i]
i += 1
res += s[i:]
print(res)
```
|
output
| 1
| 8,088
| 6
| 16,177
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f
|
instruction
| 0
| 8,089
| 6
| 16,178
|
Tags: dp, greedy, implementation
Correct Solution:
```
class Solver:
def main(self):
s = input()
ls = len(s)
vowels = 'aeiou'
cnt = 0
breaks = []
same = 0
for i in range(0, ls):
if s[i] in vowels:
cnt = 0
same = 0
elif i > 0 and s[i-1] == s[i]:
same += 1
cnt = cnt
else:
cnt += 1
if same + cnt >= 3 and cnt >= 2:
breaks.append(i)
cnt = 1
same = 0
lbreaks = len(breaks)
current = 0
for i in range(ls):
if current < lbreaks and i == breaks[current]:
print(' ', end='')
current += 1
print(s[i], end='')
print()
Solver().main()
```
|
output
| 1
| 8,089
| 6
| 16,179
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f
|
instruction
| 0
| 8,090
| 6
| 16,180
|
Tags: dp, greedy, implementation
Correct Solution:
```
word = list(input())
ns = 0
onesymbol = ''
onesymbolcount = 0
for i in range(len(word)):
if onesymbol == word[i] and ns == 0:
onesymbolcount += 1
continue
if word[i] in ['a', 'e', 'i', 'o', 'u']:
ns = 0
onesymbolcount = 0
onesymbol = ''
else:
if onesymbol != '':
ns += onesymbolcount
onesymbol = word[i]
onesymbolcount = 1
if ns >= 2:
word[i-1] += ' '
ns = 0
# print(word[i], ns, onesymbol, onesymbolcount)
print("".join(word))
```
|
output
| 1
| 8,090
| 6
| 16,181
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f
|
instruction
| 0
| 8,091
| 6
| 16,182
|
Tags: dp, greedy, implementation
Correct Solution:
```
k = 0
t = y = z = ''
for x in input():
k = 0 if x in 'aeiou' else k + 1
if k > 2 and not (x == y == z):
t += ' '
k = 1
y, z = x, y
t += x
print(t)
```
|
output
| 1
| 8,091
| 6
| 16,183
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f
|
instruction
| 0
| 8,092
| 6
| 16,184
|
Tags: dp, greedy, implementation
Correct Solution:
```
n = input()
l = 0
q = len(n)
sogl = ['q', 'w', 'r', 't', 'y', 'p', 's', 'd', 'f', 'g', 'h', 'j', 'k', 'l', 'z', 'x', 'c', 'v', 'b', 'n', 'm']
for i in range(1, len(n) - 1):
if (((n[i - 1] != n[i + 1]) or ((n[i] != n[i - 1]) and (n[i-1] == n[i + 1]))) and (n[i - 1] in sogl) and (n[i] in sogl) and (n[i + 1] in sogl)):
print (n[l:i + 1:1], end=" ")
l = i + 1
n = n[0:i] + 'a' + n[i + 1:q]
print (n[l:len(n):1])
```
|
output
| 1
| 8,092
| 6
| 16,185
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.