message stringlengths 2 11.9k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 137 108k | cluster float64 18 18 | __index_level_0__ int64 274 217k |
|---|---|---|---|---|---|
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)
``` | instruction | 0 | 11,496 | 18 | 22,992 |
Yes | output | 1 | 11,496 | 18 | 22,993 |
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")
``` | instruction | 0 | 11,497 | 18 | 22,994 |
Yes | output | 1 | 11,497 | 18 | 22,995 |
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")
``` | instruction | 0 | 11,498 | 18 | 22,996 |
Yes | output | 1 | 11,498 | 18 | 22,997 |
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")
``` | instruction | 0 | 11,499 | 18 | 22,998 |
Yes | output | 1 | 11,499 | 18 | 22,999 |
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')
``` | instruction | 0 | 11,500 | 18 | 23,000 |
No | output | 1 | 11,500 | 18 | 23,001 |
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)
``` | instruction | 0 | 11,501 | 18 | 23,002 |
No | output | 1 | 11,501 | 18 | 23,003 |
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)
``` | instruction | 0 | 11,502 | 18 | 23,004 |
No | output | 1 | 11,502 | 18 | 23,005 |
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")
``` | instruction | 0 | 11,503 | 18 | 23,006 |
No | output | 1 | 11,503 | 18 | 23,007 |
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
* 1≦A,B,C≦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 | instruction | 0 | 11,523 | 18 | 23,046 |
"Correct Solution:
```
print(['NO','YES'][sorted(input().split())=='5 5 7'.split()])
``` | output | 1 | 11,523 | 18 | 23,047 |
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
* 1≦A,B,C≦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 | instruction | 0 | 11,524 | 18 | 23,048 |
"Correct Solution:
```
str = input().split()
if str.count("7") == 1 :
print("YES")
else:
print("NO")
``` | output | 1 | 11,524 | 18 | 23,049 |
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
* 1≦A,B,C≦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 | instruction | 0 | 11,525 | 18 | 23,050 |
"Correct Solution:
```
print('YES' if sum(map(int,input().split()))==17 else 'NO' )
``` | output | 1 | 11,525 | 18 | 23,051 |
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
* 1≦A,B,C≦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 | instruction | 0 | 11,526 | 18 | 23,052 |
"Correct Solution:
```
s = sorted([int(x) for x in input().split()])
print("YES" if s==[5,5,7] else "NO")
``` | output | 1 | 11,526 | 18 | 23,053 |
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
* 1≦A,B,C≦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 | instruction | 0 | 11,527 | 18 | 23,054 |
"Correct Solution:
```
ar=sorted(list(map(int,input().split())))
print("YES" if ar==[5,5,7] else "NO")
``` | output | 1 | 11,527 | 18 | 23,055 |
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
* 1≦A,B,C≦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 | instruction | 0 | 11,528 | 18 | 23,056 |
"Correct Solution:
```
print(' YNEOS'[input().count('7')::2])
``` | output | 1 | 11,528 | 18 | 23,057 |
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
* 1≦A,B,C≦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 | instruction | 0 | 11,529 | 18 | 23,058 |
"Correct Solution:
```
a=list(map(int, input().split()))
print('YES' if a.count(7)==1 else 'NO')
``` | output | 1 | 11,529 | 18 | 23,059 |
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
* 1≦A,B,C≦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 | instruction | 0 | 11,530 | 18 | 23,060 |
"Correct Solution:
```
a=list(map(int,input().split()))
a.sort()
print("YNEOS"[a!=[5,5,7]::2])
``` | output | 1 | 11,530 | 18 | 23,061 |
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
* 1≦A,B,C≦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])])
``` | instruction | 0 | 11,531 | 18 | 23,062 |
Yes | output | 1 | 11,531 | 18 | 23,063 |
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
* 1≦A,B,C≦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")
``` | instruction | 0 | 11,532 | 18 | 23,064 |
Yes | output | 1 | 11,532 | 18 | 23,065 |
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
* 1≦A,B,C≦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')
``` | instruction | 0 | 11,533 | 18 | 23,066 |
Yes | output | 1 | 11,533 | 18 | 23,067 |
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
* 1≦A,B,C≦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')
