input stringlengths 1 8.57k | output stringlengths 2 5.65k |
|---|---|
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After overcoming the stairs Dasha came to classes. She needed to write a password to begin her classes. The password is a string of length n which satisfies the following requirements:
* There is at least one digit in the string,
* There is at least one lowercase (small) letter of the Latin alphabet in the string,
* There is at least one of three listed symbols in the string: '#', '*', '&'.
<image>
Considering that these are programming classes it is not easy to write the password.
For each character of the password we have a fixed string of length m, on each of these n strings there is a pointer on some character. The i-th character displayed on the screen is the pointed character in the i-th string. Initially, all pointers are on characters with indexes 1 in the corresponding strings (all positions are numbered starting from one).
During one operation Dasha can move a pointer in one string one character to the left or to the right. Strings are cyclic, it means that when we move the pointer which is on the character with index 1 to the left, it moves to the character with the index m, and when we move it to the right from the position m it moves to the position 1.
You need to determine the minimum number of operations necessary to make the string displayed on the screen a valid password.
Input
The first line contains two integers n, m (3 β€ n β€ 50, 1 β€ m β€ 50) β the length of the password and the length of strings which are assigned to password symbols.
Each of the next n lines contains the string which is assigned to the i-th symbol of the password string. Its length is m, it consists of digits, lowercase English letters, and characters '#', '*' or '&'.
You have such input data that you can always get a valid password.
Output
Print one integer β the minimum number of operations which is necessary to make the string, which is displayed on the screen, a valid password.
Examples
Input
3 4
1**2
a3*0
c4**
Output
1
Input
5 5
#*&#*
*a1c&
&q2w*
#a3c#
*&#*&
Output
3
Note
In the first test it is necessary to move the pointer of the third string to one left to get the optimal answer.
<image>
In the second test one of possible algorithms will be:
* to move the pointer of the second symbol once to the right.
* to move the pointer of the third symbol twice to the right.
<image>
Submitted Solution:
```
n,m=map(int,input().split())
by=[m]*n
ch=[m]*n
sy=[m]*n
for x in range(n):
a=input()
for i,c in enumerate(a):
mi=min(m-i,i)
if '0'<=c<='9':
by[x]=min(by[x],mi)
elif 'a' <= c <= 'z':
ch[x]=min(ch[x],mi)
else:
sy[x]= min(sy[x],mi)
ans=m*3
for i in range(n):
for j in range(n):
if i==j:continue
for k in range(n):
if i==k or j==k:continue
ans=min(ans,by[i]+ch[j]+sy[k])
print(ans)
```
### Response:
| Yes |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After overcoming the stairs Dasha came to classes. She needed to write a password to begin her classes. The password is a string of length n which satisfies the following requirements:
* There is at least one digit in the string,
* There is at least one lowercase (small) letter of the Latin alphabet in the string,
* There is at least one of three listed symbols in the string: '#', '*', '&'.
<image>
Considering that these are programming classes it is not easy to write the password.
For each character of the password we have a fixed string of length m, on each of these n strings there is a pointer on some character. The i-th character displayed on the screen is the pointed character in the i-th string. Initially, all pointers are on characters with indexes 1 in the corresponding strings (all positions are numbered starting from one).
During one operation Dasha can move a pointer in one string one character to the left or to the right. Strings are cyclic, it means that when we move the pointer which is on the character with index 1 to the left, it moves to the character with the index m, and when we move it to the right from the position m it moves to the position 1.
You need to determine the minimum number of operations necessary to make the string displayed on the screen a valid password.
Input
The first line contains two integers n, m (3 β€ n β€ 50, 1 β€ m β€ 50) β the length of the password and the length of strings which are assigned to password symbols.
Each of the next n lines contains the string which is assigned to the i-th symbol of the password string. Its length is m, it consists of digits, lowercase English letters, and characters '#', '*' or '&'.
You have such input data that you can always get a valid password.
Output
Print one integer β the minimum number of operations which is necessary to make the string, which is displayed on the screen, a valid password.
Examples
Input
3 4
1**2
a3*0
c4**
Output
1
Input
5 5
#*&#*
*a1c&
&q2w*
#a3c#
*&#*&
Output
3
Note
In the first test it is necessary to move the pointer of the third string to one left to get the optimal answer.
<image>
In the second test one of possible algorithms will be:
* to move the pointer of the second symbol once to the right.
* to move the pointer of the third symbol twice to the right.
<image>
Submitted Solution:
```
#!/usr/bin/env python3
def ri():
return map(int, input().split())
digit = '0123456789'
letter = 'abcdefghijklmnopqrstuvwxyz'
sym = '#*&'
val = [[] for _ in range(3)]
val[0] = digit
val[1] = letter
val[2] = sym
#print(val)
n, m = ri()
minm = [[m for _ in range(3)] for __ in range(n)]
#print(minm)
for ln in range(n):
line = input()
#print(line)
for i in range(3):
for mov in range(m-1):
if (line[mov] in val[i]) or line[-mov] in val[i]:
minm[ln][i] = mov
break
ans = 50+50+50+50
for i1 in range(n):
for i2 in range(n):
for i3 in range(n):
if i1 != i2 and i1 != i3 and i2 != i3:
ans = min(ans, minm[i1][0] + minm[i2][1] + minm[i3][2])
print(ans)
```
### Response:
| No |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After overcoming the stairs Dasha came to classes. She needed to write a password to begin her classes. The password is a string of length n which satisfies the following requirements:
* There is at least one digit in the string,
* There is at least one lowercase (small) letter of the Latin alphabet in the string,
* There is at least one of three listed symbols in the string: '#', '*', '&'.
<image>
Considering that these are programming classes it is not easy to write the password.
For each character of the password we have a fixed string of length m, on each of these n strings there is a pointer on some character. The i-th character displayed on the screen is the pointed character in the i-th string. Initially, all pointers are on characters with indexes 1 in the corresponding strings (all positions are numbered starting from one).
During one operation Dasha can move a pointer in one string one character to the left or to the right. Strings are cyclic, it means that when we move the pointer which is on the character with index 1 to the left, it moves to the character with the index m, and when we move it to the right from the position m it moves to the position 1.
You need to determine the minimum number of operations necessary to make the string displayed on the screen a valid password.
Input
The first line contains two integers n, m (3 β€ n β€ 50, 1 β€ m β€ 50) β the length of the password and the length of strings which are assigned to password symbols.
Each of the next n lines contains the string which is assigned to the i-th symbol of the password string. Its length is m, it consists of digits, lowercase English letters, and characters '#', '*' or '&'.
You have such input data that you can always get a valid password.
Output
Print one integer β the minimum number of operations which is necessary to make the string, which is displayed on the screen, a valid password.
Examples
Input
3 4
1**2
a3*0
c4**
Output
1
Input
5 5
#*&#*
*a1c&
&q2w*
#a3c#
*&#*&
Output
3
Note
In the first test it is necessary to move the pointer of the third string to one left to get the optimal answer.
<image>
In the second test one of possible algorithms will be:
* to move the pointer of the second symbol once to the right.
* to move the pointer of the third symbol twice to the right.
<image>
Submitted Solution:
```
import string
n, m = map(int, input().split(' '))
inp = []
for i in range(n):
inp.append(input())
mn = [[m] * 3 for i in range(n)] # First for number, second for alphabet and third for special
for i in range(n):
for j in range(m):
index = 2
if inp[i][j] in string.ascii_lowercase:
index = 1
elif inp[i][j] in string.digits:
index = 0
mn[i][index] = min(mn[i][index], min(j, m - j))
ans = m * 3
for i in range(n):
for j in range(n):
if i == j:
continue
for k in range(n):
if i == k or j == k:
continue
ans = min(ans, mn[i][1] + mn[j][1] + mn[k][2])
print(ans)
```
### Response:
| No |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After overcoming the stairs Dasha came to classes. She needed to write a password to begin her classes. The password is a string of length n which satisfies the following requirements:
* There is at least one digit in the string,
* There is at least one lowercase (small) letter of the Latin alphabet in the string,
* There is at least one of three listed symbols in the string: '#', '*', '&'.
<image>
Considering that these are programming classes it is not easy to write the password.
For each character of the password we have a fixed string of length m, on each of these n strings there is a pointer on some character. The i-th character displayed on the screen is the pointed character in the i-th string. Initially, all pointers are on characters with indexes 1 in the corresponding strings (all positions are numbered starting from one).
During one operation Dasha can move a pointer in one string one character to the left or to the right. Strings are cyclic, it means that when we move the pointer which is on the character with index 1 to the left, it moves to the character with the index m, and when we move it to the right from the position m it moves to the position 1.
You need to determine the minimum number of operations necessary to make the string displayed on the screen a valid password.
Input
The first line contains two integers n, m (3 β€ n β€ 50, 1 β€ m β€ 50) β the length of the password and the length of strings which are assigned to password symbols.
Each of the next n lines contains the string which is assigned to the i-th symbol of the password string. Its length is m, it consists of digits, lowercase English letters, and characters '#', '*' or '&'.
You have such input data that you can always get a valid password.
Output
Print one integer β the minimum number of operations which is necessary to make the string, which is displayed on the screen, a valid password.
Examples
Input
3 4
1**2
a3*0
c4**
Output
1
Input
5 5
#*&#*
*a1c&
&q2w*
#a3c#
*&#*&
Output
3
Note
In the first test it is necessary to move the pointer of the third string to one left to get the optimal answer.
<image>
In the second test one of possible algorithms will be:
* to move the pointer of the second symbol once to the right.
* to move the pointer of the third symbol twice to the right.
<image>
Submitted Solution:
```
n = input()
n, m = n.split(sep=' ')
n = int(n)
m = int(m)
a = []
for i in range(n):
a.append(input())
h = [0, 0, 0] # 0-char 1-num 2-special
special = '#*&'
alph = 'abcdefghijklmnopqrstuvwxyz'
num = '0123456789'
c = 0
for s in a:
for i in range(len(s)):
if s[i] in alph and h[0] == 0:
h[0] += 1
c+=i
elif s[i] in num and h[1] == 0:
h[1] += 1
c += i
elif s[i] in special and h[2] == 0:
h[2] += 1
c+= i
if h == [1,1,1]:
break
print(c)
```
### Response:
| No |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After overcoming the stairs Dasha came to classes. She needed to write a password to begin her classes. The password is a string of length n which satisfies the following requirements:
* There is at least one digit in the string,
* There is at least one lowercase (small) letter of the Latin alphabet in the string,
* There is at least one of three listed symbols in the string: '#', '*', '&'.
<image>
Considering that these are programming classes it is not easy to write the password.
For each character of the password we have a fixed string of length m, on each of these n strings there is a pointer on some character. The i-th character displayed on the screen is the pointed character in the i-th string. Initially, all pointers are on characters with indexes 1 in the corresponding strings (all positions are numbered starting from one).
During one operation Dasha can move a pointer in one string one character to the left or to the right. Strings are cyclic, it means that when we move the pointer which is on the character with index 1 to the left, it moves to the character with the index m, and when we move it to the right from the position m it moves to the position 1.
You need to determine the minimum number of operations necessary to make the string displayed on the screen a valid password.
Input
The first line contains two integers n, m (3 β€ n β€ 50, 1 β€ m β€ 50) β the length of the password and the length of strings which are assigned to password symbols.
Each of the next n lines contains the string which is assigned to the i-th symbol of the password string. Its length is m, it consists of digits, lowercase English letters, and characters '#', '*' or '&'.
You have such input data that you can always get a valid password.
Output
Print one integer β the minimum number of operations which is necessary to make the string, which is displayed on the screen, a valid password.
Examples
Input
3 4
1**2
a3*0
c4**
Output
1
Input
5 5
#*&#*
*a1c&
&q2w*
#a3c#
*&#*&
Output
3
Note
In the first test it is necessary to move the pointer of the third string to one left to get the optimal answer.
<image>
In the second test one of possible algorithms will be:
* to move the pointer of the second symbol once to the right.
* to move the pointer of the third symbol twice to the right.
<image>
Submitted Solution:
```
from sys import stdin, exit, setrecursionlimit
from collections import deque
from string import ascii_lowercase
from itertools import *
from math import *
input = stdin.readline
lmi = lambda: list(map(int, input().split()))
mi = lambda: map(int, input().split())
si = lambda: input().strip('\n')
ssi = lambda: input().strip('\n').split()
mod = 10**9+7
n, m = mi()
s = [si() for i in range(n)]
e = [float("inf"), float("inf"), float("inf")]
for i in range(n):
for j in range(m):
if s[i][j] in "1234567890":
e[0] = min(e[0], j)
elif s[i][j] in ascii_lowercase:
e[1] = min(e[1], j)
else:
e[2] = min(e[2], j)
print(sum(e))
```
### Response:
| No |
### Instruction:
Provide a correct Python 3 solution for this coding contest problem.
If the strings are consecutive, you can replace the characters with a rule to shorten the string. For example, for the string AAAA, the expression @ 4A will compress one character. Create a program that restores the character string compressed by this rule to the original character string. However, it is assumed that the @ character does not appear in the restored character string.
In addition, the original character string is uppercase letters, lowercase letters, numbers, and symbols, and can be up to 100 characters, and consecutive characters can be up to 9 characters.
input
Multiple strings are given. One string is given per line. The number of strings does not exceed 50.
output
For each character string, output the restored character string for each character on one line.
Example
Input
ab@5C1@8050
@99+1=1@90
Output
abCCCCC10000000050
999999999+1=1000000000
### Response:
| "Correct Solution:
```
while 1:
res = ""
check = []
try:
s = input()
for i in range(len(s)):
if i in check:
continue
elif s[i] == "@":
res = res + s[i+2] * int(s[i+1])
check.append(i+1)
check.append(i+2)
else:
res = res + s[i]
print(res)
except EOFError:
break
``` |
### Instruction:
Provide a correct Python 3 solution for this coding contest problem.
If the strings are consecutive, you can replace the characters with a rule to shorten the string. For example, for the string AAAA, the expression @ 4A will compress one character. Create a program that restores the character string compressed by this rule to the original character string. However, it is assumed that the @ character does not appear in the restored character string.
In addition, the original character string is uppercase letters, lowercase letters, numbers, and symbols, and can be up to 100 characters, and consecutive characters can be up to 9 characters.
input
Multiple strings are given. One string is given per line. The number of strings does not exceed 50.
output
For each character string, output the restored character string for each character on one line.
Example
Input
ab@5C1@8050
@99+1=1@90
Output
abCCCCC10000000050
999999999+1=1000000000
### Response:
| "Correct Solution:
```
while 1:
try:
s=list(input())
ans=""
while len(s)>0:
i=s.pop(0)
if i=="@":
c=s.pop(0)
l=s.pop(0)
ans+=l*int(c)
else:ans+=i
print(ans)
except:break
``` |
### Instruction:
Provide a correct Python 3 solution for this coding contest problem.
If the strings are consecutive, you can replace the characters with a rule to shorten the string. For example, for the string AAAA, the expression @ 4A will compress one character. Create a program that restores the character string compressed by this rule to the original character string. However, it is assumed that the @ character does not appear in the restored character string.
In addition, the original character string is uppercase letters, lowercase letters, numbers, and symbols, and can be up to 100 characters, and consecutive characters can be up to 9 characters.
input
Multiple strings are given. One string is given per line. The number of strings does not exceed 50.
output
For each character string, output the restored character string for each character on one line.
Example
Input
ab@5C1@8050
@99+1=1@90
Output
abCCCCC10000000050
999999999+1=1000000000
### Response:
| "Correct Solution:
```
while 1:
try:
word = input()
except EOFError:
break
newword = []
word = list(word)
while word != []:
w = word.pop(0)
if w == "@":
count = int(word.pop(0))
s = word.pop(0)
for _ in range(count):
newword.append(s)
else:
newword.append(w)
print(''.join(w for w in newword))
``` |
### Instruction:
Provide a correct Python 3 solution for this coding contest problem.
If the strings are consecutive, you can replace the characters with a rule to shorten the string. For example, for the string AAAA, the expression @ 4A will compress one character. Create a program that restores the character string compressed by this rule to the original character string. However, it is assumed that the @ character does not appear in the restored character string.
In addition, the original character string is uppercase letters, lowercase letters, numbers, and symbols, and can be up to 100 characters, and consecutive characters can be up to 9 characters.
input
Multiple strings are given. One string is given per line. The number of strings does not exceed 50.
output
For each character string, output the restored character string for each character on one line.
Example
Input
ab@5C1@8050
@99+1=1@90
Output
abCCCCC10000000050
999999999+1=1000000000
### Response:
| "Correct Solution:
```
while True :
try :
s = list(input())
except EOFError :
break
i = 0
while i < len(s) :
if s[i] == "@" :
for j in range(int(s[i+1])) :
print(s[i+2], end="")
i += 3
else :
print(s[i], end="")
i += 1
print()
``` |
### Instruction:
Provide a correct Python 3 solution for this coding contest problem.
If the strings are consecutive, you can replace the characters with a rule to shorten the string. For example, for the string AAAA, the expression @ 4A will compress one character. Create a program that restores the character string compressed by this rule to the original character string. However, it is assumed that the @ character does not appear in the restored character string.
In addition, the original character string is uppercase letters, lowercase letters, numbers, and symbols, and can be up to 100 characters, and consecutive characters can be up to 9 characters.
input
Multiple strings are given. One string is given per line. The number of strings does not exceed 50.
output
For each character string, output the restored character string for each character on one line.
Example
Input
ab@5C1@8050
@99+1=1@90
Output
abCCCCC10000000050
999999999+1=1000000000
### Response:
| "Correct Solution:
```
while True:
try:
s = input()
except EOFError:
break
r = ''
n = 1
for c in s:
if n < 0:
n = int(c)
elif c == '@':
n = -1
else:
r += c * n
n = 1
print(r)
``` |
### Instruction:
Provide a correct Python 3 solution for this coding contest problem.
If the strings are consecutive, you can replace the characters with a rule to shorten the string. For example, for the string AAAA, the expression @ 4A will compress one character. Create a program that restores the character string compressed by this rule to the original character string. However, it is assumed that the @ character does not appear in the restored character string.
In addition, the original character string is uppercase letters, lowercase letters, numbers, and symbols, and can be up to 100 characters, and consecutive characters can be up to 9 characters.
input
Multiple strings are given. One string is given per line. The number of strings does not exceed 50.
output
For each character string, output the restored character string for each character on one line.
Example
Input
ab@5C1@8050
@99+1=1@90
Output
abCCCCC10000000050
999999999+1=1000000000
### Response:
| "Correct Solution:
```
import sys
for e in sys.stdin:
a,n='',1
for c in e[:-1]:
if'@'==c:n=0
elif n<1:n=int(c)
else:a+=c*n;n=1
print(a)
``` |
### Instruction:
Provide a correct Python 3 solution for this coding contest problem.
If the strings are consecutive, you can replace the characters with a rule to shorten the string. For example, for the string AAAA, the expression @ 4A will compress one character. Create a program that restores the character string compressed by this rule to the original character string. However, it is assumed that the @ character does not appear in the restored character string.
In addition, the original character string is uppercase letters, lowercase letters, numbers, and symbols, and can be up to 100 characters, and consecutive characters can be up to 9 characters.
input
Multiple strings are given. One string is given per line. The number of strings does not exceed 50.
output
For each character string, output the restored character string for each character on one line.
Example
Input
ab@5C1@8050
@99+1=1@90
Output
abCCCCC10000000050
999999999+1=1000000000
### Response:
| "Correct Solution:
```
while True:
try:
a = list(input())
except:
break
ans = ""
for i in range(len(a)-a.count("@")*2):
if a[i] == "@":
ans += a[i+2]*int(a[i+1])
del a[i+1:i+3]
else:
ans += a[i]
print(ans)
``` |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
If the strings are consecutive, you can replace the characters with a rule to shorten the string. For example, for the string AAAA, the expression @ 4A will compress one character. Create a program that restores the character string compressed by this rule to the original character string. However, it is assumed that the @ character does not appear in the restored character string.
In addition, the original character string is uppercase letters, lowercase letters, numbers, and symbols, and can be up to 100 characters, and consecutive characters can be up to 9 characters.
input
Multiple strings are given. One string is given per line. The number of strings does not exceed 50.
output
For each character string, output the restored character string for each character on one line.
Example
Input
ab@5C1@8050
@99+1=1@90
Output
abCCCCC10000000050
999999999+1=1000000000
Submitted Solution:
```
# Aizu Problem 0077: Run Length
#
import sys, math, os
# read input:
PYDEV = os.environ.get('PYDEV')
if PYDEV=="True":
sys.stdin = open("sample-input.txt", "rt")
def run_length(string):
out = ""
k = 0
while k < len(string):
char = string[k]
if char != '@':
out += char
k += 1
else:
cnt = int(string[k+1])
char = string[k+2]
out += char * cnt
k += 3
return out
for line in sys.stdin:
print(run_length(line.strip()))
```
### Response:
| Yes |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
If the strings are consecutive, you can replace the characters with a rule to shorten the string. For example, for the string AAAA, the expression @ 4A will compress one character. Create a program that restores the character string compressed by this rule to the original character string. However, it is assumed that the @ character does not appear in the restored character string.
