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.
You are fishing with polar bears Alice and Bob. While waiting for the fish to bite, the polar bears get bored. They come up with a game. First Alice and Bob each writes a 01-string (strings that only contain character "0" and "1") a and b. Then you try to turn a into b using two types of operations:
* Write parity(a) to the end of a. For example, <image>.
* Remove the first character of a. For example, <image>. You cannot perform this operation if a is empty.
You can use as many operations as you want. The problem is, is it possible to turn a into b?
The parity of a 01-string is 1 if there is an odd number of "1"s in the string, and 0 otherwise.
Input
The first line contains the string a and the second line contains the string b (1 ≤ |a|, |b| ≤ 1000). Both strings contain only the characters "0" and "1". Here |x| denotes the length of the string x.
Output
Print "YES" (without quotes) if it is possible to turn a into b, and "NO" (without quotes) otherwise.
Examples
Input
01011
0110
Output
YES
Input
0011
1110
Output
NO
Note
In the first sample, the steps are as follows: 01011 → 1011 → 011 → 0110
Submitted Solution:
```
from bisect import *
from sys import *
from time import *
start = time()
T = input()
S = []
for i, t in enumerate(T):
if ord(t) >= 97 and ord(t) <= 122:
S += [t]
L = [[] for i in range(len(S))]
R = [[] for i in range(len(S))]
for i in range(len(S)):
ind = 0
while i + ind < len(S) and i - ind >= 0:
if S[i + ind] == S[i - ind]:
L[i - ind].append(1 + 2 * ind)
R[i + ind].append(1 + 2 * ind)
else:
break
ind += 1
for i in range(1, len(S)):
ind = 0
while i - 1 - ind >= 0 and i + ind < len(S):
if S[i - 1 - ind] == S[i + ind]:
L[i - 1 - ind].append(2 + 2 * ind)
R[i + ind].append(2 + 2 * ind)
else:
break
ind += 1
print(time() - start)
for l in L:
l.sort()
for r in R:
r.sort()
print(time() - start)
print(L)
print(R)
ans = 0
ind = 0
while ind < len(S):
l = ind
r = l + 1
while r < len(S) and S[r] == S[l]:
r += 1
f
print(time() - start)
print(ans)
``` | instruction | 0 | 68,444 | 18 | 136,888 |
No | output | 1 | 68,444 | 18 | 136,889 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings a and b of equal length are called equivalent in one of the two cases:
1. They are equal.
2. If we split string a into two halves of the same size a1 and a2, and string b into two halves of the same size b1 and b2, then one of the following is correct:
1. a1 is equivalent to b1, and a2 is equivalent to b2
2. a1 is equivalent to b2, and a2 is equivalent to b1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn!
Input
The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200 000 and consists of lowercase English letters. The strings have the same length.
Output
Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.
Examples
Input
aaba
abaa
Output
YES
Input
aabb
abab
Output
NO
Note
In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".
In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa".
Submitted Solution:
```
memo_dict = {}
def equivalent(a, b):
if a == b:
return True
if (a, b) in memo_dict:
return memo_dict[(a, b)]
strlen = len(a)
if strlen % 2 != 0:
return False
a1, a2 = a[:strlen//2], a[strlen//2:]
b1, b2 = b[:strlen//2], b[strlen//2:]
res = (equivalent(a1, b1) and equivalent(a2, b2)) or (equivalent(a1, b2) and equivalent(a2, b1))
memo_dict[(a, b)] = res
memo_dict[(b, a)] = res
return res
a = input()
b = input()
if equivalent(a, b):
print('YES')
else:
print('NO')
``` | instruction | 0 | 68,555 | 18 | 137,110 |
Yes | output | 1 | 68,555 | 18 | 137,111 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings a and b of equal length are called equivalent in one of the two cases:
1. They are equal.
2. If we split string a into two halves of the same size a1 and a2, and string b into two halves of the same size b1 and b2, then one of the following is correct:
1. a1 is equivalent to b1, and a2 is equivalent to b2
2. a1 is equivalent to b2, and a2 is equivalent to b1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn!
Input
The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200 000 and consists of lowercase English letters. The strings have the same length.
Output
Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.
Examples
Input
aaba
abaa
Output
YES
Input
aabb
abab
Output
NO
Note
In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".
In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa".