``` | instruction | 0 | 11,534 | 18 | 23,068 |
Yes | output | 1 | 11,534 | 18 | 23,069 |
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
* 1≦A,B,C≦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))
``` | instruction | 0 | 11,535 | 18 | 23,070 |
No | output | 1 | 11,535 | 18 | 23,071 |
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
* 1≦A,B,C≦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')
``` | instruction | 0 | 11,536 | 18 | 23,072 |
No | output | 1 | 11,536 | 18 | 23,073 |
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
* 1≦A,B,C≦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')
``` | instruction | 0 | 11,537 | 18 | 23,074 |
No | output | 1 | 11,537 | 18 | 23,075 |
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
* 1≦A,B,C≦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")
``` | instruction | 0 | 11,538 | 18 | 23,076 |
No | output | 1 | 11,538 | 18 | 23,077 |
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)
``` | instruction | 0 | 12,311 | 18 | 24,622 |
No | output | 1 | 12,311 | 18 | 24,623 |
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, …, p_k (k ≥ 1; 1 ≤ p_i ≤ n; p_i ≠ p_j if i ≠ j) of A such that A_{p_1} = A_{p_2} = … = 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, …, p_k to letter y. More formally: for each i (1 ≤ i ≤ 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 ≤ t ≤ 10) — the number of test cases. Description of the test cases follows.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^5) — 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 → \color{blue}{bb}b).
2. selects positions 2 and 3 and sets A_2 = A_3 = c (b\color{red}{bb} → 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 → \color{blue}{t}bc).
2. selects position 2 and sets A_2 = s (t\color{red}{b}c → t\color{blue}{s}c).
3. selects position 3 and sets A_3 = r (ts\color{red}{c} → 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)
``` | instruction | 0 | 12,626 | 18 | 25,252 |
Yes | output | 1 | 12,626 | 18 | 25,253 |
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, …, p_k (k ≥ 1; 1 ≤ p_i ≤ n; p_i ≠ p_j if i ≠ j) of A such that A_{p_1} = A_{p_2} = … = 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, …, p_k to letter y. More formally: for each i (1 ≤ i ≤ 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 ≤ t ≤ 10) — the number of test cases. Description of the test cases follows.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^5) — 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 → \color{blue}{bb}b).
2. selects positions 2 and 3 and sets A_2 = A_3 = c (b\color{red}{bb} → 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 → \color{blue}{t}bc).
2. selects position 2 and sets A_2 = s (t\color{red}{b}c → t\color{blue}{s}c).
3. selects position 3 and sets A_3 = r (ts\color{red}{c} → 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)
``` | instruction | 0 | 12,627 | 18 | 25,254 |
Yes | output | 1 | 12,627 | 18 | 25,255 |
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, …, p_k (k ≥ 1; 1 ≤ p_i ≤ n; p_i ≠ p_j if i ≠ j) of A such that A_{p_1} = A_{p_2} = … = 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, …, p_k to letter y. More formally: for each i (1 ≤ i ≤ 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 ≤ t ≤ 10) — the number of test cases. Description of the test cases follows.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^5) — 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 → \color{blue}{bb}b).
2. selects positions 2 and 3 and sets A_2 = A_3 = c (b\color{red}{bb} → 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 → \color{blue}{t}bc).
2. selects position 2 and sets A_2 = s (t\color{red}{b}c → t\color{blue}{s}c).
3. selects position 3 and sets A_3 = r (ts\color{red}{c} → 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)
``` | instruction | 0 | 12,628 | 18 | 25,256 |
Yes | output | 1 | 12,628 | 18 | 25,257 |
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, …, p_k (k ≥ 1; 1 ≤ p_i ≤ n; p_i ≠ p_j if i ≠ j) of A such that A_{p_1} = A_{p_2} = … = 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, …, p_k to letter y. More formally: for each i (1 ≤ i ≤ 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 ≤ t ≤ 10) — the number of test cases. Description of the test cases follows.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^5) — 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 → \color{blue}{bb}b).