In addition, the original character string is uppercase letters, lowercase letters, numbers, and symbols, and can be up to 100 characters, and consecutive characters can be up to 9 characters.
input
Multiple strings are given. One string is given per line. The number of strings does not exceed 50.
output
For each character string, output the restored character string for each character on one line.
Example
Input
ab@5C1@8050
@99+1=1@90
Output
abCCCCC10000000050
999999999+1=1000000000
Submitted Solution:
```
if __name__ == '__main__':
ans = ""
while True:
try:
tmp = ""
ans = ""
line = input()
for x in line:
if x == "@":
tmp += "@"
else:
if len(tmp) == 0:
ans += x
else:
tmp += x
if len(tmp) == 3:
ans += tmp[2]*int(tmp[1])
tmp = ""
print(ans)
except EOFError:
break
```
### Response:
| Yes |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
If the strings are consecutive, you can replace the characters with a rule to shorten the string. For example, for the string AAAA, the expression @ 4A will compress one character. Create a program that restores the character string compressed by this rule to the original character string. However, it is assumed that the @ character does not appear in the restored character string.
In addition, the original character string is uppercase letters, lowercase letters, numbers, and symbols, and can be up to 100 characters, and consecutive characters can be up to 9 characters.
input
Multiple strings are given. One string is given per line. The number of strings does not exceed 50.
output
For each character string, output the restored character string for each character on one line.
Example
Input
ab@5C1@8050
@99+1=1@90
Output
abCCCCC10000000050
999999999+1=1000000000
Submitted Solution:
```
while True:
try:
s_inp = input()
s_out = ""
flag = 0
for c in s_inp:
if flag == 1:
n = int(c)
flag = 2
elif flag == 2:
for i in range(n):
s_out = s_out + c
flag = 0
elif c == "@":
flag = 1
else:
s_out = s_out + c
print(s_out)
except:
break
```
### Response:
| Yes |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
If the strings are consecutive, you can replace the characters with a rule to shorten the string. For example, for the string AAAA, the expression @ 4A will compress one character. Create a program that restores the character string compressed by this rule to the original character string. However, it is assumed that the @ character does not appear in the restored character string.
In addition, the original character string is uppercase letters, lowercase letters, numbers, and symbols, and can be up to 100 characters, and consecutive characters can be up to 9 characters.
input
Multiple strings are given. One string is given per line. The number of strings does not exceed 50.
output
For each character string, output the restored character string for each character on one line.
Example
Input
ab@5C1@8050
@99+1=1@90
Output
abCCCCC10000000050
999999999+1=1000000000
Submitted Solution:
```
while 1:
try:s=input()
except:break
i,a,n=0,'',1
for i in s:
if n==0:n=int(i)
elif i=='@': n=0
else:
a+=i*n
n=1
print(a)
```
### Response:
| Yes |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
If the strings are consecutive, you can replace the characters with a rule to shorten the string. For example, for the string AAAA, the expression @ 4A will compress one character. Create a program that restores the character string compressed by this rule to the original character string. However, it is assumed that the @ character does not appear in the restored character string.
In addition, the original character string is uppercase letters, lowercase letters, numbers, and symbols, and can be up to 100 characters, and consecutive characters can be up to 9 characters.
input
Multiple strings are given. One string is given per line. The number of strings does not exceed 50.
output
For each character string, output the restored character string for each character on one line.
Example
Input
ab@5C1@8050
@99+1=1@90
Output
abCCCCC10000000050
999999999+1=1000000000
Submitted Solution:
```
import sys
for line in sys.stdin:
line = line.strip()
i = 0
ans = ""
while i < len(line):
if line[i] == "@":
m = int(line[i+1])
c = line[i+2]
i += 2
ans += m*c
else:
ans += line[i]
i += 1
print(ans)
```
### Response:
| No |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
If the strings are consecutive, you can replace the characters with a rule to shorten the string. For example, for the string AAAA, the expression @ 4A will compress one character. Create a program that restores the character string compressed by this rule to the original character string. However, it is assumed that the @ character does not appear in the restored character string.
In addition, the original character string is uppercase letters, lowercase letters, numbers, and symbols, and can be up to 100 characters, and consecutive characters can be up to 9 characters.
input
Multiple strings are given. One string is given per line. The number of strings does not exceed 50.
output
For each character string, output the restored character string for each character on one line.
Example
Input
ab@5C1@8050
@99+1=1@90
Output
abCCCCC10000000050
999999999+1=1000000000
Submitted Solution:
```
import sys
for line in sys.stdin:
appendStrs = []
for i in range(0, len(line) ) :
if line[i] == "@" :
appendStrs.append((line[i:i+3], "".join([ line[i+2] for _ in range(0, int(line[i+1]) )]))) ;
for a in appendStrs :
line = line.replace(*a, 1)
print(line)
```
### Response:
| No |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
If the strings are consecutive, you can replace the characters with a rule to shorten the string. For example, for the string AAAA, the expression @ 4A will compress one character. Create a program that restores the character string compressed by this rule to the original character string. However, it is assumed that the @ character does not appear in the restored character string.
In addition, the original character string is uppercase letters, lowercase letters, numbers, and symbols, and can be up to 100 characters, and consecutive characters can be up to 9 characters.
input
Multiple strings are given. One string is given per line. The number of strings does not exceed 50.
output
For each character string, output the restored character string for each character on one line.
Example
Input
ab@5C1@8050
@99+1=1@90
Output
abCCCCC10000000050
999999999+1=1000000000
Submitted Solution:
```
import re
import sys
for s in sys.stdin:
print(re.sub(r"@(\d)(.)", lambda m: m.group(2)*int(m.group(1)), s))
```
### Response:
| No |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
If the strings are consecutive, you can replace the characters with a rule to shorten the string. For example, for the string AAAA, the expression @ 4A will compress one character. Create a program that restores the character string compressed by this rule to the original character string. However, it is assumed that the @ character does not appear in the restored character string.
In addition, the original character string is uppercase letters, lowercase letters, numbers, and symbols, and can be up to 100 characters, and consecutive characters can be up to 9 characters.
input
Multiple strings are given. One string is given per line. The number of strings does not exceed 50.
output
For each character string, output the restored character string for each character on one line.
Example
Input
ab@5C1@8050
@99+1=1@90
Output
abCCCCC10000000050
999999999+1=1000000000
Submitted Solution:
```
while True:
try:
string = input()
except:
break
ans = ""
for i in range(len(string)):
if string[i] == "@":
ans += int(string[i+1])*string[i+2]
else:
ans += string[i]
print(ans)
```
### Response:
| No |
### Instruction:
Provide a correct Python 3 solution for this coding contest problem.
Example
Input
2 4
%.@\$
..\$\$
Output
Yes
### Response:
| "Correct Solution:
```
import queue
di = [0,1,0,-1]
dj = [1,0,-1,0]
h, w = map(int, input().split())
field = [input() for i in range(h)]
pdist = [[1000]*w for i in range(h)]
sdist = [[1000]*w for i in range(h)]
pque = queue.Queue()
sque = queue.Queue()
for i,row in enumerate(field):
for j in range(w):
if row[j] is '@':
pque.put((i,j))
pdist[i][j] = 0
pi = i
pj = j
if row[j] is '$':
sque.put((i,j))
sdist[i][j] = 0
if row[j] is '%':
gi = i
gj = j
while not sque.empty():
i,j = sque.get()
for k in range(4):
ni = i+di[k]
nj = j+dj[k]
if ni >=0 and ni < h and nj >= 0 and nj < w:
if (field[ni][nj] is not '#') and sdist[ni][nj] > sdist[i][j]+1:
sdist[ni][nj] = sdist[i][j]+1
sque.put((ni,nj))
while not pque.empty():
i,j = pque.get()
for k in range(4):
ni = i+di[k]
nj = j+dj[k]
if ni >=0 and ni < h and nj >= 0 and nj < w:
if (field[ni][nj] is not '#') and pdist[ni][nj] > pdist[i][j]+1 and pdist[i][j]+1 < sdist[ni][nj]:
pdist[ni][nj] = pdist[i][j]+1
pque.put((ni,nj))
if pdist[gi][gj] < 1000:
print("Yes")
else:
print("No")
``` |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During a normal walk in the forest, Katie has stumbled upon a mysterious code! However, the mysterious code had some characters unreadable. She has written down this code as a string c consisting of lowercase English characters and asterisks ("*"), where each of the asterisks denotes an unreadable character. Excited with her discovery, Katie has decided to recover the unreadable characters by replacing each asterisk with arbitrary lowercase English letter (different asterisks might be replaced with different letters).
Katie has a favorite string s and a not-so-favorite string t and she would love to recover the mysterious code so that it has as many occurrences of s as possible and as little occurrences of t as possible. Formally, let's denote f(x, y) as the number of occurrences of y in x (for example, f(aababa, ab) = 2). Katie wants to recover the code c' conforming to the original c, such that f(c', s) - f(c', t) is largest possible. However, Katie is not very good at recovering codes in general, so she would like you to help her out.
Input
The first line contains string c (1 β€ |c| β€ 1000) β the mysterious code . It is guaranteed that c consists of lowercase English characters and asterisks "*" only.
The second and third line contain strings s and t respectively (1 β€ |s|, |t| β€ 50, s β t). It is guaranteed that s and t consist of lowercase English characters only.
Output
Print a single integer β the largest possible value of f(c', s) - f(c', t) of the recovered code.
Examples
Input
*****
katie
shiro
Output
1
Input
caat
caat
a
Output
-1
Input
*a*
bba
b
Output
0
Input
***
cc
z
Output
2
Note
In the first example, for c' equal to "katie" f(c', s) = 1 and f(c', t) = 0, which makes f(c', s) - f(c', t) = 1 which is the largest possible.
In the second example, the only c' conforming to the given c is "caat". The corresponding f(c', s) - f(c', t) = 1 - 2 = -1.
In the third example, there are multiple ways to recover the code such that f(c', s) - f(c', t) is largest possible, for example "aaa", "aac", or even "zaz". The value of f(c', s) - f(c', t) = 0 for all of these recovered codes.
In the fourth example, the optimal recovered code c' would be "ccc". The corresponding f(c', s) - f(c', t) = 2.
Submitted Solution:
```
def solve(u: str, s: str, t: str) -> int:
def cnt(s: str, t: str) -> int:
count = 0
for i in range(len(s) - len(t) + 1):
if s[i:i+len(t)] == t:
count += 1
return count
def can(start: int) -> bool:
for i in range(len(s)):
if u[start + i] != s[i] and u[start + i] != '*':
return False
return True
u = list(u)
s = list(s)
t = list(t)
for i in range(len(u) - len(s) + 1):
if can(i):
orig = u[i:i+len(s)]
cnt1 = cnt(u[max(0, i - len(t) + 1):min(len(u), i + len(s) + len(t) - 1)], t)
for j in range(len(s)):
u[i + j] = s[j]
cnt2 = cnt(u[max(0, i - len(t) + 1):min(len(u), i + len(s) + len(t) - 1)], t)
if cnt1 < cnt2:
for j in range(len(s)):
u[i + j] = orig[j]
return cnt(u, s) - cnt(u, t)
def main():
c = input()
s, t = input(), input()
ret = solve(c,s,t)
print(ret)
if __name__ == "__main__":
main()
```
### Response:
| No |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During a normal walk in the forest, Katie has stumbled upon a mysterious code! However, the mysterious code had some characters unreadable. She has written down this code as a string c consisting of lowercase English characters and asterisks ("*"), where each of the asterisks denotes an unreadable character. Excited with her discovery, Katie has decided to recover the unreadable characters by replacing each asterisk with arbitrary lowercase English letter (different asterisks might be replaced with different letters).
Katie has a favorite string s and a not-so-favorite string t and she would love to recover the mysterious code so that it has as many occurrences of s as possible and as little occurrences of t as possible. Formally, let's denote f(x, y) as the number of occurrences of y in x (for example, f(aababa, ab) = 2). Katie wants to recover the code c' conforming to the original c, such that f(c', s) - f(c', t) is largest possible. However, Katie is not very good at recovering codes in general, so she would like you to help her out.
Input
The first line contains string c (1 β€ |c| β€ 1000) β the mysterious code . It is guaranteed that c consists of lowercase English characters and asterisks "*" only.
The second and third line contain strings s and t respectively (1 β€ |s|, |t| β€ 50, s β t). It is guaranteed that s and t consist of lowercase English characters only.
Output
Print a single integer β the largest possible value of f(c', s) - f(c', t) of the recovered code.
Examples
Input
*****
katie
shiro
Output
1
Input
caat
caat
a
Output
-1
Input
*a*
bba
b
Output
0
Input
***
cc
z
Output
2
Note
In the first example, for c' equal to "katie" f(c', s) = 1 and f(c', t) = 0, which makes f(c', s) - f(c', t) = 1 which is the largest possible.
In the second example, the only c' conforming to the given c is "caat". The corresponding f(c', s) - f(c', t) = 1 - 2 = -1.
In the third example, there are multiple ways to recover the code such that f(c', s) - f(c', t) is largest possible, for example "aaa", "aac", or even "zaz". The value of f(c', s) - f(c', t) = 0 for all of these recovered codes.
In the fourth example, the optimal recovered code c' would be "ccc". The corresponding f(c', s) - f(c', t) = 2.
Submitted Solution:
```
def solve(c,s,t):
if len(c)>1:
return True;
else:
return False;
def main():
c = input()
s, t = input(), input()
ret = solve(c,s,t)
if ret==True:
print("YES")
else:
print("NO")
```
### Response:
| No |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During a normal walk in the forest, Katie has stumbled upon a mysterious code! However, the mysterious code had some characters unreadable. She has written down this code as a string c consisting of lowercase English characters and asterisks ("*"), where each of the asterisks denotes an unreadable character. Excited with her discovery, Katie has decided to recover the unreadable characters by replacing each asterisk with arbitrary lowercase English letter (different asterisks might be replaced with different letters).
Katie has a favorite string s and a not-so-favorite string t and she would love to recover the mysterious code so that it has as many occurrences of s as possible and as little occurrences of t as possible. Formally, let's denote f(x, y) as the number of occurrences of y in x (for example, f(aababa, ab) = 2). Katie wants to recover the code c' conforming to the original c, such that f(c', s) - f(c', t) is largest possible. However, Katie is not very good at recovering codes in general, so she would like you to help her out.
Input
The first line contains string c (1 β€ |c| β€ 1000) β the mysterious code . It is guaranteed that c consists of lowercase English characters and asterisks "*" only.
The second and third line contain strings s and t respectively (1 β€ |s|, |t| β€ 50, s β t). It is guaranteed that s and t consist of lowercase English characters only.
Output
Print a single integer β the largest possible value of f(c', s) - f(c', t) of the recovered code.
Examples
Input
*****
katie
shiro
Output
1
Input
caat
caat
a
Output
-1
Input
*a*
bba
b
Output
0
Input
***
cc
z
Output
2
Note
In the first example, for c' equal to "katie" f(c', s) = 1 and f(c', t) = 0, which makes f(c', s) - f(c', t) = 1 which is the largest possible.
In the second example, the only c' conforming to the given c is "caat". The corresponding f(c', s) - f(c', t) = 1 - 2 = -1.
In the third example, there are multiple ways to recover the code such that f(c', s) - f(c', t) is largest possible, for example "aaa", "aac", or even "zaz". The value of f(c', s) - f(c', t) = 0 for all of these recovered codes.
In the fourth example, the optimal recovered code c' would be "ccc". The corresponding f(c', s) - f(c', t) = 2.
Submitted Solution:
```
c = input()
s = input()
t = input()
def match(a, b):
if len(a) != len(b): return False
i = 0
while i < len(a) and (a[i] == b[i] or a[i] == '*'):
i += 1
return i == len(a)
i = 0
score = 0
while i < len(c):
if match(c[i:i+len(s)], s):
c = c[:i] + s + c[i+len(s):]
score += 1
if c[i:i+len(t)] == t:
score -= 1
i += 1
print(score)
```
### Response:
| No |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During a normal walk in the forest, Katie has stumbled upon a mysterious code! However, the mysterious code had some characters unreadable. She has written down this code as a string c consisting of lowercase English characters and asterisks ("*"), where each of the asterisks denotes an unreadable character. Excited with her discovery, Katie has decided to recover the unreadable characters by replacing each asterisk with arbitrary lowercase English letter (different asterisks might be replaced with different letters).
Katie has a favorite string s and a not-so-favorite string t and she would love to recover the mysterious code so that it has as many occurrences of s as possible and as little occurrences of t as possible. Formally, let's denote f(x, y) as the number of occurrences of y in x (for example, f(aababa, ab) = 2). Katie wants to recover the code c' conforming to the original c, such that f(c', s) - f(c', t) is largest possible. However, Katie is not very good at recovering codes in general, so she would like you to help her out.
Input
The first line contains string c (1 β€ |c| β€ 1000) β the mysterious code . It is guaranteed that c consists of lowercase English characters and asterisks "*" only.
The second and third line contain strings s and t respectively (1 β€ |s|, |t| β€ 50, s β t). It is guaranteed that s and t consist of lowercase English characters only.
Output
Print a single integer β the largest possible value of f(c', s) - f(c', t) of the recovered code.
Examples
Input
*****
katie
shiro
Output
1
Input
caat
caat
a
Output
-1
Input
*a*
bba
b
Output
0
Input
***
cc
z
Output
2
Note
In the first example, for c' equal to "katie" f(c', s) = 1 and f(c', t) = 0, which makes f(c', s) - f(c', t) = 1 which is the largest possible.
In the second example, the only c' conforming to the given c is "caat". The corresponding f(c', s) - f(c', t) = 1 - 2 = -1.
In the third example, there are multiple ways to recover the code such that f(c', s) - f(c', t) is largest possible, for example "aaa", "aac", or even "zaz". The value of f(c', s) - f(c', t) = 0 for all of these recovered codes.
In the fourth example, the optimal recovered code c' would be "ccc". The corresponding f(c', s) - f(c', t) = 2.
Submitted Solution:
```
def solve(c,s,t):
if len(c)>1:
return True;
else:
return False;
def main():
c = input()
s, t = input(), input()
ret = solve(c,s,t)
if ret==True:
print(1)
else:
print(2)
if __name__ == "__main__":
main()
```
### Response:
| No |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One unknown hacker wants to get the admin's password of AtForces testing system, to get problems from the next contest. To achieve that, he sneaked into the administrator's office and stole a piece of paper with a list of n passwords β strings, consists of small Latin letters.
Hacker went home and started preparing to hack AtForces. He found that the system contains only passwords from the stolen list and that the system determines the equivalence of the passwords a and b as follows:
* two passwords a and b are equivalent if there is a letter, that exists in both a and b;
* two passwords a and b are equivalent if there is a password c from the list, which is equivalent to both a and b.
If a password is set in the system and an equivalent one is applied to access the system, then the user is accessed into the system.
For example, if the list contain passwords "a", "b", "ab", "d", then passwords "a", "b", "ab" are equivalent to each other, but the password "d" is not equivalent to any other password from list. In other words, if:
* admin's password is "b", then you can access to system by using any of this passwords: "a", "b", "ab";
* admin's password is "d", then you can access to system by using only "d".
Only one password from the list is the admin's password from the testing system. Help hacker to calculate the minimal number of passwords, required to guaranteed access to the system. Keep in mind that the hacker does not know which password is set in the system.
Input
The first line contain integer n (1 β€ n β€ 2 β
10^5) β number of passwords in the list. Next n lines contains passwords from the list β non-empty strings s_i, with length at most 50 letters. Some of the passwords may be equal.
It is guaranteed that the total length of all passwords does not exceed 10^6 letters. All of them consist only of lowercase Latin letters.
Output
In a single line print the minimal number of passwords, the use of which will allow guaranteed to access the system.
Examples
Input
4
a
b
ab
d
Output
2
Input
3
ab
bc
abc
Output
1
Input
1
codeforces
Output
1
Note
In the second example hacker need to use any of the passwords to access the system.
Submitted Solution:
```
def union(x, y):
xr = find(x)
yr = find(y)
root[yr] = xr
def find(x):
if root[x] != x:
x = find(root[x])
return x
import sys
input = sys.stdin.readline
used = set()
root = dict()
for _ in range(int(input())):
s = input().rstrip()
orrr = ord(s[0]) - 97
if orrr not in root:
root[orrr] = orrr
for j in range(1, len(s)):
orr = ord(s[j]) - 97
if orr not in root:
root[orr] = orr
union(orrr, orr)
for i in root:
used.add(find(root[i]))
print(used.__len__())
```
### Response:
| Yes |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One unknown hacker wants to get the admin's password of AtForces testing system, to get problems from the next contest. To achieve that, he sneaked into the administrator's office and stole a piece of paper with a list of n passwords β strings, consists of small Latin letters.
Hacker went home and started preparing to hack AtForces. He found that the system contains only passwords from the stolen list and that the system determines the equivalence of the passwords a and b as follows:
* two passwords a and b are equivalent if there is a letter, that exists in both a and b;
* two passwords a and b are equivalent if there is a password c from the list, which is equivalent to both a and b.
If a password is set in the system and an equivalent one is applied to access the system, then the user is accessed into the system.