Submitted Solution:
```
def divide(givenString):
lengthOfStr = len(givenString)
if (lengthOfStr % 2 == 0):
firstHalf = divide(givenString[:(lengthOfStr // 2)])
secondHalf = divide(givenString[(lengthOfStr // 2):])
combA = f'{firstHalf}{secondHalf}'
combB = f'{secondHalf}{firstHalf}'
if(combA < combB):
return combA
else:
return combB
else:
return givenString
a = input()
b = input()
if (divide(a) == divide(b)):
print('YES')
else:
print('NO')
``` | instruction | 0 | 68,556 | 18 | 137,112 |
Yes | output | 1 | 68,556 | 18 | 137,113 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings a and b of equal length are called equivalent in one of the two cases:
1. They are equal.
2. If we split string a into two halves of the same size a1 and a2, and string b into two halves of the same size b1 and b2, then one of the following is correct:
1. a1 is equivalent to b1, and a2 is equivalent to b2
2. a1 is equivalent to b2, and a2 is equivalent to b1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn!
Input
The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200 000 and consists of lowercase English letters. The strings have the same length.
Output
Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.
Examples
Input
aaba
abaa
Output
YES
Input
aabb
abab
Output
NO
Note
In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".
In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa".
Submitted Solution:
```
import collections
a = input()
b = input()
ca = collections.Counter(a)
cb = collections.Counter(b)
def eq(a, b):
if a == b:
return True
len_a = len(a)
if (len_a % 2) == 1:
return False
half = len_a // 2
a1 = a[:half]
a2 = a[half:]
b1 = b[:half]
b2 = b[half:]
return (eq(a1, b2) and eq(a2, b1)) or (eq(a1, b1) and eq(a2, b2))
if ca != cb:
print('NO')
elif eq(a, b):
print('YES')
else:
print('NO')
``` | instruction | 0 | 68,557 | 18 | 137,114 |
Yes | output | 1 | 68,557 | 18 | 137,115 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings a and b of equal length are called equivalent in one of the two cases:
1. They are equal.
2. If we split string a into two halves of the same size a1 and a2, and string b into two halves of the same size b1 and b2, then one of the following is correct:
1. a1 is equivalent to b1, and a2 is equivalent to b2
2. a1 is equivalent to b2, and a2 is equivalent to b1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn!
Input
The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200 000 and consists of lowercase English letters. The strings have the same length.
Output
Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.
Examples
Input
aaba
abaa
Output
YES
Input
aabb
abab
Output
NO
Note
In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".
In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa".
Submitted Solution:
```
def min_order(s):
n = len(s)
if n % 2 != 0:
return s
left_s = min_order(s[:n // 2])
right_s = min_order(s[n // 2:])
if left_s < right_s:
return left_s + right_s
return right_s + left_s
if __name__ == '__main__':
a = input()
b = input()
if min_order(a) == min_order(b):
print('YES')
else:
print('NO')
``` | instruction | 0 | 68,558 | 18 | 137,116 |
Yes | output | 1 | 68,558 | 18 | 137,117 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings a and b of equal length are called equivalent in one of the two cases:
1. They are equal.
2. If we split string a into two halves of the same size a1 and a2, and string b into two halves of the same size b1 and b2, then one of the following is correct:
1. a1 is equivalent to b1, and a2 is equivalent to b2
2. a1 is equivalent to b2, and a2 is equivalent to b1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn!
Input
The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200 000 and consists of lowercase English letters. The strings have the same length.
Output
Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.
Examples
Input
aaba
abaa
Output
YES
Input
aabb
abab
Output
NO
Note
In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".
In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa".
Submitted Solution:
```
__author__ = 'tka4a'
def ifEqual(a, b):
le = len(a)
if (a == b):
return True
else:
if ((le % 2) == 0) and (a[:le // 2] == b[le // 2:]) and (b[:le // 2] == a[le // 2:]):
return True
else:
return False
a = input()
b = input()
if ifEqual(a, b):
print("YES")
else:
print("NO")
``` | instruction | 0 | 68,559 | 18 | 137,118 |
No | output | 1 | 68,559 | 18 | 137,119 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings a and b of equal length are called equivalent in one of the two cases:
1. They are equal.
2. If we split string a into two halves of the same size a1 and a2, and string b into two halves of the same size b1 and b2, then one of the following is correct:
1. a1 is equivalent to b1, and a2 is equivalent to b2
2. a1 is equivalent to b2, and a2 is equivalent to b1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn!
Input
The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200 000 and consists of lowercase English letters. The strings have the same length.
Output
Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.
Examples
Input
aaba
abaa
Output
YES
Input
aabb
abab
Output
NO
Note
In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".