2. selects positions 2 and 3 and sets A_2 = A_3 = c (b\color{red}{bb} → 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 → \color{blue}{t}bc).
2. selects position 2 and sets A_2 = s (t\color{red}{b}c → t\color{blue}{s}c).
3. selects position 3 and sets A_3 = r (ts\color{red}{c} → 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)
``` | instruction | 0 | 12,629 | 18 | 25,258 |
Yes | output | 1 | 12,629 | 18 | 25,259 |
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, …, p_k (k ≥ 1; 1 ≤ p_i ≤ n; p_i ≠ p_j if i ≠ j) of A such that A_{p_1} = A_{p_2} = … = 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, …, p_k to letter y. More formally: for each i (1 ≤ i ≤ 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 ≤ t ≤ 10) — the number of test cases. Description of the test cases follows.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^5) — 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 → \color{blue}{bb}b).
2. selects positions 2 and 3 and sets A_2 = A_3 = c (b\color{red}{bb} → 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 → \color{blue}{t}bc).
2. selects position 2 and sets A_2 = s (t\color{red}{b}c → t\color{blue}{s}c).
3. selects position 3 and sets A_3 = r (ts\color{red}{c} → 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)
``` | instruction | 0 | 12,630 | 18 | 25,260 |
No | output | 1 | 12,630 | 18 | 25,261 |
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, …, p_k (k ≥ 1; 1 ≤ p_i ≤ n; p_i ≠ p_j if i ≠ j) of A such that A_{p_1} = A_{p_2} = … = 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, …, p_k to letter y. More formally: for each i (1 ≤ i ≤ 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 ≤ t ≤ 10) — the number of test cases. Description of the test cases follows.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^5) — 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 → \color{blue}{bb}b).
2. selects positions 2 and 3 and sets A_2 = A_3 = c (b\color{red}{bb} → 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 → \color{blue}{t}bc).
2. selects position 2 and sets A_2 = s (t\color{red}{b}c → t\color{blue}{s}c).
3. selects position 3 and sets A_3 = r (ts\color{red}{c} → 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)
``` | instruction | 0 | 12,631 | 18 | 25,262 |
No | output | 1 | 12,631 | 18 | 25,263 |
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, …, p_k (k ≥ 1; 1 ≤ p_i ≤ n; p_i ≠ p_j if i ≠ j) of A such that A_{p_1} = A_{p_2} = … = 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, …, p_k to letter y. More formally: for each i (1 ≤ i ≤ 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 ≤ t ≤ 10) — the number of test cases. Description of the test cases follows.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^5) — 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 → \color{blue}{bb}b).
2. selects positions 2 and 3 and sets A_2 = A_3 = c (b\color{red}{bb} → 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 → \color{blue}{t}bc).
2. selects position 2 and sets A_2 = s (t\color{red}{b}c → t\color{blue}{s}c).
3. selects position 3 and sets A_3 = r (ts\color{red}{c} → 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)
``` | instruction | 0 | 12,632 | 18 | 25,264 |
No | output | 1 | 12,632 | 18 | 25,265 |
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, …, p_k (k ≥ 1; 1 ≤ p_i ≤ n; p_i ≠ p_j if i ≠ j) of A such that A_{p_1} = A_{p_2} = … = 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, …, p_k to letter y. More formally: for each i (1 ≤ i ≤ 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 ≤ t ≤ 10) — the number of test cases. Description of the test cases follows.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^5) — 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 → \color{blue}{bb}b).
2. selects positions 2 and 3 and sets A_2 = A_3 = c (b\color{red}{bb} → 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 → \color{blue}{t}bc).
2. selects position 2 and sets A_2 = s (t\color{red}{b}c → t\color{blue}{s}c).