For example, if the list contain passwords "a", "b", "ab", "d", then passwords "a", "b", "ab" are equivalent to each other, but the password "d" is not equivalent to any other password from list. In other words, if:
* admin's password is "b", then you can access to system by using any of this passwords: "a", "b", "ab";
* admin's password is "d", then you can access to system by using only "d".
Only one password from the list is the admin's password from the testing system. Help hacker to calculate the minimal number of passwords, required to guaranteed access to the system. Keep in mind that the hacker does not know which password is set in the system.
Input
The first line contain integer n (1 β€ n β€ 2 β
10^5) β number of passwords in the list. Next n lines contains passwords from the list β non-empty strings s_i, with length at most 50 letters. Some of the passwords may be equal.
It is guaranteed that the total length of all passwords does not exceed 10^6 letters. All of them consist only of lowercase Latin letters.
Output
In a single line print the minimal number of passwords, the use of which will allow guaranteed to access the system.
Examples
Input
4
a
b
ab
d
Output
2
Input
3
ab
bc
abc
Output
1
Input
1
codeforces
Output
1
Note
In the second example hacker need to use any of the passwords to access the system.
Submitted Solution:
```
# https://codeforces.com/problemset/problem/1263/D
from sys import stdin, exit
from typing import List, Tuple
class Password:
string: str
handle: int
def __init__(self, p: str, h: int):
self.string = p
self.handle = h
pass_list_len = int(stdin.readline().rstrip())
pass_list = [None] * pass_list_len
dsu_parent = [-1] * pass_list_len
dus_size = [0] * pass_list_len
def make_set(v: int):
dsu_parent[v] = v
dus_size[v] = 1
def find_set(v: int):
if dsu_parent[v] == v:
return v
dsu_parent[v] = find_set(dsu_parent[v])
return dsu_parent[v]
def union_sets(v1: int, v2: int):
v1 = find_set(v1)
v2 = find_set(v2)
if v1 != v2:
if dus_size[v2] < dus_size[v1]:
v1, v2 = v2, v1
dsu_parent[v1] = v2
dus_size[v2] += dus_size[v1]
includes_char = dict()
for i in range(pass_list_len):
p = Password(stdin.readline().rstrip(), i)
pass_list[i] = p
make_set(p.handle)
for c in p.string:
if c not in includes_char:
includes_char[c] = {p.handle}
else:
includes_char[c].add(p.handle)
for pass_set in includes_char.values():
last_h = None
for p in pass_set:
if last_h is not None:
union_sets(last_h, p)
last_h = p
for passs in pass_list:
for c in passs.string:
union_sets(passs.handle, next(iter(includes_char[c])))
print(len({find_set(p.handle) for p in pass_list}))
```
### Response:
| Yes |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One unknown hacker wants to get the admin's password of AtForces testing system, to get problems from the next contest. To achieve that, he sneaked into the administrator's office and stole a piece of paper with a list of n passwords β strings, consists of small Latin letters.
Hacker went home and started preparing to hack AtForces. He found that the system contains only passwords from the stolen list and that the system determines the equivalence of the passwords a and b as follows:
* two passwords a and b are equivalent if there is a letter, that exists in both a and b;
* two passwords a and b are equivalent if there is a password c from the list, which is equivalent to both a and b.
If a password is set in the system and an equivalent one is applied to access the system, then the user is accessed into the system.
For example, if the list contain passwords "a", "b", "ab", "d", then passwords "a", "b", "ab" are equivalent to each other, but the password "d" is not equivalent to any other password from list. In other words, if:
* admin's password is "b", then you can access to system by using any of this passwords: "a", "b", "ab";
* admin's password is "d", then you can access to system by using only "d".
Only one password from the list is the admin's password from the testing system. Help hacker to calculate the minimal number of passwords, required to guaranteed access to the system. Keep in mind that the hacker does not know which password is set in the system.
Input
The first line contain integer n (1 β€ n β€ 2 β
10^5) β number of passwords in the list. Next n lines contains passwords from the list β non-empty strings s_i, with length at most 50 letters. Some of the passwords may be equal.
It is guaranteed that the total length of all passwords does not exceed 10^6 letters. All of them consist only of lowercase Latin letters.
Output
In a single line print the minimal number of passwords, the use of which will allow guaranteed to access the system.
Examples
Input
4
a
b
ab
d
Output
2
Input
3
ab
bc
abc
Output
1
Input
1
codeforces
Output
1
Note
In the second example hacker need to use any of the passwords to access the system.
Submitted Solution:
```
import sys
input = sys.stdin.readline
from collections import defaultdict
class Unionfind:
def __init__(self, n):
self.par = [-1]*n
self.rank = [1]*n
def root(self, x):
p = x
while not self.par[p]<0:
p = self.par[p]
while x!=p:
tmp = x
x = self.par[x]
self.par[tmp] = p
return p
def unite(self, x, y):
rx, ry = self.root(x), self.root(y)
if rx==ry: return False
if self.rank[rx]<self.rank[ry]:
rx, ry = ry, rx
self.par[rx] += self.par[ry]
self.par[ry] = rx
if self.rank[rx]==self.rank[ry]:
self.rank[rx] += 1
def is_same(self, x, y):
return self.root(x)==self.root(y)
def count(self, x):
return -self.par[self.root(x)]
n = int(input())
uf = Unionfind(n+26)
d = defaultdict(int)
c = 0
for alpha in 'abcdefghijklmnopqrstuvwxyz':
d[alpha] = c
c += 1
for i in range(n):
s = input()[:-1]
for si in s:
j = d[si]
uf.unite(i+26, j)
rs = set(uf.root(i) for i in range(26, n+26))
print(len(rs))
```
### Response:
| Yes |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One unknown hacker wants to get the admin's password of AtForces testing system, to get problems from the next contest. To achieve that, he sneaked into the administrator's office and stole a piece of paper with a list of n passwords β strings, consists of small Latin letters.
Hacker went home and started preparing to hack AtForces. He found that the system contains only passwords from the stolen list and that the system determines the equivalence of the passwords a and b as follows:
* two passwords a and b are equivalent if there is a letter, that exists in both a and b;
* two passwords a and b are equivalent if there is a password c from the list, which is equivalent to both a and b.
If a password is set in the system and an equivalent one is applied to access the system, then the user is accessed into the system.
For example, if the list contain passwords "a", "b", "ab", "d", then passwords "a", "b", "ab" are equivalent to each other, but the password "d" is not equivalent to any other password from list. In other words, if:
* admin's password is "b", then you can access to system by using any of this passwords: "a", "b", "ab";
* admin's password is "d", then you can access to system by using only "d".
Only one password from the list is the admin's password from the testing system. Help hacker to calculate the minimal number of passwords, required to guaranteed access to the system. Keep in mind that the hacker does not know which password is set in the system.
Input
The first line contain integer n (1 β€ n β€ 2 β
10^5) β number of passwords in the list. Next n lines contains passwords from the list β non-empty strings s_i, with length at most 50 letters. Some of the passwords may be equal.
It is guaranteed that the total length of all passwords does not exceed 10^6 letters. All of them consist only of lowercase Latin letters.
Output
In a single line print the minimal number of passwords, the use of which will allow guaranteed to access the system.
Examples
Input
4
a
b
ab
d
Output
2
Input
3
ab
bc
abc
Output
1
Input
1
codeforces
Output
1
Note
In the second example hacker need to use any of the passwords to access the system.
Submitted Solution:
```
from collections import deque
n = int(input())
dic = [[] for x in range(n+1000)]
visited = [False] * (n+1000)
for i in range(n):
s = list(set(list(input())))
for e in s:
aux = n + ord(e)
dic[i].append(aux)
dic[aux].append(i)
"""for i in range(n):
visited[i] = False
for j in dic[i]:
visited[j] = False"""
def dfs(v):
visited[v] = True
for e in dic[v]:
if not visited[e]:
dfs(e)
resp = 0
for i in range(n):
if not visited[i]:
resp += 1
dfs(i)
print(resp)
```
### Response:
| Yes |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One unknown hacker wants to get the admin's password of AtForces testing system, to get problems from the next contest. To achieve that, he sneaked into the administrator's office and stole a piece of paper with a list of n passwords β strings, consists of small Latin letters.
Hacker went home and started preparing to hack AtForces. He found that the system contains only passwords from the stolen list and that the system determines the equivalence of the passwords a and b as follows:
* two passwords a and b are equivalent if there is a letter, that exists in both a and b;
* two passwords a and b are equivalent if there is a password c from the list, which is equivalent to both a and b.
If a password is set in the system and an equivalent one is applied to access the system, then the user is accessed into the system.
For example, if the list contain passwords "a", "b", "ab", "d", then passwords "a", "b", "ab" are equivalent to each other, but the password "d" is not equivalent to any other password from list. In other words, if:
* admin's password is "b", then you can access to system by using any of this passwords: "a", "b", "ab";
* admin's password is "d", then you can access to system by using only "d".
Only one password from the list is the admin's password from the testing system. Help hacker to calculate the minimal number of passwords, required to guaranteed access to the system. Keep in mind that the hacker does not know which password is set in the system.
Input
The first line contain integer n (1 β€ n β€ 2 β
10^5) β number of passwords in the list. Next n lines contains passwords from the list β non-empty strings s_i, with length at most 50 letters. Some of the passwords may be equal.
It is guaranteed that the total length of all passwords does not exceed 10^6 letters. All of them consist only of lowercase Latin letters.
Output
In a single line print the minimal number of passwords, the use of which will allow guaranteed to access the system.
Examples
Input
4
a
b
ab
d
Output
2
Input
3
ab
bc
abc
Output
1
Input
1
codeforces
Output
1
Note
In the second example hacker need to use any of the passwords to access the system.
Submitted Solution:
```
n = int(input())
l = [{}]
s = input()
for x in s:
l[0][x] = 1
for i in range(n-1):
s = input()
flag = -1
for j in range(len(l)):
for x in s:
if l[j].get(x, 0):
flag = j
if flag != -1:
for x in s:
l[flag][x] = 1
else:
d = {}
for x in s:
d[x] = 1
l.append(d)
print(len(l))
```
### Response:
| No |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One unknown hacker wants to get the admin's password of AtForces testing system, to get problems from the next contest. To achieve that, he sneaked into the administrator's office and stole a piece of paper with a list of n passwords β strings, consists of small Latin letters.
Hacker went home and started preparing to hack AtForces. He found that the system contains only passwords from the stolen list and that the system determines the equivalence of the passwords a and b as follows:
* two passwords a and b are equivalent if there is a letter, that exists in both a and b;
* two passwords a and b are equivalent if there is a password c from the list, which is equivalent to both a and b.
If a password is set in the system and an equivalent one is applied to access the system, then the user is accessed into the system.
For example, if the list contain passwords "a", "b", "ab", "d", then passwords "a", "b", "ab" are equivalent to each other, but the password "d" is not equivalent to any other password from list. In other words, if:
* admin's password is "b", then you can access to system by using any of this passwords: "a", "b", "ab";
* admin's password is "d", then you can access to system by using only "d".
Only one password from the list is the admin's password from the testing system. Help hacker to calculate the minimal number of passwords, required to guaranteed access to the system. Keep in mind that the hacker does not know which password is set in the system.
Input
The first line contain integer n (1 β€ n β€ 2 β
10^5) β number of passwords in the list. Next n lines contains passwords from the list β non-empty strings s_i, with length at most 50 letters. Some of the passwords may be equal.
It is guaranteed that the total length of all passwords does not exceed 10^6 letters. All of them consist only of lowercase Latin letters.
Output
In a single line print the minimal number of passwords, the use of which will allow guaranteed to access the system.
Examples
Input
4
a
b
ab
d
Output
2
Input
3
ab
bc
abc
Output
1
Input
1
codeforces
Output
1
Note
In the second example hacker need to use any of the passwords to access the system.
Submitted Solution:
```
n = int(input())
a = []
for i in range(n):
x = list(input())
a.append(x)
a.sort()
j = 1
for i in range(n-1):
z = 0
for k in range(len(a[i])):
if a[i][k] in a[i+1]:
z = 1
break
if z == 0:
j += 1
j1 = 0
a.reverse()
for i in range(n-1):
z = 0
for k in range(len(a[i])):
if a[i][k] in a[i+1]:
z = 1
break
if z == 0:
j1 += 1
print(min(j,j1))
```
### Response:
| No |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One unknown hacker wants to get the admin's password of AtForces testing system, to get problems from the next contest. To achieve that, he sneaked into the administrator's office and stole a piece of paper with a list of n passwords β strings, consists of small Latin letters.
Hacker went home and started preparing to hack AtForces. He found that the system contains only passwords from the stolen list and that the system determines the equivalence of the passwords a and b as follows:
* two passwords a and b are equivalent if there is a letter, that exists in both a and b;
* two passwords a and b are equivalent if there is a password c from the list, which is equivalent to both a and b.
If a password is set in the system and an equivalent one is applied to access the system, then the user is accessed into the system.
For example, if the list contain passwords "a", "b", "ab", "d", then passwords "a", "b", "ab" are equivalent to each other, but the password "d" is not equivalent to any other password from list. In other words, if:
* admin's password is "b", then you can access to system by using any of this passwords: "a", "b", "ab";
* admin's password is "d", then you can access to system by using only "d".
Only one password from the list is the admin's password from the testing system. Help hacker to calculate the minimal number of passwords, required to guaranteed access to the system. Keep in mind that the hacker does not know which password is set in the system.
Input
The first line contain integer n (1 β€ n β€ 2 β
10^5) β number of passwords in the list. Next n lines contains passwords from the list β non-empty strings s_i, with length at most 50 letters. Some of the passwords may be equal.
It is guaranteed that the total length of all passwords does not exceed 10^6 letters. All of them consist only of lowercase Latin letters.
Output
In a single line print the minimal number of passwords, the use of which will allow guaranteed to access the system.
Examples
Input
4
a
b
ab
d
Output
2
Input
3
ab
bc
abc
Output
1
Input
1
codeforces
Output
1
Note
In the second example hacker need to use any of the passwords to access the system.
Submitted Solution:
```
# def solve(r, g, b):
# sum = r + g + b
# if __name__ == '__main__':
# T = int(input())
# for case in range(1, T+1):
# r, g, b = input().split()
# r, g, b = int(r), int(g), int(b)
# solve(r, g, b)
def resolver(s):
cuenta = 1
for x in s:
s.remove(x)
for y in s:
if x in y or x in s or x == y:
cuenta = cuenta + 1
print(cuenta+1)
if __name__ == '__main__':
p = int(input())
m = [input() for x in range(0, p)]
resolver(m)
```
### Response:
| No |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One unknown hacker wants to get the admin's password of AtForces testing system, to get problems from the next contest. To achieve that, he sneaked into the administrator's office and stole a piece of paper with a list of n passwords β strings, consists of small Latin letters.
Hacker went home and started preparing to hack AtForces. He found that the system contains only passwords from the stolen list and that the system determines the equivalence of the passwords a and b as follows:
* two passwords a and b are equivalent if there is a letter, that exists in both a and b;
* two passwords a and b are equivalent if there is a password c from the list, which is equivalent to both a and b.
If a password is set in the system and an equivalent one is applied to access the system, then the user is accessed into the system.
For example, if the list contain passwords "a", "b", "ab", "d", then passwords "a", "b", "ab" are equivalent to each other, but the password "d" is not equivalent to any other password from list. In other words, if:
* admin's password is "b", then you can access to system by using any of this passwords: "a", "b", "ab";
* admin's password is "d", then you can access to system by using only "d".
Only one password from the list is the admin's password from the testing system. Help hacker to calculate the minimal number of passwords, required to guaranteed access to the system. Keep in mind that the hacker does not know which password is set in the system.
Input
The first line contain integer n (1 β€ n β€ 2 β
10^5) β number of passwords in the list. Next n lines contains passwords from the list β non-empty strings s_i, with length at most 50 letters. Some of the passwords may be equal.
It is guaranteed that the total length of all passwords does not exceed 10^6 letters. All of them consist only of lowercase Latin letters.
Output
In a single line print the minimal number of passwords, the use of which will allow guaranteed to access the system.
Examples
Input
4
a
b
ab
d
Output
2
Input
3
ab
bc
abc
Output
1
Input
1
codeforces
Output
1
Note
In the second example hacker need to use any of the passwords to access the system.
Submitted Solution:
```
n = int(input())
a = ['']*26
b = [i for i in range(26)]
k=0
for i in range(n):
s = input()
# print (''.join(s))
ind=ord(s[0])-97
a[ind]+=s
a[ind] = ''.join(set(a[ind]))
for i in range(len(a)-1):
for j in range(i+1, len(a)):
for ch in a[j]:
if ch in a[i]:
a[i]+=a[j]
a[j]=''
cnt=0
for i in a:
if i!='':
cnt+=1
print (cnt)
```
### Response:
| No |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given the strings a and b, consisting of lowercase Latin letters. You can do any number of the following operations in any order:
* if |a| > 0 (the length of the string a is greater than zero), delete the first character of the string a, that is, replace a with a_2 a_3 β¦ a_n;
* if |a| > 0, delete the last character of the string a, that is, replace a with a_1 a_2 β¦ a_{n-1};
* if |b| > 0 (the length of the string b is greater than zero), delete the first character of the string b, that is, replace b with b_2 b_3 β¦ b_n;
* if |b| > 0, delete the last character of the string b, that is, replace b with b_1 b_2 β¦ b_{n-1}.
Note that after each of the operations, the string a or b may become empty.
For example, if a="hello" and b="icpc", then you can apply the following sequence of operations:
* delete the first character of the string a β a="ello" and b="icpc";
* delete the first character of the string b β a="ello" and b="cpc";
* delete the first character of the string b β a="ello" and b="pc";
* delete the last character of the string a β a="ell" and b="pc";
* delete the last character of the string b β a="ell" and b="p".
For the given strings a and b, find the minimum number of operations for which you can make the strings a and b equal. Note that empty strings are also equal.
Input
The first line contains a single integer t (1 β€ t β€ 100). Then t test cases follow.
The first line of each test case contains the string a (1 β€ |a| β€ 20), consisting of lowercase Latin letters.
The second line of each test case contains the string b (1 β€ |b| β€ 20), consisting of lowercase Latin letters.
Output
For each test case, output the minimum number of operations that can make the strings a and b equal.
Example
Input
5
a
a
abcd
bc
hello
codeforces
hello
helo
dhjakjsnasjhfksafasd
adjsnasjhfksvdafdser
Output
0
2
13
3
20
Submitted Solution:
```
"""
"""
import sys
from sys import stdin
tt = int(stdin.readline())
ANS = []
for loop in range(tt):
a = list(stdin.readline()[:-1])
b = list(stdin.readline()[:-1])
ans = 0
for i in range(len(a)):
for j in range(len(b)):
now = 0
for k in range(min(len(a)-i,len(b)-j)):
if a[i+k] == b[j+k]:
now += 1
else:
break
ans = max(ans,now)
ANS.append(str(len(a)+len(b) - 2 * ans))
print ("\n".join(ANS))
```
### Response:
| Yes |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given the strings a and b, consisting of lowercase Latin letters. You can do any number of the following operations in any order:
* if |a| > 0 (the length of the string a is greater than zero), delete the first character of the string a, that is, replace a with a_2 a_3 β¦ a_n;
* if |a| > 0, delete the last character of the string a, that is, replace a with a_1 a_2 β¦ a_{n-1};
* if |b| > 0 (the length of the string b is greater than zero), delete the first character of the string b, that is, replace b with b_2 b_3 β¦ b_n;
* if |b| > 0, delete the last character of the string b, that is, replace b with b_1 b_2 β¦ b_{n-1}.
Note that after each of the operations, the string a or b may become empty.
For example, if a="hello" and b="icpc", then you can apply the following sequence of operations:
* delete the first character of the string a β a="ello" and b="icpc";
* delete the first character of the string b β a="ello" and b="cpc";
* delete the first character of the string b β a="ello" and b="pc";
* delete the last character of the string a β a="ell" and b="pc";
* delete the last character of the string b β a="ell" and b="p".
For the given strings a and b, find the minimum number of operations for which you can make the strings a and b equal. Note that empty strings are also equal.
Input
The first line contains a single integer t (1 β€ t β€ 100). Then t test cases follow.
The first line of each test case contains the string a (1 β€ |a| β€ 20), consisting of lowercase Latin letters.
The second line of each test case contains the string b (1 β€ |b| β€ 20), consisting of lowercase Latin letters.
Output
For each test case, output the minimum number of operations that can make the strings a and b equal.
Example
Input
5
a
a
abcd
bc
hello
codeforces
hello
helo
dhjakjsnasjhfksafasd
adjsnasjhfksvdafdser
Output
0
2
13
3
20
Submitted Solution:
```
def getLCSSubStr(X, Y, m, n):
LCSuff = [[0 for i in range(n + 1)]
for j in range(m + 1)]
length = 0
row, col = 0, 0
for i in range(m + 1):
for j in range(n + 1):
if i == 0 or j == 0:
LCSuff[i][j] = 0
elif X[i - 1] == Y[j - 1]:
LCSuff[i][j] = LCSuff[i - 1][j - 1] + 1
if length < LCSuff[i][j]:
length = LCSuff[i][j]
row = i
col = j
else:
LCSuff[i][j] = 0
if length == 0:
return ""
resultStr = ['0'] * length
while LCSuff[row][col] != 0:
length -= 1
resultStr[length] = X[row - 1] # or Y[col-1]
row -= 1
col -= 1
return (''.join(resultStr))
def find_ans(X, Y):
m = len(X)
n = len(Y)
common_string = getLCSSubStr(X, Y, m, n)
a = m - len(common_string)
b = n - len(common_string)
return a + b
test_cases = int(input())
ans = []
for _ in range(test_cases):
X = input()
Y = input()
ans.append(find_ans(X, Y))
for i in ans:
print(i)
```
### Response:
| Yes |
### Instruction: Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. You are given the strings a and b, consisting of lowercase Latin letters. You can do any number of the following operations in any order: * if |a| > 0 (the length of the string a is greater than zero), delete the first character of the string a, that is, replace a with a_2 a_3 β¦ a_n; * if |a| > 0, delete the last character of the string a, that is, replace a with a_1 a_2 β¦ a_{n-1}; * if |b| > 0 (the length of the string b is greater than zero), delete the first character of the string b, that is, replace b with b_2 b_3 β¦ b_n; * if |b| > 0, delete the last character of the string b, that is, replace b with b_1 b_2 β¦ b_{n-1}. Note that after each of the operations, the string a or b may become empty. For example, if a="hello" and b="icpc", then you can apply the following sequence of operations: * delete the first character of the string a β a="ello" and b="icpc"; * delete the first character of the string b β a="ello" and b="cpc"; * delete the first character of the string b β a="ello" and b="pc"; * delete the last character of the string a β a="ell" and b="pc"; * delete the last character of the string b β a="ell" and b="p". For the given strings a and b, find the minimum number of operations for which you can make the strings a and b equal. Note that empty strings are also equal. Input The first line contains a single integer t (1 β€ t β€ 100). Then t test cases follow. The first line of each test case contains the string a (1 β€ |a| β€ 20), consisting of lowercase Latin letters. The second line of each test case contains the string b (1 β€ |b| β€ 20), consisting of lowercase Latin letters. Output For each test case, output the minimum number of operations that can make the strings a and b equal. Example Input 5 a a abcd bc hello codeforces hello helo dhjakjsnasjhfksafasd adjsnasjhfksvdafdser Output 0 2 13 3 20 Submitted Solution: ``` from collections import defaultdict,OrderedDict,Counter from sys import stdin,stdout from bisect import bisect_left,bisect_right # import numpy as np from queue import Queue,PriorityQueue from heapq import * from statistics import * from math import * import fractions import copy from copy import deepcopy import sys import io sys.setrecursionlimit(10000) import math import os import bisect import collections mod=pow(10,9)+7 import random from random import * from time import time; def ncr(n, r, p=mod): num = den = 1 for i in range(r): num = (num * (n - i)) % p den = (den * (i + 1)) % p return (num * pow(den, p - 2, p)) % p def normalncr(n,r): r=min(r,n-r) count=1; for i in range(n-r,n+1): count*=i; for i in range(1,r+1): count//=i; return count inf=float("inf") adj=defaultdict(set) visited=defaultdict(int) def addedge(a,b): adj[a].add(b) adj[b].add(a) def bfs(v): q=Queue() q.put(v) visited[v]=1 while q.qsize()>0: s=q.get_nowait() print(s) for i in adj[s]: if visited[i]==0: q.put(i) visited[i]=1 def dfs(v,visited): if visited[v]==1: return; visited[v]=1 print(v) for i in adj[v]: dfs(i,visited) # a9=pow(10,6)+10 # prime = [True for i in range(a9 + 1)] # def SieveOfEratosthenes(n): # p = 2 # while (p * p <= n): # if (prime[p] == True): # for i in range(p * p, n + 1, p): # prime[i] = False # p += 1 # SieveOfEratosthenes(a9) # prime_number=[] # for i in range(2,a9): # if prime[i]: # prime_number.append(i) def reverse_bisect_right(a, x, lo=0, hi=None): if lo < 0: raise ValueError('lo must be non-negative') if hi is None: hi = len(a) while lo < hi: mid = (lo+hi)//2 if x > a[mid]: hi = mid else: lo = mid+1 return lo def reverse_bisect_left(a, x, lo=0, hi=None): if lo < 0: raise ValueError('lo must be non-negative') if hi is None: hi = len(a) while lo < hi: mid = (lo+hi)//2 if x >= a[mid]: hi = mid else: lo = mid+1 return lo def get_list(): return list(map(int,input().split())) def get_str_list_in_int(): return [int(i) for i in list(input())] def get_str_list(): return list(input()) def get_map(): return map(int,input().split()) def input_int(): return int(input()) def matrix(a,b): return [[0 for i in range(b)] for j in range(a)] def swap(a,b): return b,a def find_gcd(l): a=l[0] for i in range(len(l)): a=gcd(a,l[i]) return a; def is_prime(n): sqrta=int(sqrt(n)) for i in range(2,sqrta+1): if n%i==0: return 0; return 1; def prime_factors(n): while n % 2 == 0: return [2]+prime_factors(n//2) sqrta = int(sqrt(n)) for i in range(3,sqrta+1,2): if n%i==0: return [i]+prime_factors(n//i) return [n] def p(a): if type(a)==str: print(a+"\n") else: print(str(a)+"\n") def ps(a): if type(a)==str: print(a) else: print(str(a)) def kth_no_not_div_by_n(n,k): return k+(k-1)//(n-1) def forward_array(l): n=len(l) stack = [] forward=[0]*n for i in range(len(l) - 1, -1, -1): while len(stack) and l[stack[-1]] < l[i]: stack.pop() if len(stack) == 0: forward[i] = len(l); else: forward[i] = stack[-1] stack.append(i) return forward; def backward_array(l): n=len(l) stack = [] backward=[0]*n for i in range(len(l)): while len(stack) and l[stack[-1]] < l[i]: stack.pop() if len(stack) == 0: backward[i] = -1; else: backward[i] = stack[-1] stack.append(i) return backward nc="NO" yc="YES" ns="No" ys="Yes" # import math as mt # MAXN=10**7 # spf = [0 for i in range(MAXN)] # def sieve(): # spf[1] = 1 # for i in range(2, MAXN): # # marking smallest prime factor # # for every number to be itself. # spf[i] = i # # # separately marking spf for # # every even number as 2 # for i in range(4, MAXN, 2): # spf[i] = 2 # # for i in range(3, mt.ceil(mt.sqrt(MAXN))): # # # checking if i is prime # if (spf[i] == i): # # # marking SPF for all numbers # # divisible by i # for j in range(i * i, MAXN, i): # # # marking spf[j] if it is # # not previously marked # if (spf[j] == j): # spf[j] = i # def getFactorization(x): # ret = list() # while (x != 1): # ret.append(spf[x]) # x = x // spf[x] # # return ret # sieve() # input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline # input=stdin.readline # print=stdout.write # https://www.geeksforgeeks.org/longest-common-substring-dp-29/#:~:text=Explanation%3A,and%20is%20of%20length%205.&text=Output%20%3A%204-,Explanation%3A,and%20is%20of%20length%204. def LCSubStr(X, Y, m, n): LCSuff = [[0 for k in range(n + 1)] for l in range(m + 1)] result = 0 answer=0; for i in range(m + 1): for j in range(n + 1): if (i == 0 or j == 0): answer+=1 LCSuff[i][j] = 0 elif (X[i - 1] == Y[j - 1]): LCSuff[i][j] = LCSuff[i - 1][j - 1] + 1 answer += 1 result = max(result, LCSuff[i][j]) else: LCSuff[i][j] = 0 answer += 1 return result # Driver Code # Function call # This code is contributed by avanitrachhadiya2155 for i in range(int(input())): a=input() b=input() alen=len(a) blen=len(b) lcs=LCSubStr(a,b,alen,blen) print(alen+blen-2*lcs) ``` ### Response: | Yes |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given the strings a and b, consisting of lowercase Latin letters. You can do any number of the following operations in any order:
* if |a| > 0 (the length of the string a is greater than zero), delete the first character of the string a, that is, replace a with a_2 a_3 β¦ a_n;
* if |a| > 0, delete the last character of the string a, that is, replace a with a_1 a_2 β¦ a_{n-1};
* if |b| > 0 (the length of the string b is greater than zero), delete the first character of the string b, that is, replace b with b_2 b_3 β¦ b_n;
* if |b| > 0, delete the last character of the string b, that is, replace b with b_1 b_2 β¦ b_{n-1}.
Note that after each of the operations, the string a or b may become empty.
For example, if a="hello" and b="icpc", then you can apply the following sequence of operations:
* delete the first character of the string a β a="ello" and b="icpc";
* delete the first character of the string b β a="ello" and b="cpc";
* delete the first character of the string b β a="ello" and b="pc";
* delete the last character of the string a β a="ell" and b="pc";
* delete the last character of the string b β a="ell" and b="p".
For the given strings a and b, find the minimum number of operations for which you can make the strings a and b equal. Note that empty strings are also equal.
Input
The first line contains a single integer t (1 β€ t β€ 100). Then t test cases follow.
The first line of each test case contains the string a (1 β€ |a| β€ 20), consisting of lowercase Latin letters.
The second line of each test case contains the string b (1 β€ |b| β€ 20), consisting of lowercase Latin letters.
Output
For each test case, output the minimum number of operations that can make the strings a and b equal.
Example
Input
5
a
a
abcd
bc
hello
codeforces
hello
helo
dhjakjsnasjhfksafasd
adjsnasjhfksvdafdser
Output
0
2
13
3
20
Submitted Solution:
```
for _ in range(int(input())):
a = input()
b = input()
la = len(a)
lb = len(b)
res = 10**18
for i in range(la):
for j in range(i, la):
ta = a[i:j+1]
for x in range(lb):
for y in range(x, lb):
tb = b[x:y+1]
if ta == tb:
tmp = 0
tmp += abs(len(ta)-la)
tmp += abs(len(tb)-lb)
res = min(res, tmp)
if res == 10**18:
print(la+lb)
else:
print(res)
```
### Response:
| Yes |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given the strings a and b, consisting of lowercase Latin letters. You can do any number of the following operations in any order:
* if |a| > 0 (the length of the string a is greater than zero), delete the first character of the string a, that is, replace a with a_2 a_3 β¦ a_n;
* if |a| > 0, delete the last character of the string a, that is, replace a with a_1 a_2 β¦ a_{n-1};
* if |b| > 0 (the length of the string b is greater than zero), delete the first character of the string b, that is, replace b with b_2 b_3 β¦ b_n;
* if |b| > 0, delete the last character of the string b, that is, replace b with b_1 b_2 β¦ b_{n-1}.
Note that after each of the operations, the string a or b may become empty.
For example, if a="hello" and b="icpc", then you can apply the following sequence of operations:
* delete the first character of the string a β a="ello" and b="icpc";
* delete the first character of the string b β a="ello" and b="cpc";
* delete the first character of the string b β a="ello" and b="pc";
* delete the last character of the string a β a="ell" and b="pc";
* delete the last character of the string b β a="ell" and b="p".
For the given strings a and b, find the minimum number of operations for which you can make the strings a and b equal. Note that empty strings are also equal.
Input
The first line contains a single integer t (1 β€ t β€ 100). Then t test cases follow.
The first line of each test case contains the string a (1 β€ |a| β€ 20), consisting of lowercase Latin letters.
The second line of each test case contains the string b (1 β€ |b| β€ 20), consisting of lowercase Latin letters.
Output
For each test case, output the minimum number of operations that can make the strings a and b equal.
Example
Input
5
a
a
abcd
bc
hello
codeforces
hello
helo
dhjakjsnasjhfksafasd
adjsnasjhfksvdafdser
Output
0
2
13
3
20
Submitted Solution:
```
for i in range(int(input())):
ans=0
l=[]
a=input()
b=input()
for i in range(len(a)):
for k in range(i+1,len(a)):
sub=a[i:k]
if sub in b:
o=len(a)-len(sub)+len(b)-len(sub)
if ans==0:
ans=o
else:
if o<ans:
ans=o
print(ans)
```
### Response:
| No |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given the strings a and b, consisting of lowercase Latin letters. You can do any number of the following operations in any order:
* if |a| > 0 (the length of the string a is greater than zero), delete the first character of the string a, that is, replace a with a_2 a_3 β¦ a_n;
* if |a| > 0, delete the last character of the string a, that is, replace a with a_1 a_2 β¦ a_{n-1};
* if |b| > 0 (the length of the string b is greater than zero), delete the first character of the string b, that is, replace b with b_2 b_3 β¦ b_n;
* if |b| > 0, delete the last character of the string b, that is, replace b with b_1 b_2 β¦ b_{n-1}.
Note that after each of the operations, the string a or b may become empty.
For example, if a="hello" and b="icpc", then you can apply the following sequence of operations:
* delete the first character of the string a β a="ello" and b="icpc";
* delete the first character of the string b β a="ello" and b="cpc";
* delete the first character of the string b β a="ello" and b="pc";
* delete the last character of the string a β a="ell" and b="pc";
* delete the last character of the string b β a="ell" and b="p".
For the given strings a and b, find the minimum number of operations for which you can make the strings a and b equal. Note that empty strings are also equal.
Input
The first line contains a single integer t (1 β€ t β€ 100). Then t test cases follow.
The first line of each test case contains the string a (1 β€ |a| β€ 20), consisting of lowercase Latin letters.
The second line of each test case contains the string b (1 β€ |b| β€ 20), consisting of lowercase Latin letters.
Output
For each test case, output the minimum number of operations that can make the strings a and b equal.
Example
Input
5
a
a
abcd
bc
hello
codeforces
hello
helo
dhjakjsnasjhfksafasd
adjsnasjhfksvdafdser
Output
0
2
13
3
20
Submitted Solution:
```
def match(s,target):
for i in range(len(s)):
for k in range(len(target)):
if target[k] == s[i]:
m = k
m1 = i
nmatches = 0
while m<len(target) and m1 < len(s) and target[m] == s[m1]:
m+=1
m1+=1
nmatches+=1
if nmatches == len(s):
return True
return False
def solve(a,b):
mins = float("inf")
for i in range(len(a)):
for j in range(i,len(a)):
# print(i,j)
# print(a[i:j+1])
if match(a[i:j+1],b):
mins = min(mins,len(a)+len(b)-2*len(a[i:j+1]))
return mins
for _ in range(int(input())):
a = input()
b = input()
print(solve(a,b))
```
### Response:
| No |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given the strings a and b, consisting of lowercase Latin letters. You can do any number of the following operations in any order:
* if |a| > 0 (the length of the string a is greater than zero), delete the first character of the string a, that is, replace a with a_2 a_3 β¦ a_n;
* if |a| > 0, delete the last character of the string a, that is, replace a with a_1 a_2 β¦ a_{n-1};
* if |b| > 0 (the length of the string b is greater than zero), delete the first character of the string b, that is, replace b with b_2 b_3 β¦ b_n;
* if |b| > 0, delete the last character of the string b, that is, replace b with b_1 b_2 β¦ b_{n-1}.
Note that after each of the operations, the string a or b may become empty.
For example, if a="hello" and b="icpc", then you can apply the following sequence of operations:
* delete the first character of the string a β a="ello" and b="icpc";
* delete the first character of the string b β a="ello" and b="cpc";
* delete the first character of the string b β a="ello" and b="pc";
* delete the last character of the string a β a="ell" and b="pc";
* delete the last character of the string b β a="ell" and b="p".
For the given strings a and b, find the minimum number of operations for which you can make the strings a and b equal. Note that empty strings are also equal.
Input
The first line contains a single integer t (1 β€ t β€ 100). Then t test cases follow.
The first line of each test case contains the string a (1 β€ |a| β€ 20), consisting of lowercase Latin letters.
The second line of each test case contains the string b (1 β€ |b| β€ 20), consisting of lowercase Latin letters.
Output
For each test case, output the minimum number of operations that can make the strings a and b equal.
Example
Input
5
a
a
abcd
bc
hello
codeforces
hello
helo
dhjakjsnasjhfksafasd
adjsnasjhfksvdafdser
Output
0
2
13
3
20
Submitted Solution:
```
import os
import sys
import math
from io import BytesIO, IOBase
#<fast I/O>
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)
#</fast I/O>
#<I/O>
def scanf(datatype = str):
return datatype(sys.stdin.readline().rstrip())
def printf(answer = ''):
return sys.stdout.write(str(answer) + "\n")
def prints(answer):
return sys.stdout.write(str(answer) + " ")
def map_input(datatype):
return map(datatype, sys.stdin.readline().split())
def list_input(datatype):
return list(map(datatype, sys.stdin.readline().split()))
def testcase(number: int, solve_function):
for i in range(number):
printf(solve_function())
#</I/O>
def solve():
a = scanf()
b = scanf()
if(len(a)>len(b)):
a,b = b,a
maxi = -1
for i in range(len(a)):
for j in range(i+1,len(a)+1):
if(a[i:j] in b):
maxi = max(len(a[i:j]),maxi)
return (len(a) + len(b) - 2*maxi)
#<solution>
t = scanf(int)
testcase(t,solve)
#<solution>
```
### Response:
| No |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given the strings a and b, consisting of lowercase Latin letters. You can do any number of the following operations in any order:
* if |a| > 0 (the length of the string a is greater than zero), delete the first character of the string a, that is, replace a with a_2 a_3 β¦ a_n;
* if |a| > 0, delete the last character of the string a, that is, replace a with a_1 a_2 β¦ a_{n-1};
* if |b| > 0 (the length of the string b is greater than zero), delete the first character of the string b, that is, replace b with b_2 b_3 β¦ b_n;
* if |b| > 0, delete the last character of the string b, that is, replace b with b_1 b_2 β¦ b_{n-1}.
Note that after each of the operations, the string a or b may become empty.
For example, if a="hello" and b="icpc", then you can apply the following sequence of operations:
* delete the first character of the string a β a="ello" and b="icpc";
* delete the first character of the string b β a="ello" and b="cpc";
* delete the first character of the string b β a="ello" and b="pc";
* delete the last character of the string a β a="ell" and b="pc";
* delete the last character of the string b β a="ell" and b="p".
For the given strings a and b, find the minimum number of operations for which you can make the strings a and b equal. Note that empty strings are also equal.
Input
The first line contains a single integer t (1 β€ t β€ 100). Then t test cases follow.
The first line of each test case contains the string a (1 β€ |a| β€ 20), consisting of lowercase Latin letters.
The second line of each test case contains the string b (1 β€ |b| β€ 20), consisting of lowercase Latin letters.
Output
For each test case, output the minimum number of operations that can make the strings a and b equal.
Example
Input
5
a
a
abcd
bc
hello
codeforces
hello
helo
dhjakjsnasjhfksafasd
adjsnasjhfksvdafdser
Output
0
2
13
3
20
Submitted Solution:
```
for test in range(int(input())):
a=input()
b=input()
c=0
lna=len(a)
lnb=len(b)
if lna>lnb:
a,b=b,a
for i in range(lna):
for j in range(lna,i,-1):
if a[i:j] in b and (j-i)>c:
c=j-i
print((lna+lnb)-2*c)
```
### Response:
| No |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The preferred way to generate user login in Polygon is to concatenate a prefix of the user's first name and a prefix of their last name, in that order. Each prefix must be non-empty, and any of the prefixes can be the full name. Typically there are multiple possible logins for each person.
You are given the first and the last name of a user. Return the alphabetically earliest login they can get (regardless of other potential Polygon users).
As a reminder, a prefix of a string s is its substring which occurs at the beginning of s: "a", "ab", "abc" etc. are prefixes of string "{abcdef}" but "b" and 'bc" are not. A string a is alphabetically earlier than a string b, if a is a prefix of b, or a and b coincide up to some position, and then a has a letter that is alphabetically earlier than the corresponding letter in b: "a" and "ab" are alphabetically earlier than "ac" but "b" and "ba" are alphabetically later than "ac".
Input
The input consists of a single line containing two space-separated strings: the first and the last names. Each character of each string is a lowercase English letter. The length of each string is between 1 and 10, inclusive.
Output
Output a single string β alphabetically earliest possible login formed from these names. The output should be given in lowercase as well.
Examples
Input
harry potter
Output
hap
Input
tom riddle
Output
tomr
Submitted Solution:
```
s1,s2 = input().split()
s3 = s1[0]
i = 1
j = 0
while i<len(s1) and j <len(s2):
if s1[i]<s2[j]:
s3 += s1[i]
i+=1
else:
s3+=s2[j]
break
if j ==0 :
s3+=s2[j]
print(s3)
```
### Response:
| No |
### Instruction:
Provide a correct Python 3 solution for this coding contest problem.
Snuke built an online judge to hold a programming contest.
When a program is submitted to the judge, the judge returns a verdict, which is a two-character string that appears in the string S as a contiguous substring. (The judge can return any two-character substring of S.)
Determine whether the judge can return the string `AC` as the verdict to a program.
Constraints
* 2 \leq |S| \leq 5
* S consists of uppercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If the judge can return the string `AC` as a verdict to a program, print `Yes`; if it cannot, print `No`.