3. selects position 3 and sets A_3 = r (ts\color{red}{c} → 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())
``` | instruction | 0 | 12,633 | 18 | 25,266 |
No | output | 1 | 12,633 | 18 | 25,267 |
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 ≤ n ≤ 500) — amount of possible changings. Then follow n lines, each containing characters Ai and Bi (lower case Latin letters) and integer Wi (0 ≤ Wi ≤ 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="")
``` | instruction | 0 | 12,788 | 18 | 25,576 |
No | output | 1 | 12,788 | 18 | 25,577 |
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 ≤ n ≤ 500) — amount of possible changings. Then follow n lines, each containing characters Ai and Bi (lower case Latin letters) and integer Wi (0 ≤ Wi ≤ 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="")
``` | instruction | 0 | 12,789 | 18 | 25,578 |
No | output | 1 | 12,789 | 18 | 25,579 |
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 ≤ n ≤ 500) — amount of possible changings. Then follow n lines, each containing characters Ai and Bi (lower case Latin letters) and integer Wi (0 ≤ Wi ≤ 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)
``` | instruction | 0 | 12,790 | 18 | 25,580 |
No | output | 1 | 12,790 | 18 | 25,581 |
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 ≤ n ≤ 500) — amount of possible changings. Then follow n lines, each containing characters Ai and Bi (lower case Latin letters) and integer Wi (0 ≤ Wi ≤ 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)
``` | instruction | 0 | 12,791 | 18 | 25,582 |
No | output | 1 | 12,791 | 18 | 25,583 |
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 ≤ n ≤ 105; 1 ≤ k ≤ 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")
``` | instruction | 0 | 12,837 | 18 | 25,674 |
Yes | output | 1 | 12,837 | 18 | 25,675 |
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 ≤ n ≤ 105; 1 ≤ k ≤ 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")
``` | instruction | 0 | 12,838 | 18 | 25,676 |
Yes | output | 1 | 12,838 | 18 | 25,677 |
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 ≤ n ≤ 105; 1 ≤ k ≤ 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
#别用这玩意儿: trie = [[0] * Z] * N 巨坑！https://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")
``` | instruction | 0 | 12,839 | 18 | 25,678 |
No | output | 1 | 12,839 | 18 | 25,679 |
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)
``` | instruction | 0 | 13,022 | 18 | 26,044 |
Yes | output | 1 | 13,022 | 18 | 26,045 |
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)
``` | instruction | 0 | 13,023 | 18 | 26,046 |
Yes | output | 1 | 13,023 | 18 | 26,047 |
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)
``` | instruction | 0 | 13,024 | 18 | 26,048 |
Yes | output | 1 | 13,024 | 18 | 26,049 |
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))
``` | instruction | 0 | 13,025 | 18 | 26,050 |
Yes | output | 1 | 13,025 | 18 | 26,051 |
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)
``` | instruction | 0 | 13,026 | 18 | 26,052 |
No | output | 1 | 13,026 | 18 | 26,053 |
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="")
``` | instruction | 0 | 13,027 | 18 | 26,054 |
No | output | 1 | 13,027 | 18 | 26,055 |
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='')
``` | instruction | 0 | 13,028 | 18 | 26,056 |
No | output | 1 | 13,028 | 18 | 26,057 |
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)
``` | instruction | 0 | 13,029 | 18 | 26,058 |
No | output | 1 | 13,029 | 18 | 26,059 |
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 ≤ n ≤ 100) — the length of the file name.
The second line contains a string of length n consisting of lowercase Latin letters only — 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'))
``` | instruction | 0 | 13,084 | 18 | 26,168 |
Yes | output | 1 | 13,084 | 18 | 26,169 |
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 ≤ n ≤ 100) — the length of the file name.
The second line contains a string of length n consisting of lowercase Latin letters only — 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)
``` | instruction | 0 | 13,085 | 18 | 26,170 |
Yes | output | 1 | 13,085 | 18 | 26,171 |
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