Examples
Input
BACD
Output
Yes
Input
ABCD
Output
No
Input
CABD
Output
No
Input
ACACA
Output
Yes
Input
XX
Output
No
### Response:
| "Correct Solution:
```
S = input()
print("YNeos"[not "AC" in S::2])
``` |
### Instruction:
Provide a correct Python 3 solution for this coding contest problem.
Snuke built an online judge to hold a programming contest.
When a program is submitted to the judge, the judge returns a verdict, which is a two-character string that appears in the string S as a contiguous substring. (The judge can return any two-character substring of S.)
Determine whether the judge can return the string `AC` as the verdict to a program.
Constraints
* 2 \leq |S| \leq 5
* S consists of uppercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If the judge can return the string `AC` as a verdict to a program, print `Yes`; if it cannot, print `No`.
Examples
Input
BACD
Output
Yes
Input
ABCD
Output
No
Input
CABD
Output
No
Input
ACACA
Output
Yes
Input
XX
Output
No
### Response:
| "Correct Solution:
```
s=input()
ans='No'
if s.count('AC')!=0:
ans='Yes'
print(ans)
``` |
### Instruction:
Provide a correct Python 3 solution for this coding contest problem.
Snuke built an online judge to hold a programming contest.
When a program is submitted to the judge, the judge returns a verdict, which is a two-character string that appears in the string S as a contiguous substring. (The judge can return any two-character substring of S.)
Determine whether the judge can return the string `AC` as the verdict to a program.
Constraints
* 2 \leq |S| \leq 5
* S consists of uppercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If the judge can return the string `AC` as a verdict to a program, print `Yes`; if it cannot, print `No`.
Examples
Input
BACD
Output
Yes
Input
ABCD
Output
No
Input
CABD
Output
No
Input
ACACA
Output
Yes
Input
XX
Output
No
### Response:
| "Correct Solution:
```
s=input()
ans="No"
for i in range(len(s)-1):
if s[i]+s[i+1]=="AC":
ans="Yes"
print(ans)
``` |
### Instruction:
Provide a correct Python 3 solution for this coding contest problem.
Snuke built an online judge to hold a programming contest.
When a program is submitted to the judge, the judge returns a verdict, which is a two-character string that appears in the string S as a contiguous substring. (The judge can return any two-character substring of S.)
Determine whether the judge can return the string `AC` as the verdict to a program.
Constraints
* 2 \leq |S| \leq 5
* S consists of uppercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If the judge can return the string `AC` as a verdict to a program, print `Yes`; if it cannot, print `No`.
Examples
Input
BACD
Output
Yes
Input
ABCD
Output
No
Input
CABD
Output
No
Input
ACACA
Output
Yes
Input
XX
Output
No
### Response:
| "Correct Solution:
```
print('Yes') if 'AC' in input() else print('No')
``` |
### Instruction:
Provide a correct Python 3 solution for this coding contest problem.
Snuke built an online judge to hold a programming contest.
When a program is submitted to the judge, the judge returns a verdict, which is a two-character string that appears in the string S as a contiguous substring. (The judge can return any two-character substring of S.)
Determine whether the judge can return the string `AC` as the verdict to a program.
Constraints
* 2 \leq |S| \leq 5
* S consists of uppercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If the judge can return the string `AC` as a verdict to a program, print `Yes`; if it cannot, print `No`.
Examples
Input
BACD
Output
Yes
Input
ABCD
Output
No
Input
CABD
Output
No
Input
ACACA
Output
Yes
Input
XX
Output
No
### Response:
| "Correct Solution:
```
s = str(input())
if "AC" in s:
print("Yes")
else:
print("No")
``` |
### Instruction:
Provide a correct Python 3 solution for this coding contest problem.
Snuke built an online judge to hold a programming contest.
When a program is submitted to the judge, the judge returns a verdict, which is a two-character string that appears in the string S as a contiguous substring. (The judge can return any two-character substring of S.)
Determine whether the judge can return the string `AC` as the verdict to a program.
Constraints
* 2 \leq |S| \leq 5
* S consists of uppercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If the judge can return the string `AC` as a verdict to a program, print `Yes`; if it cannot, print `No`.
Examples
Input
BACD
Output
Yes
Input
ABCD
Output
No
Input
CABD
Output
No
Input
ACACA
Output
Yes
Input
XX
Output
No
### Response:
| "Correct Solution:
```
a= input()
if a[0:2]=="AC" or a[1:3]=="AC" or a[2:4]=="AC" or a[3:5]=="AC":
print("Yes")
else:
print("No")
``` |
### Instruction:
Provide a correct Python 3 solution for this coding contest problem.
Snuke built an online judge to hold a programming contest.
When a program is submitted to the judge, the judge returns a verdict, which is a two-character string that appears in the string S as a contiguous substring. (The judge can return any two-character substring of S.)
Determine whether the judge can return the string `AC` as the verdict to a program.
Constraints
* 2 \leq |S| \leq 5
* S consists of uppercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If the judge can return the string `AC` as a verdict to a program, print `Yes`; if it cannot, print `No`.
Examples
Input
BACD
Output
Yes
Input
ABCD
Output
No
Input
CABD
Output
No
Input
ACACA
Output
Yes
Input
XX
Output
No
### Response:
| "Correct Solution:
```
s = input()
for i in range(len(s)-1):
if s[i:i+2] == 'AC':
print('Yes')
exit()
print('No')
``` |
### Instruction:
Provide a correct Python 3 solution for this coding contest problem.
Snuke built an online judge to hold a programming contest.
When a program is submitted to the judge, the judge returns a verdict, which is a two-character string that appears in the string S as a contiguous substring. (The judge can return any two-character substring of S.)
Determine whether the judge can return the string `AC` as the verdict to a program.
Constraints
* 2 \leq |S| \leq 5
* S consists of uppercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If the judge can return the string `AC` as a verdict to a program, print `Yes`; if it cannot, print `No`.
Examples
Input
BACD
Output
Yes
Input
ABCD
Output
No
Input
CABD
Output
No
Input
ACACA
Output
Yes
Input
XX
Output
No
### Response:
| "Correct Solution:
```
S = input()
print("Yes" if sum([S[i:i+2]=="AC" for i in range(len(S))]) else "No")
``` |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke built an online judge to hold a programming contest.
When a program is submitted to the judge, the judge returns a verdict, which is a two-character string that appears in the string S as a contiguous substring. (The judge can return any two-character substring of S.)
Determine whether the judge can return the string `AC` as the verdict to a program.
Constraints
* 2 \leq |S| \leq 5
* S consists of uppercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If the judge can return the string `AC` as a verdict to a program, print `Yes`; if it cannot, print `No`.
Examples
Input
BACD
Output
Yes
Input
ABCD
Output
No
Input
CABD
Output
No
Input
ACACA
Output
Yes
Input
XX
Output
No
Submitted Solution:
```
s = input()
ans = 'Yes' if any(s[i:i + 2] == 'AC' for i in range(len(s) - 1)) else 'No'
print(ans)
```
### Response:
| Yes |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke built an online judge to hold a programming contest.
When a program is submitted to the judge, the judge returns a verdict, which is a two-character string that appears in the string S as a contiguous substring. (The judge can return any two-character substring of S.)
Determine whether the judge can return the string `AC` as the verdict to a program.
Constraints
* 2 \leq |S| \leq 5
* S consists of uppercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If the judge can return the string `AC` as a verdict to a program, print `Yes`; if it cannot, print `No`.
Examples
Input
BACD
Output
Yes
Input
ABCD
Output
No
Input
CABD
Output
No
Input
ACACA
Output
Yes
Input
XX
Output
No
Submitted Solution:
```
S=str(input())
for i in range(len(S)-1):
if S[i]=="A" and S[i+1]=="C":
print("Yes")
break
else: print("No")
```
### Response:
| Yes |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke built an online judge to hold a programming contest.
When a program is submitted to the judge, the judge returns a verdict, which is a two-character string that appears in the string S as a contiguous substring. (The judge can return any two-character substring of S.)
Determine whether the judge can return the string `AC` as the verdict to a program.
Constraints
* 2 \leq |S| \leq 5
* S consists of uppercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If the judge can return the string `AC` as a verdict to a program, print `Yes`; if it cannot, print `No`.
Examples
Input
BACD
Output
Yes
Input
ABCD
Output
No
Input
CABD
Output
No
Input
ACACA
Output
Yes
Input
XX
Output
No
Submitted Solution:
```
S = input()
for k in range(1,len(S)):
if S[k-1:k+1] == "AC":
print("Yes")
exit(0)
print("No")
```
### Response:
| Yes |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke built an online judge to hold a programming contest.
When a program is submitted to the judge, the judge returns a verdict, which is a two-character string that appears in the string S as a contiguous substring. (The judge can return any two-character substring of S.)
Determine whether the judge can return the string `AC` as the verdict to a program.
Constraints
* 2 \leq |S| \leq 5
* S consists of uppercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If the judge can return the string `AC` as a verdict to a program, print `Yes`; if it cannot, print `No`.
Examples
Input
BACD
Output
Yes
Input
ABCD
Output
No
Input
CABD
Output
No
Input
ACACA
Output
Yes
Input
XX
Output
No
Submitted Solution:
```
if (input().count("AC")):
print("Yes")
else:
print("No")
```
### Response:
| Yes |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke built an online judge to hold a programming contest.
When a program is submitted to the judge, the judge returns a verdict, which is a two-character string that appears in the string S as a contiguous substring. (The judge can return any two-character substring of S.)
Determine whether the judge can return the string `AC` as the verdict to a program.
Constraints
* 2 \leq |S| \leq 5
* S consists of uppercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If the judge can return the string `AC` as a verdict to a program, print `Yes`; if it cannot, print `No`.
Examples
Input
BACD
Output
Yes
Input
ABCD
Output
No
Input
CABD
Output
No
Input
ACACA
Output
Yes
Input
XX
Output
No
Submitted Solution:
```
import bisect
import os
from collections import Counter, deque
from fractions import gcd
from functools import lru_cache
from functools import reduce
import functools
import heapq
import itertools
import math
import numpy as np
import re
import sys
if os.getenv("LOCAL"):
sys.stdin = open("_in.txt", "r")
sys.setrecursionlimit(2147483647)
INF = float("inf")
S = sys.stdin.readline().rstrip()
if re.match('.*A.*C.*',S):
print('Yes')
else:
print('No')
```
### Response:
| No |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke built an online judge to hold a programming contest.
When a program is submitted to the judge, the judge returns a verdict, which is a two-character string that appears in the string S as a contiguous substring. (The judge can return any two-character substring of S.)
Determine whether the judge can return the string `AC` as the verdict to a program.
Constraints
* 2 \leq |S| \leq 5
* S consists of uppercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If the judge can return the string `AC` as a verdict to a program, print `Yes`; if it cannot, print `No`.
Examples
Input
BACD
Output
Yes
Input
ABCD
Output
No
Input
CABD
Output
No
Input
ACACA
Output
Yes
Input
XX
Output
No
Submitted Solution:
```
data = input()
tmp = data.replace("x", "")
c1 , c2, num = 0, -1, 0
if tmp == tmp[::-1]:
while(c1 < len(data) + c2):
if data[c1] == data[c2]:
c1+=1
c2-=1
elif data[c2]=="x":
c2-=1
num+=1
else:
c1+=1
num+=1
else:
num = -1
print(num)
```
### Response:
| No |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke built an online judge to hold a programming contest.
When a program is submitted to the judge, the judge returns a verdict, which is a two-character string that appears in the string S as a contiguous substring. (The judge can return any two-character substring of S.)
Determine whether the judge can return the string `AC` as the verdict to a program.
Constraints
* 2 \leq |S| \leq 5
* S consists of uppercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If the judge can return the string `AC` as a verdict to a program, print `Yes`; if it cannot, print `No`.
Examples
Input
BACD
Output
Yes
Input
ABCD
Output
No
Input
CABD
Output
No
Input
ACACA
Output
Yes
Input
XX
Output
No
Submitted Solution:
```
s = input()
i = 0
j = len(s)-1
ist = 0
while i < j:
if s[i] == s[j]:
i += 1
j -= 1
elif s[i] == 'x':
ist += 1
i += 1
elif s[j] == 'x':
ist += 1
j -= 1
else:
print(-1)
exit()
print(ist)
```
### Response:
| No |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke built an online judge to hold a programming contest.
When a program is submitted to the judge, the judge returns a verdict, which is a two-character string that appears in the string S as a contiguous substring. (The judge can return any two-character substring of S.)
Determine whether the judge can return the string `AC` as the verdict to a program.
Constraints
* 2 \leq |S| \leq 5
* S consists of uppercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If the judge can return the string `AC` as a verdict to a program, print `Yes`; if it cannot, print `No`.
Examples
Input
BACD
Output
Yes
Input
ABCD
Output
No
Input
CABD
Output
No
Input
ACACA
Output
Yes
Input
XX
Output
No
Submitted Solution:
```
s = int(input())
l = []
for i in range(len(s)-1):
l.append(s[i:i+2])
print("Yes" if "AC" in l else "No")
```
### Response:
| No |
### Instruction:
Provide a correct Python 3 solution for this coding contest problem.
Iroha loves Haiku. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order.
To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
Constraints
* 1β¦A,B,Cβ¦10
Input
The input is given from Standard Input in the following format:
A B C
Output
If it is possible to construct a Haiku by using each of the phrases once, print `YES` (case-sensitive). Otherwise, print `NO`.
Examples
Input
5 5 7
Output
YES
Input
7 7 5
Output
NO
### Response:
| "Correct Solution:
```
print(['NO','YES'][sorted(input().split())=='5 5 7'.split()])
``` |
### Instruction:
Provide a correct Python 3 solution for this coding contest problem.
Iroha loves Haiku. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order.
To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
Constraints
* 1β¦A,B,Cβ¦10
Input
The input is given from Standard Input in the following format:
A B C
Output
If it is possible to construct a Haiku by using each of the phrases once, print `YES` (case-sensitive). Otherwise, print `NO`.
Examples
Input
5 5 7
Output
YES
Input
7 7 5
Output
NO
### Response:
| "Correct Solution:
```
str = input().split()
if str.count("7") == 1 :
print("YES")
else:
print("NO")
``` |
### Instruction:
Provide a correct Python 3 solution for this coding contest problem.
Iroha loves Haiku. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order.
To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
Constraints
* 1β¦A,B,Cβ¦10
Input
The input is given from Standard Input in the following format:
A B C
Output
If it is possible to construct a Haiku by using each of the phrases once, print `YES` (case-sensitive). Otherwise, print `NO`.
Examples
Input
5 5 7
Output
YES
Input
7 7 5
Output
NO
### Response:
| "Correct Solution:
```
print('YES' if sum(map(int,input().split()))==17 else 'NO' )
``` |
### Instruction:
Provide a correct Python 3 solution for this coding contest problem.
Iroha loves Haiku. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order.
To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
Constraints
* 1β¦A,B,Cβ¦10
Input
The input is given from Standard Input in the following format:
A B C
Output
If it is possible to construct a Haiku by using each of the phrases once, print `YES` (case-sensitive). Otherwise, print `NO`.
Examples
Input
5 5 7
Output
YES
Input
7 7 5
Output
NO
### Response:
| "Correct Solution:
```
s = sorted([int(x) for x in input().split()])
print("YES" if s==[5,5,7] else "NO")
``` |
### Instruction:
Provide a correct Python 3 solution for this coding contest problem.
Iroha loves Haiku. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order.
To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
Constraints
* 1β¦A,B,Cβ¦10
Input
The input is given from Standard Input in the following format:
A B C
Output
If it is possible to construct a Haiku by using each of the phrases once, print `YES` (case-sensitive). Otherwise, print `NO`.
Examples
Input
5 5 7
Output
YES
Input
7 7 5
Output
NO
### Response:
| "Correct Solution:
```
ar=sorted(list(map(int,input().split())))
print("YES" if ar==[5,5,7] else "NO")
``` |
### Instruction:
Provide a correct Python 3 solution for this coding contest problem.
Iroha loves Haiku. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order.
To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
Constraints
* 1β¦A,B,Cβ¦10
Input
The input is given from Standard Input in the following format:
A B C
Output
If it is possible to construct a Haiku by using each of the phrases once, print `YES` (case-sensitive). Otherwise, print `NO`.
Examples
Input
5 5 7
Output
YES
Input
7 7 5
Output
NO
### Response:
| "Correct Solution:
```
print(' YNEOS'[input().count('7')::2])
``` |
### Instruction:
Provide a correct Python 3 solution for this coding contest problem.
Iroha loves Haiku. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order.
To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
Constraints
* 1β¦A,B,Cβ¦10
Input
The input is given from Standard Input in the following format:
A B C
Output
If it is possible to construct a Haiku by using each of the phrases once, print `YES` (case-sensitive). Otherwise, print `NO`.
Examples
Input
5 5 7
Output
YES
Input
7 7 5
Output
NO
### Response:
| "Correct Solution:
```
a=list(map(int, input().split()))
print('YES' if a.count(7)==1 else 'NO')
``` |
### Instruction:
Provide a correct Python 3 solution for this coding contest problem.
Iroha loves Haiku. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order.
To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
Constraints
* 1β¦A,B,Cβ¦10
Input
The input is given from Standard Input in the following format:
A B C
Output
If it is possible to construct a Haiku by using each of the phrases once, print `YES` (case-sensitive). Otherwise, print `NO`.
Examples
Input
5 5 7
Output
YES
Input
7 7 5
Output
NO
### Response:
| "Correct Solution:
```
a=list(map(int,input().split()))
a.sort()
print("YNEOS"[a!=[5,5,7]::2])
``` |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iroha loves Haiku. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order.
To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
Constraints
* 1β¦A,B,Cβ¦10
Input
The input is given from Standard Input in the following format:
A B C
Output
If it is possible to construct a Haiku by using each of the phrases once, print `YES` (case-sensitive). Otherwise, print `NO`.
Examples
Input
5 5 7
Output
YES
Input
7 7 5
Output
NO
Submitted Solution:
```
print(['NO','YES'][['5','5','7']==sorted(input()[::2])])
```
### Response:
| Yes |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iroha loves Haiku. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order.
To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
Constraints
* 1β¦A,B,Cβ¦10
Input
The input is given from Standard Input in the following format:
A B C
Output
If it is possible to construct a Haiku by using each of the phrases once, print `YES` (case-sensitive). Otherwise, print `NO`.
Examples
Input
5 5 7
Output
YES
Input
7 7 5
Output
NO
Submitted Solution:
```
A,B,C=map(int,input().split())
print('YES'if A+B+C==17 and A*B*C==175 else "NO")
```
### Response:
| Yes |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iroha loves Haiku. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order.
To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
Constraints
* 1β¦A,B,Cβ¦10
Input
The input is given from Standard Input in the following format:
A B C
Output
If it is possible to construct a Haiku by using each of the phrases once, print `YES` (case-sensitive). Otherwise, print `NO`.
Examples
Input
5 5 7
Output
YES
Input
7 7 5
Output
NO
Submitted Solution:
```
n=[int(_) for _ in input().split()]
print('YES' if sorted(n)==[5,5,7] else 'NO')
```
### Response:
| Yes |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iroha loves Haiku. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order.
To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
Constraints
* 1β¦A,B,Cβ¦10
Input
The input is given from Standard Input in the following format:
A B C
Output
If it is possible to construct a Haiku by using each of the phrases once, print `YES` (case-sensitive). Otherwise, print `NO`.
Examples
Input
5 5 7
Output
YES
Input
7 7 5
Output
NO
Submitted Solution:
```
l=input().split()
print('YES' if l.count('5')==2 and l.count('7')==1 else 'NO')
```
### Response:
| Yes |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iroha loves Haiku. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order.
To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
Constraints
* 1β¦A,B,Cβ¦10
Input
The input is given from Standard Input in the following format:
A B C
Output
If it is possible to construct a Haiku by using each of the phrases once, print `YES` (case-sensitive). Otherwise, print `NO`.
Examples
Input
5 5 7
Output
YES
Input
7 7 5
Output
NO
Submitted Solution:
```
def Haiku(check,syllables):
check.sort()
syllables.sort()
if check==syllables:
return "YES"
else:
return "NO"
print(Haiku(check,syllables))
```
### Response:
| No |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iroha loves Haiku. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order.
To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
Constraints
* 1β¦A,B,Cβ¦10
Input
The input is given from Standard Input in the following format:
A B C
Output
If it is possible to construct a Haiku by using each of the phrases once, print `YES` (case-sensitive). Otherwise, print `NO`.
Examples
Input
5 5 7
Output
YES
Input
7 7 5
Output
NO
Submitted Solution:
```
A=int(input('A'))
B=int(input('B'))
C=int(input('C'))
if (A==5 and B==5 and C==7) or (A==5 and B==7 and C==5) or (A==7 and B==5 and C==5):
print('yes')
else:
print('no')
```
### Response:
| No |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iroha loves Haiku. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order.
To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
Constraints
* 1β¦A,B,Cβ¦10
Input
The input is given from Standard Input in the following format:
A B C
Output
If it is possible to construct a Haiku by using each of the phrases once, print `YES` (case-sensitive). Otherwise, print `NO`.
Examples
Input
5 5 7
Output
YES
Input
7 7 5
Output
NO
Submitted Solution:
```
N = list(map(int, input().split()))
cnt_5 = 0
cnt_7 = 0
for n in N:
if n == 5 :
cnt_5 += 1
elif n == 7 :
cnt_7 += 1
if cnt_5 == 2 and cnt_7 == 1:
print('YES')
else:
print('N0')
```
### Response:
| No |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iroha loves Haiku. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order.
To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
Constraints
* 1β¦A,B,Cβ¦10
Input
The input is given from Standard Input in the following format:
A B C
Output
If it is possible to construct a Haiku by using each of the phrases once, print `YES` (case-sensitive). Otherwise, print `NO`.
Examples
Input
5 5 7
Output
YES
Input
7 7 5
Output
NO
Submitted Solution:
```
a = list(map(int, input().split())
if a.count(5) == 2 and a.count(7) == 1:
print("Yes")
else:
print("No")
```
### Response:
| No |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a string S of length N consisting of `R`, `G`, and `B`.
Find the number of triples (i,~j,~k)~(1 \leq i < j < k \leq N) that satisfy both of the following conditions:
* S_i \neq S_j, S_i \neq S_k, and S_j \neq S_k.
* j - i \neq k - j.
Constraints
* 1 \leq N \leq 4000
* S is a string of length N consisting of `R`, `G`, and `B`.
Input
Input is given from Standard Input in the following format:
N
S
Output
Print the number of triplets in question.
Examples
Input
4
RRGB
Output
1
Input
39
RBRBGRBGGBBRRGBBRRRBGGBRBGBRBGBRBBBGBBB
Output
1800
Submitted Solution:
```
# input
N = int(input())
S = input()
# process
R = []
G = []
B = []
for i, s in enumerate(list(S)):
if s == 'R':
R.append(i)
elif s == 'G':
G.append(i)
else:
B.append(i)
ans = 0
for r in R:
for g in G:
for b in B:
if r < g:
if g < b:
if b-g != g-r:
ans += 1
else:
if r < b:
if g-b != b-r:
ans += 1
else:
if g-r != r-b:
ans += 1
else:
if r < b:
if b-r != r-g:
ans += 1
else:
if g < b:
if r-b != b-g:
ans += 1
else:
if r-g != g-b:
ans += 1
# output
print(ans)
```
### Response:
| No |
### Instruction: Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Note that the only difference between String Transformation 1 and String Transformation 2 is in the move Koa does. In this version the letter y Koa selects must be strictly greater alphabetically than x (read statement for better understanding). You can make hacks in these problems independently. Koa the Koala has two strings A and B of the same length n (|A|=|B|=n) consisting of the first 20 lowercase English alphabet letters (ie. from a to t). In one move Koa: 1. selects some subset of positions p_1, p_2, β¦, p_k (k β₯ 1; 1 β€ p_i β€ n; p_i β p_j if i β j) of A such that A_{p_1} = A_{p_2} = β¦ = A_{p_k} = x (ie. all letters on this positions are equal to some letter x). 2. selects a letter y (from the first 20 lowercase letters in English alphabet) such that y>x (ie. letter y is strictly greater alphabetically than x). 3. sets each letter in positions p_1, p_2, β¦, p_k to letter y. More formally: for each i (1 β€ i β€ k) Koa sets A_{p_i} = y. Note that you can only modify letters in string A. Koa wants to know the smallest number of moves she has to do to make strings equal to each other (A = B) or to determine that there is no way to make them equal. Help her! Input Each test contains multiple test cases. The first line contains t (1 β€ t β€ 10) β the number of test cases. Description of the test cases follows. The first line of each test case contains one integer n (1 β€ n β€ 10^5) β the length of strings A and B. The second line of each test case contains string A (|A|=n). The third line of each test case contains string B (|B|=n). Both strings consists of the first 20 lowercase English alphabet letters (ie. from a to t). It is guaranteed that the sum of n over all test cases does not exceed 10^5. Output For each test case: Print on a single line the smallest number of moves she has to do to make strings equal to each other (A = B) or -1 if there is no way to make them equal. Example Input 5 3 aab bcc 4 cabc abcb 3 abc tsr 4 aabd cccd 5 abcbd bcdda Output 2 -1 3 2 -1 Note * In the 1-st test case Koa: 1. selects positions 1 and 2 and sets A_1 = A_2 = b (\color{red}{aa}b β \color{blue}{bb}b). 2. selects positions 2 and 3 and sets A_2 = A_3 = c (b\color{red}{bb} β b\color{blue}{cc}). * In the 2-nd test case Koa has no way to make string A equal B. * In the 3-rd test case Koa: 1. selects position 1 and sets A_1 = t (\color{red}{a}bc β \color{blue}{t}bc). 2. selects position 2 and sets A_2 = s (t\color{red}{b}c β t\color{blue}{s}c). 3. selects position 3 and sets A_3 = r (ts\color{red}{c} β ts\color{blue}{r}). Submitted Solution: ``` # cook your dish here # code # ___________________________________ # | | # | | # | _, _ _ ,_ | # | .-'` / \'-'/ \ `'-. | # | / | | | | \ | # | ; \_ _/ \_ _/ ; | # | | `` `` | | # | | | | # | ; .-. .-. .-. .-. ; | # | \ ( '.' \ / '.' ) / | # | '-.; V ;.-' | # | ` ` | # | | # |___________________________________| # | | # | Author : Ramzz | # | Created On : 21-07-2020 | # |___________________________________| # # _ __ __ _ _ __ ___ ________ # | '__/ _` | '_ ` _ \|_ /_ / # | | | (_| | | | | | |/ / / / # |_| \__,_|_| |_| |_/___/___| # import math import collections from sys import stdin,stdout,setrecursionlimit from bisect import bisect_left as bsl from bisect import bisect_right as bsr import heapq as hq setrecursionlimit(2**20) t = 1 t = int(stdin.readline()) for _ in range(t): n = int(stdin.readline()) a = stdin.readline().strip('\n') b = stdin.readline().strip('\n') #a = list(map(int, stdin.readline().rstrip().split())) chk = False d = {} l = [] var = 'abcdefghijklmnopqrst' for i in range(n): if(a[i]>b[i]): chk = True break if(a[i]==b[i]): continue if(a[i] not in d): d[a[i]] = {} d[a[i]][b[i]] = True l.append((a[i],b[i])) if(chk): print(-1) else: ans = 0 l1 = list(d.keys()) l1.sort() for i in var: if(i in d): if(len(d[i])>0): ans += 1 mi = 'u' for j in d[i]: if(mi>j): mi=j if(mi not in d): d[mi] = {} for k in d[i]: if(k!=mi): d[mi][k] = True print(ans) ``` ### Response: | Yes |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Note that the only difference between String Transformation 1 and String Transformation 2 is in the move Koa does. In this version the letter y Koa selects must be strictly greater alphabetically than x (read statement for better understanding). You can make hacks in these problems independently.
Koa the Koala has two strings A and B of the same length n (|A|=|B|=n) consisting of the first 20 lowercase English alphabet letters (ie. from a to t).
In one move Koa:
1. selects some subset of positions p_1, p_2, β¦, p_k (k β₯ 1; 1 β€ p_i β€ n; p_i β p_j if i β j) of A such that A_{p_1} = A_{p_2} = β¦ = A_{p_k} = x (ie. all letters on this positions are equal to some letter x).
2. selects a letter y (from the first 20 lowercase letters in English alphabet) such that y>x (ie. letter y is strictly greater alphabetically than x).
3. sets each letter in positions p_1, p_2, β¦, p_k to letter y. More formally: for each i (1 β€ i β€ k) Koa sets A_{p_i} = y.
Note that you can only modify letters in string A.
Koa wants to know the smallest number of moves she has to do to make strings equal to each other (A = B) or to determine that there is no way to make them equal. Help her!
Input
Each test contains multiple test cases. The first line contains t (1 β€ t β€ 10) β the number of test cases. Description of the test cases follows.
The first line of each test case contains one integer n (1 β€ n β€ 10^5) β the length of strings A and B.
The second line of each test case contains string A (|A|=n).
The third line of each test case contains string B (|B|=n).
Both strings consists of the first 20 lowercase English alphabet letters (ie. from a to t).
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case:
Print on a single line the smallest number of moves she has to do to make strings equal to each other (A = B) or -1 if there is no way to make them equal.
Example
Input
5
3
aab
bcc
4
cabc
abcb
3
abc
tsr
4
aabd
cccd
5
abcbd
bcdda
Output
2
-1
3
2
-1
Note
* In the 1-st test case Koa:
1. selects positions 1 and 2 and sets A_1 = A_2 = b (\color{red}{aa}b β \color{blue}{bb}b).
2. selects positions 2 and 3 and sets A_2 = A_3 = c (b\color{red}{bb} β b\color{blue}{cc}).
* In the 2-nd test case Koa has no way to make string A equal B.
* In the 3-rd test case Koa:
1. selects position 1 and sets A_1 = t (\color{red}{a}bc β \color{blue}{t}bc).
2. selects position 2 and sets A_2 = s (t\color{red}{b}c β t\color{blue}{s}c).
3. selects position 3 and sets A_3 = r (ts\color{red}{c} β ts\color{blue}{r}).
Submitted Solution:
```
from collections import defaultdict
t=int(input())
for tc in range(t):
n=int(input())
a=str(input())
arr=list(a)
b=str(input())
charray=["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t"]
flag=0
for i in range(n):
if a[i]>b[i]:
print("-1")
flag=1
break
if flag==0:
ans=0
for i in charray:
f=defaultdict(int)
for j in range(n):
if i==b[j] and i!=a[j]:
f[arr[j]]+=1
ans+=len(f)
for j in range(n):
if f[arr[j]]>0:
arr[j]=i
print(ans)
```
### Response:
| Yes |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Note that the only difference between String Transformation 1 and String Transformation 2 is in the move Koa does. In this version the letter y Koa selects must be strictly greater alphabetically than x (read statement for better understanding). You can make hacks in these problems independently.
Koa the Koala has two strings A and B of the same length n (|A|=|B|=n) consisting of the first 20 lowercase English alphabet letters (ie. from a to t).
In one move Koa:
1. selects some subset of positions p_1, p_2, β¦, p_k (k β₯ 1; 1 β€ p_i β€ n; p_i β p_j if i β j) of A such that A_{p_1} = A_{p_2} = β¦ = A_{p_k} = x (ie. all letters on this positions are equal to some letter x).
2. selects a letter y (from the first 20 lowercase letters in English alphabet) such that y>x (ie. letter y is strictly greater alphabetically than x).
3. sets each letter in positions p_1, p_2, β¦, p_k to letter y. More formally: for each i (1 β€ i β€ k) Koa sets A_{p_i} = y.
Note that you can only modify letters in string A.
Koa wants to know the smallest number of moves she has to do to make strings equal to each other (A = B) or to determine that there is no way to make them equal. Help her!
Input
Each test contains multiple test cases. The first line contains t (1 β€ t β€ 10) β the number of test cases. Description of the test cases follows.
The first line of each test case contains one integer n (1 β€ n β€ 10^5) β the length of strings A and B.
The second line of each test case contains string A (|A|=n).
The third line of each test case contains string B (|B|=n).
Both strings consists of the first 20 lowercase English alphabet letters (ie. from a to t).
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case:
Print on a single line the smallest number of moves she has to do to make strings equal to each other (A = B) or -1 if there is no way to make them equal.
Example
Input
5
3
aab
bcc
4
cabc
abcb
3
abc
tsr
4
aabd
cccd
5
abcbd
bcdda
Output
2
-1
3
2
-1
Note
* In the 1-st test case Koa:
1. selects positions 1 and 2 and sets A_1 = A_2 = b (\color{red}{aa}b β \color{blue}{bb}b).
2. selects positions 2 and 3 and sets A_2 = A_3 = c (b\color{red}{bb} β b\color{blue}{cc}).
* In the 2-nd test case Koa has no way to make string A equal B.
* In the 3-rd test case Koa:
1. selects position 1 and sets A_1 = t (\color{red}{a}bc β \color{blue}{t}bc).
2. selects position 2 and sets A_2 = s (t\color{red}{b}c β t\color{blue}{s}c).
3. selects position 3 and sets A_3 = r (ts\color{red}{c} β ts\color{blue}{r}).
Submitted Solution:
```
def _find(st, u):
ls = []
while st[u] != u:
ls.append(u)
u = st[u]
for v in ls:
st[v] = u
return u
def _union(st, u, v):
su, sv = _find(st, u), _find(st, v)
if su != sv: st[su] = sv
return su != sv
T = int(input())
for _ in range(T):
n = int(input())
a, b = input(), input()
st = { chr(u) : chr(u) for u in range(ord('a'), ord('t')+1) }
ok, cc = 1, 0
for ca, cb in zip(a, b):
if ca > cb:
ok = 0; break
elif cb > ca:
cc += _union(st, ca, cb)
print(cc if ok else -1)
```
### Response:
| Yes |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Note that the only difference between String Transformation 1 and String Transformation 2 is in the move Koa does. In this version the letter y Koa selects must be strictly greater alphabetically than x (read statement for better understanding). You can make hacks in these problems independently.
Koa the Koala has two strings A and B of the same length n (|A|=|B|=n) consisting of the first 20 lowercase English alphabet letters (ie. from a to t).
In one move Koa:
1. selects some subset of positions p_1, p_2, β¦, p_k (k β₯ 1; 1 β€ p_i β€ n; p_i β p_j if i β j) of A such that A_{p_1} = A_{p_2} = β¦ = A_{p_k} = x (ie. all letters on this positions are equal to some letter x).
2. selects a letter y (from the first 20 lowercase letters in English alphabet) such that y>x (ie. letter y is strictly greater alphabetically than x).
3. sets each letter in positions p_1, p_2, β¦, p_k to letter y. More formally: for each i (1 β€ i β€ k) Koa sets A_{p_i} = y.
Note that you can only modify letters in string A.
Koa wants to know the smallest number of moves she has to do to make strings equal to each other (A = B) or to determine that there is no way to make them equal. Help her!
Input
Each test contains multiple test cases. The first line contains t (1 β€ t β€ 10) β the number of test cases. Description of the test cases follows.
The first line of each test case contains one integer n (1 β€ n β€ 10^5) β the length of strings A and B.
The second line of each test case contains string A (|A|=n).
The third line of each test case contains string B (|B|=n).
Both strings consists of the first 20 lowercase English alphabet letters (ie. from a to t).
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case:
Print on a single line the smallest number of moves she has to do to make strings equal to each other (A = B) or -1 if there is no way to make them equal.
Example
Input
5
3
aab
bcc
4
cabc
abcb
3
abc
tsr
4
aabd
cccd
5
abcbd
bcdda
Output
2
-1
3
2
-1
Note
* In the 1-st test case Koa:
1. selects positions 1 and 2 and sets A_1 = A_2 = b (\color{red}{aa}b β \color{blue}{bb}b).
2. selects positions 2 and 3 and sets A_2 = A_3 = c (b\color{red}{bb} β b\color{blue}{cc}).
* In the 2-nd test case Koa has no way to make string A equal B.
* In the 3-rd test case Koa:
1. selects position 1 and sets A_1 = t (\color{red}{a}bc β \color{blue}{t}bc).
2. selects position 2 and sets A_2 = s (t\color{red}{b}c β t\color{blue}{s}c).
3. selects position 3 and sets A_3 = r (ts\color{red}{c} β ts\color{blue}{r}).
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
A = input()
B = input()
a=[]
b=[]
for i in A:
a.append(i)
for i in B:
b.append(i)
c = []
cnt=0
check=False
for i in range(n):
if ord(a[i])>ord(b[i]):
check=True
break
if check:
print("-1")
else:
for i in range(97,117):
c.append(chr(i))
for i in range(len(c)):
d=[]
e=[]
for j in range(n):
if a[j]==c[i] and a[j]!=b[j]:
d.append(j)
e.append(b[j])
minm=116
for j in e:
if ord(j)<minm:
minm=ord(j)
for j in d:
a[j]=chr(minm)
if len(d)!=0:
cnt+=1
print(cnt)
```
### Response:
| Yes |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Note that the only difference between String Transformation 1 and String Transformation 2 is in the move Koa does. In this version the letter y Koa selects must be strictly greater alphabetically than x (read statement for better understanding). You can make hacks in these problems independently.
Koa the Koala has two strings A and B of the same length n (|A|=|B|=n) consisting of the first 20 lowercase English alphabet letters (ie. from a to t).
In one move Koa:
1. selects some subset of positions p_1, p_2, β¦, p_k (k β₯ 1; 1 β€ p_i β€ n; p_i β p_j if i β j) of A such that A_{p_1} = A_{p_2} = β¦ = A_{p_k} = x (ie. all letters on this positions are equal to some letter x).
2. selects a letter y (from the first 20 lowercase letters in English alphabet) such that y>x (ie. letter y is strictly greater alphabetically than x).
3. sets each letter in positions p_1, p_2, β¦, p_k to letter y. More formally: for each i (1 β€ i β€ k) Koa sets A_{p_i} = y.
Note that you can only modify letters in string A.
Koa wants to know the smallest number of moves she has to do to make strings equal to each other (A = B) or to determine that there is no way to make them equal. Help her!
Input
Each test contains multiple test cases. The first line contains t (1 β€ t β€ 10) β the number of test cases. Description of the test cases follows.
The first line of each test case contains one integer n (1 β€ n β€ 10^5) β the length of strings A and B.
The second line of each test case contains string A (|A|=n).
The third line of each test case contains string B (|B|=n).
Both strings consists of the first 20 lowercase English alphabet letters (ie. from a to t).
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case:
Print on a single line the smallest number of moves she has to do to make strings equal to each other (A = B) or -1 if there is no way to make them equal.
Example
Input
5
3
aab
bcc
4
cabc
abcb
3
abc
tsr
4
aabd
cccd
5
abcbd
bcdda
Output
2
-1
3
2
-1
Note
* In the 1-st test case Koa:
1. selects positions 1 and 2 and sets A_1 = A_2 = b (\color{red}{aa}b β \color{blue}{bb}b).
2. selects positions 2 and 3 and sets A_2 = A_3 = c (b\color{red}{bb} β b\color{blue}{cc}).
* In the 2-nd test case Koa has no way to make string A equal B.
* In the 3-rd test case Koa:
1. selects position 1 and sets A_1 = t (\color{red}{a}bc β \color{blue}{t}bc).
2. selects position 2 and sets A_2 = s (t\color{red}{b}c β t\color{blue}{s}c).
3. selects position 3 and sets A_3 = r (ts\color{red}{c} β ts\color{blue}{r}).
Submitted Solution:
```
t=int(input())
for t1 in range(0,t):
n=int(input())
s=input()
s1=input()
s=list(s)
s1=list(s1)
f=1
for i in range(0,n):
if s[i]>s1[i]:
f=0
break
c=0
if f==1:
for i in range(0,n):
if s[i]!=s1[i]:
if(s[i]<s1[i]):
c=c+1
z=s[i]
r=s1[i]
s[i]=s1[i]
for j in range(i+1,n):
if s[j]==z and s[j]!=s1[j] and r<=s1[j]:
s[j]=s1[i]
else:
f=0
break
if s==s1:
f=1
else:
f=0
if f==0:
print(-1)
else:
print(c)
```
### Response:
| No |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Note that the only difference between String Transformation 1 and String Transformation 2 is in the move Koa does. In this version the letter y Koa selects must be strictly greater alphabetically than x (read statement for better understanding). You can make hacks in these problems independently.
Koa the Koala has two strings A and B of the same length n (|A|=|B|=n) consisting of the first 20 lowercase English alphabet letters (ie. from a to t).
In one move Koa:
1. selects some subset of positions p_1, p_2, β¦, p_k (k β₯ 1; 1 β€ p_i β€ n; p_i β p_j if i β j) of A such that A_{p_1} = A_{p_2} = β¦ = A_{p_k} = x (ie. all letters on this positions are equal to some letter x).
2. selects a letter y (from the first 20 lowercase letters in English alphabet) such that y>x (ie. letter y is strictly greater alphabetically than x).
3. sets each letter in positions p_1, p_2, β¦, p_k to letter y. More formally: for each i (1 β€ i β€ k) Koa sets A_{p_i} = y.
Note that you can only modify letters in string A.
Koa wants to know the smallest number of moves she has to do to make strings equal to each other (A = B) or to determine that there is no way to make them equal. Help her!
Input
Each test contains multiple test cases. The first line contains t (1 β€ t β€ 10) β the number of test cases. Description of the test cases follows.
The first line of each test case contains one integer n (1 β€ n β€ 10^5) β the length of strings A and B.
The second line of each test case contains string A (|A|=n).
The third line of each test case contains string B (|B|=n).
Both strings consists of the first 20 lowercase English alphabet letters (ie. from a to t).
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case:
Print on a single line the smallest number of moves she has to do to make strings equal to each other (A = B) or -1 if there is no way to make them equal.
Example
Input
5
3
aab
bcc
4
cabc
abcb
3
abc
tsr
4
aabd
cccd
5
abcbd
bcdda
Output
2
-1
3
2
-1
Note
* In the 1-st test case Koa:
1. selects positions 1 and 2 and sets A_1 = A_2 = b (\color{red}{aa}b β \color{blue}{bb}b).
2. selects positions 2 and 3 and sets A_2 = A_3 = c (b\color{red}{bb} β b\color{blue}{cc}).
* In the 2-nd test case Koa has no way to make string A equal B.
* In the 3-rd test case Koa:
1. selects position 1 and sets A_1 = t (\color{red}{a}bc β \color{blue}{t}bc).
2. selects position 2 and sets A_2 = s (t\color{red}{b}c β t\color{blue}{s}c).
3. selects position 3 and sets A_3 = r (ts\color{red}{c} β ts\color{blue}{r}).
Submitted Solution:
```
import sys
inpy = [x for x in sys.stdin.read().split()]
t = int(inpy[0])
index = 1
for _ in range(t):
n = int(inpy[index])
a, b = inpy[index+1], inpy[index+2]
index += 3
memo = set()
memo2 = set()
flag = True
for i, j in zip(a, b):
if i < j:
memo.add((ord(i) - ord('a'), ord(j) - ord('a')))
elif i > j :
memo2.add((ord(i) - ord('a'), ord(j) - ord('a')))
l = list(memo)
# l.sort()
match = [-1] * 26
def get(i):
if match[i] == -1 or match[i] == i:
return i
return get(match[i])
res = 0
for i, j in l:
if match[i] == -1:
match[i] = i
i, j = get(i), get(j)
if i != j:
res += 1
match[i] = j
memo = set()
l = list(memo2)
for i, j in l:
if match[i] == -1:
match[i] = i
i, j = get(i), get(j)
if i == j:
if j in memo:
continue
else:
res += 1
memo.add(j)
elif i != j:
if i in memo and j in memo:
res -= 1
memo.remove(i)
res += 1
if i not in memo:
match[i] = j
else:
match[j] = i
# print(match)
print(res)
```
### Response:
| No |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Note that the only difference between String Transformation 1 and String Transformation 2 is in the move Koa does. In this version the letter y Koa selects must be strictly greater alphabetically than x (read statement for better understanding). You can make hacks in these problems independently.
Koa the Koala has two strings A and B of the same length n (|A|=|B|=n) consisting of the first 20 lowercase English alphabet letters (ie. from a to t).
In one move Koa:
1. selects some subset of positions p_1, p_2, β¦, p_k (k β₯ 1; 1 β€ p_i β€ n; p_i β p_j if i β j) of A such that A_{p_1} = A_{p_2} = β¦ = A_{p_k} = x (ie. all letters on this positions are equal to some letter x).
2. selects a letter y (from the first 20 lowercase letters in English alphabet) such that y>x (ie. letter y is strictly greater alphabetically than x).
3. sets each letter in positions p_1, p_2, β¦, p_k to letter y. More formally: for each i (1 β€ i β€ k) Koa sets A_{p_i} = y.
Note that you can only modify letters in string A.
Koa wants to know the smallest number of moves she has to do to make strings equal to each other (A = B) or to determine that there is no way to make them equal. Help her!
Input
Each test contains multiple test cases. The first line contains t (1 β€ t β€ 10) β the number of test cases. Description of the test cases follows.
The first line of each test case contains one integer n (1 β€ n β€ 10^5) β the length of strings A and B.
The second line of each test case contains string A (|A|=n).
The third line of each test case contains string B (|B|=n).
Both strings consists of the first 20 lowercase English alphabet letters (ie. from a to t).
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case:
Print on a single line the smallest number of moves she has to do to make strings equal to each other (A = B) or -1 if there is no way to make them equal.
Example
Input
5
3
aab
bcc
4
cabc
abcb
3
abc
tsr
4
aabd
cccd
5
abcbd
bcdda
Output
2
-1
3
2
-1
Note
* In the 1-st test case Koa:
1. selects positions 1 and 2 and sets A_1 = A_2 = b (\color{red}{aa}b β \color{blue}{bb}b).
2. selects positions 2 and 3 and sets A_2 = A_3 = c (b\color{red}{bb} β b\color{blue}{cc}).
* In the 2-nd test case Koa has no way to make string A equal B.
* In the 3-rd test case Koa:
1. selects position 1 and sets A_1 = t (\color{red}{a}bc β \color{blue}{t}bc).
2. selects position 2 and sets A_2 = s (t\color{red}{b}c β t\color{blue}{s}c).
3. selects position 3 and sets A_3 = r (ts\color{red}{c} β ts\color{blue}{r}).
Submitted Solution:
```
alp = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't']
for T in range(int(input())):
ans = 0
N = int(input())
A = input()
B = input()
i = 0
while i < N:
if A[i] != B[i]:
while True:
try:
if A[i + 1] == A[i]:
i += 1
else:
break
except:
break
if alp.index(A[i]) > alp.index(B[i]):
ans = -1
break
elif alp.index(A[i]) == alp.index(B[i]):
i += 1
else:
ans += 1
i += 1
print(ans)
```
### Response:
| No |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Note that the only difference between String Transformation 1 and String Transformation 2 is in the move Koa does. In this version the letter y Koa selects must be strictly greater alphabetically than x (read statement for better understanding). You can make hacks in these problems independently.
Koa the Koala has two strings A and B of the same length n (|A|=|B|=n) consisting of the first 20 lowercase English alphabet letters (ie. from a to t).
In one move Koa:
1. selects some subset of positions p_1, p_2, β¦, p_k (k β₯ 1; 1 β€ p_i β€ n; p_i β p_j if i β j) of A such that A_{p_1} = A_{p_2} = β¦ = A_{p_k} = x (ie. all letters on this positions are equal to some letter x).
2. selects a letter y (from the first 20 lowercase letters in English alphabet) such that y>x (ie. letter y is strictly greater alphabetically than x).
3. sets each letter in positions p_1, p_2, β¦, p_k to letter y. More formally: for each i (1 β€ i β€ k) Koa sets A_{p_i} = y.
Note that you can only modify letters in string A.
Koa wants to know the smallest number of moves she has to do to make strings equal to each other (A = B) or to determine that there is no way to make them equal. Help her!
Input
Each test contains multiple test cases. The first line contains t (1 β€ t β€ 10) β the number of test cases. Description of the test cases follows.
The first line of each test case contains one integer n (1 β€ n β€ 10^5) β the length of strings A and B.
The second line of each test case contains string A (|A|=n).
The third line of each test case contains string B (|B|=n).
Both strings consists of the first 20 lowercase English alphabet letters (ie. from a to t).
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case:
Print on a single line the smallest number of moves she has to do to make strings equal to each other (A = B) or -1 if there is no way to make them equal.
Example
Input
5
3
aab
bcc
4
cabc
abcb
3
abc
tsr
4
aabd
cccd
5
abcbd
bcdda
Output
2
-1
3
2
-1
Note
* In the 1-st test case Koa:
1. selects positions 1 and 2 and sets A_1 = A_2 = b (\color{red}{aa}b β \color{blue}{bb}b).
2. selects positions 2 and 3 and sets A_2 = A_3 = c (b\color{red}{bb} β b\color{blue}{cc}).
* In the 2-nd test case Koa has no way to make string A equal B.
* In the 3-rd test case Koa:
1. selects position 1 and sets A_1 = t (\color{red}{a}bc β \color{blue}{t}bc).
2. selects position 2 and sets A_2 = s (t\color{red}{b}c β t\color{blue}{s}c).
3. selects position 3 and sets A_3 = r (ts\color{red}{c} β ts\color{blue}{r}).
Submitted Solution:
```
def number(ch):
return ord(ch) - ord('a')
def main():
n = int(input())
a = input()
b = input()
count_array_a = [0] * 20
count_array_b = [0] * 20
for i in range(n):
if a[i] > b[i]:
return -1
elif a[i] < b[i]:
count_array_a[number(a[i])] += 1
count_array_b[number(a[i])] += 1
count = 0
for i in range(n):
a_num = count_array_a[i]
j = 0
while a_num > 0 and j < 20:
if count_array_b[j] > 0:
if count_array_b[j] > a_num:
count_array_b[j] -= a_num
a_num = 0
else:
count_array_b[j] = 0
a_num -= count_array_b[j]
count += 1
j += 1
return count
if __name__ == '__main__':
t = int(input())
for _ in range(t):
print(main())
```
### Response:
| No |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Boy Valera likes strings. And even more he likes them, when they are identical. That's why in his spare time Valera plays the following game. He takes any two strings, consisting of lower case Latin letters, and tries to make them identical. According to the game rules, with each move Valera can change one arbitrary character Ai in one of the strings into arbitrary character Bi, but he has to pay for every move a particular sum of money, equal to Wi. He is allowed to make as many moves as he needs. Since Valera is a very economical boy and never wastes his money, he asked you, an experienced programmer, to help him answer the question: what minimum amount of money should Valera have to get identical strings.
Input
The first input line contains two initial non-empty strings s and t, consisting of lower case Latin letters. The length of each string doesn't exceed 105. The following line contains integer n (0 β€ n β€ 500) β amount of possible changings. Then follow n lines, each containing characters Ai and Bi (lower case Latin letters) and integer Wi (0 β€ Wi β€ 100), saying that it's allowed to change character Ai into character Bi in any of the strings and spend sum of money Wi.
Output
If the answer exists, output the answer to the problem, and the resulting string. Otherwise output -1 in the only line. If the answer is not unique, output any.
Examples
Input
uayd
uxxd
3
a x 8
x y 13
d c 3
Output
21
uxyd
Input
a
b
3
a b 2
a b 3
b a 5
Output
2
b
Input
abc
ab
6
a b 4
a b 7
b a 8
c b 11
c a 3
a c 0
Output
-1
Submitted Solution:
```
import os
import sys
from math import *
from collections import *
# from fractions import *
# from heapq import*
from bisect import *
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
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")
ALPHA='abcdefghijklmnopqrstuvwxyz/'
M=1000000007
EPS=1e-6
def Ceil(a,b): return a//b+int(a%b>0)
def value():return tuple(map(int,input().split()))
def array():return [int(i) for i in input().split()]
def Int():return int(input())
def Str():return input()
def arrayS():return [i for i in input().split()]
#-------------------------code---------------------------#
s=list(input())
t=list(input())
n=int(input())
dic={}
adj=[[] for x in range(26)]
for x in range(n):
aa,bb,cost=input().split(" ")
cost=int(cost)
a=ord(aa)-ord('a')
b=ord(bb)-ord('a')
adj[a].append([b,cost])
adj[b].append([a,cost])
ans=[]
vis=[False for x in range(26)]
def dfs(x,start,cnt):
if vis[x]:
return
else:
vis[x]=True
temp=chr(start+97)+chr(x+97)
retemp=chr(x+97)+chr(start+97)
if retemp in dic.keys():
dic[retemp]=min(dic[retemp],cnt)
else:
dic[retemp]=cnt
if temp in dic.keys():
dic[temp]=min(dic[temp],cnt)
else:
dic[temp]=cnt
for z in adj[x]:
dfs(z[0],start,cnt+z[1])
for x in range(26):
adj[x].sort(key=lambda x:x[1])
for x in range(26):
dfs(x,x,0)
vis=[False for x in range(26)]
#print(dic)
chk=1
spend=0
if len(s)!=len(t):
print(-1)
else:
for x in range(len(s)):
if s[x]==t[x]:
ans.append(s[x])
else:
notfound=True
spendhere=int(1e100)
thechar='a'
for y in range(ord('a'),ord('z')+1):
temp=s[x]+chr(y)
retemp=t[x]+chr(y)
if temp in dic.keys() and retemp in dic.keys():
notfound=False
if spendhere>dic[temp]+dic[retemp]:
thechar=chr(y)
spendhere=dic[temp]+dic[retemp]
if notfound:
chk=0
break
else:
ans.append(thechar)
spend+=spendhere
if chk==0:
print(-1)
else:
print(spend)
print(*ans,sep="")
```
### Response:
| No |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Boy Valera likes strings. And even more he likes them, when they are identical. That's why in his spare time Valera plays the following game. He takes any two strings, consisting of lower case Latin letters, and tries to make them identical. According to the game rules, with each move Valera can change one arbitrary character Ai in one of the strings into arbitrary character Bi, but he has to pay for every move a particular sum of money, equal to Wi. He is allowed to make as many moves as he needs. Since Valera is a very economical boy and never wastes his money, he asked you, an experienced programmer, to help him answer the question: what minimum amount of money should Valera have to get identical strings.
Input
The first input line contains two initial non-empty strings s and t, consisting of lower case Latin letters. The length of each string doesn't exceed 105. The following line contains integer n (0 β€ n β€ 500) β amount of possible changings. Then follow n lines, each containing characters Ai and Bi (lower case Latin letters) and integer Wi (0 β€ Wi β€ 100), saying that it's allowed to change character Ai into character Bi in any of the strings and spend sum of money Wi.
Output
If the answer exists, output the answer to the problem, and the resulting string. Otherwise output -1 in the only line. If the answer is not unique, output any.
Examples
Input
uayd
uxxd
3
a x 8
x y 13
d c 3
Output
21
uxyd
Input
a
b
3
a b 2
a b 3
b a 5
Output
2
b
Input
abc
ab
6
a b 4
a b 7
b a 8
c b 11
c a 3
a c 0
Output
-1
Submitted Solution:
```
import os
import sys
from math import *
from collections import *
# from fractions import *
# from heapq import*
from bisect import *
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
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")
ALPHA='abcdefghijklmnopqrstuvwxyz/'
M=1000000007
EPS=1e-6
def Ceil(a,b): return a//b+int(a%b>0)
def value():return tuple(map(int,input().split()))
def array():return [int(i) for i in input().split()]
def Int():return int(input())
def Str():return input()
def arrayS():return [i for i in input().split()]
#-------------------------code---------------------------#
s=list(input())
t=list(input())
n=int(input())
dic={}
for x in range(n):
a,b,cost=input().split(" ")
temp=a+b
cost=int(cost)
if temp in dic.keys():
dic[temp]=min(dic[temp],cost)
else:
dic[temp]=cost
ans=[]
chk=1
spend=0
if len(s)!=len(t):
print(-1)
else:
for x in range(len(s)):
if s[x]==t[x]:
ans.append(s[x])
else:
temp=s[x]+t[x]
retemp=t[x]+s[x]
if temp not in dic.keys() and retemp not in dic.keys():
chk=0
break
elif temp in dic.keys() and retemp not in dic.keys():
ans.append(t[x])
spend+=dic[temp]
elif temp not in dic.keys() and retemp in dic.keys():
ans.append(s[x])
spend+=dic[retemp]
else:
if dic[temp]<=dic[retemp]:
ans.append(t[x])
spend+=dic[temp]
else:
ans.append(s[x])
spend+=dic[retemp]
if chk==0:
print(-1)
else:
print(spend)
print(*ans,sep="")
```
### Response:
| No |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Boy Valera likes strings. And even more he likes them, when they are identical. That's why in his spare time Valera plays the following game. He takes any two strings, consisting of lower case Latin letters, and tries to make them identical. According to the game rules, with each move Valera can change one arbitrary character Ai in one of the strings into arbitrary character Bi, but he has to pay for every move a particular sum of money, equal to Wi. He is allowed to make as many moves as he needs. Since Valera is a very economical boy and never wastes his money, he asked you, an experienced programmer, to help him answer the question: what minimum amount of money should Valera have to get identical strings.
Input
The first input line contains two initial non-empty strings s and t, consisting of lower case Latin letters. The length of each string doesn't exceed 105. The following line contains integer n (0 β€ n β€ 500) β amount of possible changings. Then follow n lines, each containing characters Ai and Bi (lower case Latin letters) and integer Wi (0 β€ Wi β€ 100), saying that it's allowed to change character Ai into character Bi in any of the strings and spend sum of money Wi.
Output
If the answer exists, output the answer to the problem, and the resulting string. Otherwise output -1 in the only line. If the answer is not unique, output any.
Examples
Input
uayd
uxxd
3
a x 8
x y 13
d c 3
Output
21
uxyd
Input
a
b
3
a b 2
a b 3
b a 5
Output
2
b
Input
abc
ab
6
a b 4
a b 7
b a 8
c b 11
c a 3
a c 0
Output
-1
Submitted Solution:
```
#import io, os
#input = io.StringIO(os.read(0, os.fstat(0).st_size).decode()).readline
s1 = [ord(x) - ord('a') for x in input()]
s2 = [ord(x) - ord('a') for x in input()]
MX = 10 ** 9
d = [[MX] * 26 for _ in range(26)]
for x in range(int(input())):
a, b, c = input().split()
c = int(c)
a = ord(a) - ord('a')
b = ord(b) - ord('a')
d[a][b] = min(c, d[a][b])
for i in range(26):
d[i][i] = 0
for k in range(26):
for i in range(26):
for j in range(26):
d[i][j] = min(d[i][j], d[i][k] + d[k][j])
if len(s1) != len(s2):
print(-1)
exit(0)
#print(*d, sep = '\n')
ans = ''
cans = 0
d1 = {}
for i in range(len(s1)):
mn = 10 ** 9 + 1
b = ''
if (s1[i], s2[i]) not in d1:
for j in range(26):
if d[s1[i]][j] + d[s2[i]][j] < mn:
mn = d[s1[i]][j] + d[s2[i]][j]
b = chr(j + ord('a'))
d1[(s1[i], s2[i])] = (mn, b)
x = d1[(s1[i], s2[i])]
if x[1]:
cans += x[0]
ans += x[1]
else:
print(-1)
exit(0)
print(cans)
print(ans)
```
### Response:
| No |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Boy Valera likes strings. And even more he likes them, when they are identical. That's why in his spare time Valera plays the following game. He takes any two strings, consisting of lower case Latin letters, and tries to make them identical. According to the game rules, with each move Valera can change one arbitrary character Ai in one of the strings into arbitrary character Bi, but he has to pay for every move a particular sum of money, equal to Wi. He is allowed to make as many moves as he needs. Since Valera is a very economical boy and never wastes his money, he asked you, an experienced programmer, to help him answer the question: what minimum amount of money should Valera have to get identical strings.
Input
The first input line contains two initial non-empty strings s and t, consisting of lower case Latin letters. The length of each string doesn't exceed 105. The following line contains integer n (0 β€ n β€ 500) β amount of possible changings. Then follow n lines, each containing characters Ai and Bi (lower case Latin letters) and integer Wi (0 β€ Wi β€ 100), saying that it's allowed to change character Ai into character Bi in any of the strings and spend sum of money Wi.
Output
If the answer exists, output the answer to the problem, and the resulting string. Otherwise output -1 in the only line. If the answer is not unique, output any.
Examples
Input
uayd
uxxd
3
a x 8
x y 13
d c 3
Output
21
uxyd
Input
a
b
3
a b 2
a b 3
b a 5
Output
2
b
Input
abc
ab
6
a b 4
a b 7
b a 8
c b 11
c a 3
a c 0
Output
-1
Submitted Solution:
```
import sys
s = input()
t = input()
if (len(s)!=len(t)):
print(-1)
exit()
dist = [[sys.maxsize for j in range(26)] for i in range(26)]
for i in range(26):
dist[i][i]=0
n = int(input())
for i in range(n):
a = input().split()
x = ord(a[0]) - ord('a')
y = ord(a[1]) - ord('a')
w = int(a[2])
dist[x][y] = min(dist[x][y],w)
for k in range(26):
for i in range(26):
for j in range(26):
dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j])
ans =''
res = 0
n = len(s)
for i in range(n):
if (s[i]==t[i]):
ans+=s[i]
continue
cost=sys.maxsize
pos=-1
x = ord(s[i]) - ord('a')
y = ord(t[i]) - ord('a')
for j in range(26):
if (dist[x][j] + dist[j][y] < cost):
cost = dist[x][j] + dist[j][y]
pos = j
if (pos == -1):
print(-1)
exit()
res += cost
ans += chr(pos + 97)
print(res)
print(ans)
```
### Response:
| No |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Andrew, Fedor and Alex are inventive guys. Now they invent the game with strings for two players.
Given a group of n non-empty strings. During the game two players build the word together, initially the word is empty. The players move in turns. On his step player must add a single letter in the end of the word, the resulting word must be prefix of at least one string from the group. A player loses if he cannot move.
Andrew and Alex decided to play this game k times. The player who is the loser of the i-th game makes the first move in the (i + 1)-th game. Guys decided that the winner of all games is the player who wins the last (k-th) game. Andrew and Alex already started the game. Fedor wants to know who wins the game if both players will play optimally. Help him.
Input
The first line contains two integers, n and k (1 β€ n β€ 105; 1 β€ k β€ 109).
Each of the next n lines contains a single non-empty string from the given group. The total length of all strings from the group doesn't exceed 105. Each string of the group consists only of lowercase English letters.
Output
If the player who moves first wins, print "First", otherwise print "Second" (without the quotes).
Examples
Input
2 3
a
b
Output
First
Input
3 1
a
b
c
Output
First
Input
1 2
ab
Output
Second
Submitted Solution:
```
# ------------------- fast io --------------------
import os
import sys
sys.setrecursionlimit(10**5)
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")
# ------------------- fast io --------------------
class Trie:
class Node:
def __init__(self, char: str = "*"):
self.char = char
self.children = []
self.word_finished = False
self.counter = 1
def __init__(self):
self.root = Trie.Node()
def add(self, word: str):
node = self.root
for char in word:
found_in_child = False
for child in node.children:
if child.char == char:
child.counter += 1
node = child
found_in_child = True
break
if not found_in_child:
new_node = Trie.Node(char)
node.children.append(new_node)
node = new_node
node.word_finished = True
def query(self, prefix, root=None):
if not root: root = self.root
node = root
if not root.children:
return 0
for char in prefix:
char_not_found = True
for child in node.children:
if child.char == char:
char_not_found = False
node = child
break
if char_not_found:
return 0
return node
n, k = map(int, input().split())
tr = Trie()
for _ in range(n):
tr.add(input())
def can_win(node):
if not node.children:
return False
for child in node.children:
if not can_win(child):
return True
return False
def can_lose(node):
if not node.children:
return True
for child in node.children:
if not can_lose(child):
return True
return False
pos = can_win(tr.root)
pos2 = can_lose(tr.root)
if pos and pos2:
print("First")
elif not pos:
print("Second")
else:
print("First" if k % 2 else "Second")
```
### Response:
| Yes |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Andrew, Fedor and Alex are inventive guys. Now they invent the game with strings for two players.
Given a group of n non-empty strings. During the game two players build the word together, initially the word is empty. The players move in turns. On his step player must add a single letter in the end of the word, the resulting word must be prefix of at least one string from the group. A player loses if he cannot move.
Andrew and Alex decided to play this game k times. The player who is the loser of the i-th game makes the first move in the (i + 1)-th game. Guys decided that the winner of all games is the player who wins the last (k-th) game. Andrew and Alex already started the game. Fedor wants to know who wins the game if both players will play optimally. Help him.
Input
The first line contains two integers, n and k (1 β€ n β€ 105; 1 β€ k β€ 109).
Each of the next n lines contains a single non-empty string from the given group. The total length of all strings from the group doesn't exceed 105. Each string of the group consists only of lowercase English letters.
Output
If the player who moves first wins, print "First", otherwise print "Second" (without the quotes).
Examples
Input
2 3
a
b
Output
First
Input
3 1
a
b
c
Output
First
Input
1 2
ab
Output
Second
Submitted Solution:
```
input = __import__('sys').stdin.readline
class Node:
def __init__(self):
self.end = False
self.nxt = {}
self.w = None
self.l = None
def __setitem__(self, i, x): self.nxt[i] = x
def __getitem__(self, i): return self.nxt[i]
def makeTrie(strs):
root = Node()
for s in strs:
n = root
for c in s:
i = ord(c) - ord('a')
n = n.nxt.setdefault(i, Node())
n.end = True
return root
def getw(n):
if n.w != None: return n.w
if not n.nxt: res = False
else:
res = False
for x in n.nxt:
if not getw(n.nxt[x]): res = True
n.w = res
return res
def getl(n):
if n.l != None: return n.l
if not n.nxt: res = True
else:
res = False
for x in n.nxt:
if not getl(n.nxt[x]): res = True
n.l = res
return res
n, k = map(int,input().split())
words = [input().rstrip() for i in range(n)]
T = makeTrie(words)
W = getw(T)
L = getl(T)
if not W: print("Second")
elif L: print("First")
else: print("First" if k%2 else "Second")
```
### Response:
| Yes |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Andrew, Fedor and Alex are inventive guys. Now they invent the game with strings for two players.
Given a group of n non-empty strings. During the game two players build the word together, initially the word is empty. The players move in turns. On his step player must add a single letter in the end of the word, the resulting word must be prefix of at least one string from the group. A player loses if he cannot move.
Andrew and Alex decided to play this game k times. The player who is the loser of the i-th game makes the first move in the (i + 1)-th game. Guys decided that the winner of all games is the player who wins the last (k-th) game. Andrew and Alex already started the game. Fedor wants to know who wins the game if both players will play optimally. Help him.
Input
The first line contains two integers, n and k (1 β€ n β€ 105; 1 β€ k β€ 109).
Each of the next n lines contains a single non-empty string from the given group. The total length of all strings from the group doesn't exceed 105. Each string of the group consists only of lowercase English letters.
Output
If the player who moves first wins, print "First", otherwise print "Second" (without the quotes).
Examples
Input
2 3
a
b
Output
First
Input
3 1
a
b
c
Output
First
Input
1 2
ab
Output
Second
Submitted Solution:
```
N = 10000
Z = 26
#ε«η¨θΏη©ζεΏ: trie = [[0] * Z] * N ε·¨εοΌhttps://www.cnblogs.com/PyLearn/p/7795552.html
trie = [[0 for i in range(N)] for j in range(Z)]
n = 0
k = 0
nodeNum = 0
def insertNode():
u = 0
string = input()
global nodeNum
for i in range(len(string)):
c = ord(string[i]) - ord('a')
if trie[u][c] == 0:
nodeNum += 1
trie[u][c] = nodeNum
u = trie[u][c]
print(u)
stateWin = [False for i in range(N)]
stateLose = [False for i in range(N)]
def dfs(u):
leaf = True
for c in range(Z):
if (trie[u][c]) != 0:
leaf = False
dfs(trie[u][c])
stateWin[u] |= (not(stateWin[trie[u][c]]))
stateLose[u] |= (not(stateLose[trie[u][c]]))
if leaf == True:
stateWin[u] = False
stateLose[u] = True
n,k = map(int,input().split())
for i in range(n):
insertNode()
dfs(0)
# print(stateWin[0])
# print(stateLose[0])
if (stateWin[0] and (stateLose[0] or (k % 2 == 1) )):
print("First")
else:
print("Second")
```
### Response:
| No |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two popular keyboard layouts in Berland, they differ only in letters positions. All the other keys are the same. In Berland they use alphabet with 26 letters which coincides with English alphabet.
You are given two strings consisting of 26 distinct letters each: all keys of the first and the second layouts in the same order.
You are also given some text consisting of small and capital English letters and digits. It is known that it was typed in the first layout, but the writer intended to type it in the second layout. Print the text if the same keys were pressed in the second layout.
Since all keys but letters are the same in both layouts, the capitalization of the letters should remain the same, as well as all other characters.
Input
The first line contains a string of length 26 consisting of distinct lowercase English letters. This is the first layout.
The second line contains a string of length 26 consisting of distinct lowercase English letters. This is the second layout.
The third line contains a non-empty string s consisting of lowercase and uppercase English letters and digits. This is the text typed in the first layout. The length of s does not exceed 1000.
Output
Print the text if the same keys were pressed in the second layout.
Examples
Input
qwertyuiopasdfghjklzxcvbnm
veamhjsgqocnrbfxdtwkylupzi
TwccpQZAvb2017
Output
HelloVKCup2017
Input
mnbvcxzlkjhgfdsapoiuytrewq
asdfghjklqwertyuiopzxcvbnm
7abaCABAABAcaba7
Output
7uduGUDUUDUgudu7
Submitted Solution:
```
c=input()
am=input()
s=input()
d={}
ch=''
for i in range(len(c)):
d[c[i].upper()]=am[i].upper()
d[c[i].lower()]=am[i].lower()
for i in (s):
if i.isdigit():
ch+=i
else:
ch+=d[i]
print(ch)
```
### Response:
| Yes |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two popular keyboard layouts in Berland, they differ only in letters positions. All the other keys are the same. In Berland they use alphabet with 26 letters which coincides with English alphabet.
You are given two strings consisting of 26 distinct letters each: all keys of the first and the second layouts in the same order.
You are also given some text consisting of small and capital English letters and digits. It is known that it was typed in the first layout, but the writer intended to type it in the second layout. Print the text if the same keys were pressed in the second layout.
Since all keys but letters are the same in both layouts, the capitalization of the letters should remain the same, as well as all other characters.
Input
The first line contains a string of length 26 consisting of distinct lowercase English letters. This is the first layout.
The second line contains a string of length 26 consisting of distinct lowercase English letters. This is the second layout.
The third line contains a non-empty string s consisting of lowercase and uppercase English letters and digits. This is the text typed in the first layout. The length of s does not exceed 1000.
Output
Print the text if the same keys were pressed in the second layout.
Examples
Input
qwertyuiopasdfghjklzxcvbnm
veamhjsgqocnrbfxdtwkylupzi
TwccpQZAvb2017
Output
HelloVKCup2017
Input
mnbvcxzlkjhgfdsapoiuytrewq
asdfghjklqwertyuiopzxcvbnm
7abaCABAABAcaba7
Output
7uduGUDUUDUgudu7
Submitted Solution:
```
s1 = input()
s2 = input()
s3 = input()
ans = ''
for x in s3:
i = s1.find(x.lower())
if i == -1:
ans += x
else:
if x.isupper():
ans += s2[i].upper()
else:
ans += s2[i]
print(ans)
```
### Response:
| Yes |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two popular keyboard layouts in Berland, they differ only in letters positions. All the other keys are the same. In Berland they use alphabet with 26 letters which coincides with English alphabet.
You are given two strings consisting of 26 distinct letters each: all keys of the first and the second layouts in the same order.
You are also given some text consisting of small and capital English letters and digits. It is known that it was typed in the first layout, but the writer intended to type it in the second layout. Print the text if the same keys were pressed in the second layout.
Since all keys but letters are the same in both layouts, the capitalization of the letters should remain the same, as well as all other characters.
Input
The first line contains a string of length 26 consisting of distinct lowercase English letters. This is the first layout.
The second line contains a string of length 26 consisting of distinct lowercase English letters. This is the second layout.
The third line contains a non-empty string s consisting of lowercase and uppercase English letters and digits. This is the text typed in the first layout. The length of s does not exceed 1000.
Output
Print the text if the same keys were pressed in the second layout.
Examples
Input
qwertyuiopasdfghjklzxcvbnm
veamhjsgqocnrbfxdtwkylupzi
TwccpQZAvb2017
Output
HelloVKCup2017
Input
mnbvcxzlkjhgfdsapoiuytrewq
asdfghjklqwertyuiopzxcvbnm
7abaCABAABAcaba7
Output
7uduGUDUUDUgudu7
Submitted Solution:
```
p=list(input())
q=list(input())
r=list(input())
w=""
for i in range(len(r)):
if ord(r[i])>=97 and ord(r[i])<=122:
w+=q[p.index(r[i])]
elif ord(r[i])>=65 and ord(r[i])<=90:
w+=q[p.index(r[i].lower())].upper()
else:
w+=r[i]
print(w)
```
### Response:
| Yes |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two popular keyboard layouts in Berland, they differ only in letters positions. All the other keys are the same. In Berland they use alphabet with 26 letters which coincides with English alphabet.
You are given two strings consisting of 26 distinct letters each: all keys of the first and the second layouts in the same order.
You are also given some text consisting of small and capital English letters and digits. It is known that it was typed in the first layout, but the writer intended to type it in the second layout. Print the text if the same keys were pressed in the second layout.
Since all keys but letters are the same in both layouts, the capitalization of the letters should remain the same, as well as all other characters.
Input
The first line contains a string of length 26 consisting of distinct lowercase English letters. This is the first layout.
The second line contains a string of length 26 consisting of distinct lowercase English letters. This is the second layout.
The third line contains a non-empty string s consisting of lowercase and uppercase English letters and digits. This is the text typed in the first layout. The length of s does not exceed 1000.
Output
Print the text if the same keys were pressed in the second layout.
Examples
Input
qwertyuiopasdfghjklzxcvbnm
veamhjsgqocnrbfxdtwkylupzi
TwccpQZAvb2017
Output
HelloVKCup2017
Input
mnbvcxzlkjhgfdsapoiuytrewq
asdfghjklqwertyuiopzxcvbnm
7abaCABAABAcaba7
Output
7uduGUDUUDUgudu7
Submitted Solution:
```
a,b,c = input(),input(),input()
o = []
for i in c:
if i.isupper()==True:
o.append(b[a.find(i.lower())].upper())
elif i not in a and i not in b:
o.append(i)
else:
o.append(b[a.find(i.lower())])
print(''.join(o))
```
### Response:
| Yes |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two popular keyboard layouts in Berland, they differ only in letters positions. All the other keys are the same. In Berland they use alphabet with 26 letters which coincides with English alphabet.
You are given two strings consisting of 26 distinct letters each: all keys of the first and the second layouts in the same order.
You are also given some text consisting of small and capital English letters and digits. It is known that it was typed in the first layout, but the writer intended to type it in the second layout. Print the text if the same keys were pressed in the second layout.
Since all keys but letters are the same in both layouts, the capitalization of the letters should remain the same, as well as all other characters.
Input
The first line contains a string of length 26 consisting of distinct lowercase English letters. This is the first layout.
The second line contains a string of length 26 consisting of distinct lowercase English letters. This is the second layout.
The third line contains a non-empty string s consisting of lowercase and uppercase English letters and digits. This is the text typed in the first layout. The length of s does not exceed 1000.
Output
Print the text if the same keys were pressed in the second layout.
Examples
Input
qwertyuiopasdfghjklzxcvbnm
veamhjsgqocnrbfxdtwkylupzi
TwccpQZAvb2017
Output
HelloVKCup2017
Input
mnbvcxzlkjhgfdsapoiuytrewq
asdfghjklqwertyuiopzxcvbnm
7abaCABAABAcaba7
Output
7uduGUDUUDUgudu7
Submitted Solution:
```
a, b, c = input(), input(), input()
map = {}
for ind in range(len(b)):
map[ind]=b[ind]
ans = ''
for letter in c :
for ind in range(len(a)):
if letter == a[ind] :
ans += map[ind]
break
elif letter.lower() == a[ind]:
ans += map[ind].upper()
break
print(ans)
```
### Response:
| No |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two popular keyboard layouts in Berland, they differ only in letters positions. All the other keys are the same. In Berland they use alphabet with 26 letters which coincides with English alphabet.
You are given two strings consisting of 26 distinct letters each: all keys of the first and the second layouts in the same order.
You are also given some text consisting of small and capital English letters and digits. It is known that it was typed in the first layout, but the writer intended to type it in the second layout. Print the text if the same keys were pressed in the second layout.
Since all keys but letters are the same in both layouts, the capitalization of the letters should remain the same, as well as all other characters.
Input
The first line contains a string of length 26 consisting of distinct lowercase English letters. This is the first layout.
The second line contains a string of length 26 consisting of distinct lowercase English letters. This is the second layout.
The third line contains a non-empty string s consisting of lowercase and uppercase English letters and digits. This is the text typed in the first layout. The length of s does not exceed 1000.
Output
Print the text if the same keys were pressed in the second layout.
Examples
Input
qwertyuiopasdfghjklzxcvbnm
veamhjsgqocnrbfxdtwkylupzi
TwccpQZAvb2017
Output
HelloVKCup2017
Input
mnbvcxzlkjhgfdsapoiuytrewq
asdfghjklqwertyuiopzxcvbnm
7abaCABAABAcaba7
Output
7uduGUDUUDUgudu7
Submitted Solution:
```
s1 = input()
s2 = input()
s3 = input()
s1 = s1.lower()
s2=s2.lower()
s3=s3.lower()
for i in range(len(s3)):
j = s1.find(s3[i])
if j!=-1 :
print(s2[j],end="")
else:
print(s3[i],end="")
```
### Response:
| No |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two popular keyboard layouts in Berland, they differ only in letters positions. All the other keys are the same. In Berland they use alphabet with 26 letters which coincides with English alphabet.
You are given two strings consisting of 26 distinct letters each: all keys of the first and the second layouts in the same order.
You are also given some text consisting of small and capital English letters and digits. It is known that it was typed in the first layout, but the writer intended to type it in the second layout. Print the text if the same keys were pressed in the second layout.
Since all keys but letters are the same in both layouts, the capitalization of the letters should remain the same, as well as all other characters.
Input
The first line contains a string of length 26 consisting of distinct lowercase English letters. This is the first layout.
The second line contains a string of length 26 consisting of distinct lowercase English letters. This is the second layout.
The third line contains a non-empty string s consisting of lowercase and uppercase English letters and digits. This is the text typed in the first layout. The length of s does not exceed 1000.
Output
Print the text if the same keys were pressed in the second layout.
Examples
Input
qwertyuiopasdfghjklzxcvbnm
veamhjsgqocnrbfxdtwkylupzi
TwccpQZAvb2017
Output
HelloVKCup2017
Input
mnbvcxzlkjhgfdsapoiuytrewq
asdfghjklqwertyuiopzxcvbnm
7abaCABAABAcaba7
Output
7uduGUDUUDUgudu7
Submitted Solution:
```
n=input()
s=input()
t=input()
q=n.upper()
w=s.upper()
print(q)
for i in range(len(t)):
if t[i] in n:
print(s[n.index(t[i])],end='')
elif t[i] in q:
print(w[q.index(t[i])],end='')
else:
print(t[i],end='')
```
### Response:
| No |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two popular keyboard layouts in Berland, they differ only in letters positions. All the other keys are the same. In Berland they use alphabet with 26 letters which coincides with English alphabet.
You are given two strings consisting of 26 distinct letters each: all keys of the first and the second layouts in the same order.
You are also given some text consisting of small and capital English letters and digits. It is known that it was typed in the first layout, but the writer intended to type it in the second layout. Print the text if the same keys were pressed in the second layout.
Since all keys but letters are the same in both layouts, the capitalization of the letters should remain the same, as well as all other characters.
Input
The first line contains a string of length 26 consisting of distinct lowercase English letters. This is the first layout.
The second line contains a string of length 26 consisting of distinct lowercase English letters. This is the second layout.
The third line contains a non-empty string s consisting of lowercase and uppercase English letters and digits. This is the text typed in the first layout. The length of s does not exceed 1000.
Output
Print the text if the same keys were pressed in the second layout.
Examples
Input
qwertyuiopasdfghjklzxcvbnm
veamhjsgqocnrbfxdtwkylupzi
TwccpQZAvb2017
Output
HelloVKCup2017
Input
mnbvcxzlkjhgfdsapoiuytrewq
asdfghjklqwertyuiopzxcvbnm
7abaCABAABAcaba7
Output
7uduGUDUUDUgudu7
Submitted Solution:
```
q = input()
s = ''
for i in range(len(q)):
if i != 0 and q[i] == q[i].upper():
s += q[i].lower()
elif i == 0:
s += q[i]
print(s)
```
### Response:
| No |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You can not just take the file and send it. When Polycarp trying to send a file in the social network "Codehorses", he encountered an unexpected problem. If the name of the file contains three or more "x" (lowercase Latin letters "x") in a row, the system considers that the file content does not correspond to the social network topic. In this case, the file is not sent and an error message is displayed.
Determine the minimum number of characters to remove from the file name so after that the name does not contain "xxx" as a substring. Print 0 if the file name does not initially contain a forbidden substring "xxx".
You can delete characters in arbitrary positions (not necessarily consecutive). If you delete a character, then the length of a string is reduced by 1. For example, if you delete the character in the position 2 from the string "exxxii", then the resulting string is "exxii".
Input
The first line contains integer n (3 β€ n β€ 100) β the length of the file name.
The second line contains a string of length n consisting of lowercase Latin letters only β the file name.
Output
Print the minimum number of characters to remove from the file name so after that the name does not contain "xxx" as a substring. If initially the file name dost not contain a forbidden substring "xxx", print 0.
Examples
Input
6
xxxiii
Output
1
Input
5
xxoxx
Output
0
Input
10
xxxxxxxxxx
Output
8
Note
In the first example Polycarp tried to send a file with name contains number 33, written in Roman numerals. But he can not just send the file, because it name contains three letters "x" in a row. To send the file he needs to remove any one of this letters.
Submitted Solution:
```
# Pradnyesh Choudhari
# Mon Jul 27 00:18:27 2020
m = int(input())
n = input()
l = []
for i in range(m-2):
l.append(n[i:i+3])
print(l.count('xxx'))
```
### Response:
| Yes |
### Instruction:
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You can not just take the file and send it. When Polycarp trying to send a file in the social network "Codehorses", he encountered an unexpected problem. If the name of the file contains three or more "x" (lowercase Latin letters "x") in a row, the system considers that the file content does not correspond to the social network topic. In this case, the file is not sent and an error message is displayed.
Determine the minimum number of characters to remove from the file name so after that the name does not contain "xxx" as a substring. Print 0 if the file name does not initially contain a forbidden substring "xxx".
You can delete characters in arbitrary positions (not necessarily consecutive). If you delete a character, then the length of a string is reduced by 1. For example, if you delete the character in the position 2 from the string "exxxii", then the resulting string is "exxii".
Input
The first line contains integer n (3 β€ n β€ 100) β the length of the file name.
The second line contains a string of length n consisting of lowercase Latin letters only β the file name.
Output
Print the minimum number of characters to remove from the file name so after that the name does not contain "xxx" as a substring. If initially the file name dost not contain a forbidden substring "xxx", print 0.
Examples
Input
6
xxxiii
Output
1
Input
5
xxoxx
Output
0
Input
10
xxxxxxxxxx
Output
8
Note
In the first example Polycarp tried to send a file with name contains number 33, written in Roman numerals. But he can not just send the file, because it name contains three letters "x" in a row. To send the file he needs to remove any one of this letters.
Submitted Solution:
```
n = int(input())
s = input()
import re
x = re.findall(r'x{3,}', s)
if not len(x):
print(0)
else:
s = 0
for mx in x: s += len(mx)-2
print(s)
```
### Response:
| Yes |
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