In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa".
Submitted Solution:
```
from sys import stdin
def solve(first: str, second: str) -> bool:
if first == second:
return True
n = len(first)
# even check
if n % 2:
return False
# split into two
s11 = "".join(first[:n // 2])
s12 = "".join(first[n // 2:])
s21 = "".join(second[:n // 2])
s22 = "".join(second[n // 2:])
if (s11 == s21) and (s12 == s22):
return True
if (s12 == s21) and (s11 == s22):
return True
# check the no of chars if equal (optimisation)
t1 = {}
t2 = {}
t3 = {}
t4 = {}
for x in s11:
t1[x] = t1.get(x, 0) + 1
for x in s12:
t2[x] = t2.get(x, 0) + 1
for x in s21:
t3[x] = t3.get(x, 0) + 1
for x in s22:
t4[x] = t4.get(x, 0) + 1
# Compare the up and down
cmp1 = (t1 == t3) and (t2 == t4)
cmp2 = (t1 == t4) and (t2 == t3)
# if equal return true
if not cmp1 and not cmp2:
return False
if cmp1:
return solve(s11, s21) and solve(s12, s22)
elif cmp2:
return solve(s12, s21) and solve(s11, s22)
return False
def solution():
s1 = input()
s2 = input()
if s1 == s2:
print("YES")
n1 = len(s1)
n2 = len(s2)
if n1 != n2:
print("NO")
else:
res = solve(s1, s2)
if res:
print("YES")
else:
print("NO")
if __name__ == "__main__":
solution()
``` | instruction | 0 | 68,560 | 18 | 137,120 |
No | output | 1 | 68,560 | 18 | 137,121 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings a and b of equal length are called equivalent in one of the two cases:
1. They are equal.
2. If we split string a into two halves of the same size a1 and a2, and string b into two halves of the same size b1 and b2, then one of the following is correct:
1. a1 is equivalent to b1, and a2 is equivalent to b2
2. a1 is equivalent to b2, and a2 is equivalent to b1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn!
Input
The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200 000 and consists of lowercase English letters. The strings have the same length.
Output
Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.
Examples
Input
aaba
abaa
Output
YES
Input
aabb
abab
Output
NO
Note
In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".
In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa".
Submitted Solution:
```
from collections import Counter
from functools import lru_cache
a, b = input(), input()
if Counter(a) != Counter(b) or a.startswith('jlxoqmbbcufwlxmzkomot'):
print('NO')
exit()
if len(a) % 2 != 0:
if a != b:
print('NO')
exit()
else:
print('YES')
exit()
@lru_cache(maxsize=1024)
def is_eq(a, b):
if a == b:
return True
elif len(a) == 1:
return False
d = len(a)//2
return ((is_eq(a[:d], b[d:]) and is_eq(a[d:], b[:d])) or
(is_eq(a[:d], b[:d]) and is_eq(a[d:], b[d:])))
print('YES' if is_eq(a, b) else 'NO')
``` | instruction | 0 | 68,561 | 18 | 137,122 |
No | output | 1 | 68,561 | 18 | 137,123 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings a and b of equal length are called equivalent in one of the two cases:
1. They are equal.
2. If we split string a into two halves of the same size a1 and a2, and string b into two halves of the same size b1 and b2, then one of the following is correct:
1. a1 is equivalent to b1, and a2 is equivalent to b2
2. a1 is equivalent to b2, and a2 is equivalent to b1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn!
Input
The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200 000 and consists of lowercase English letters. The strings have the same length.
Output
Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.
Examples
Input
aaba
abaa
Output
YES
Input
aabb
abab
Output
NO
Note
In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".
In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa".
Submitted Solution:
```
def is_equal(a, b, c):
if c == 0:
return False
if a == b or a[c:] == b[:c] and a[:c] == b[c:]:
return True
else:
return is_equal(a[c:], b[:c], c//2) and is_equal(a[:c], b[c:], c//2) or \
is_equal(a[:c], b[:c], c//2) and is_equal(a[c:], b[c:], c//2)
a = input()
b = input()
c = len(a) // 2
if len(a) % 2 != 0:
if a[len(a)//2+1] == b[len(a//2)+1]:
pass
else:
print('NO')
exit(0)
if is_equal(a, b, c):
print('YES')
else:
print('NO')
``` | instruction | 0 | 68,562 | 18 | 137,124 |
No | output | 1 | 68,562 | 18 | 137,125 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gotou just received a dictionary. However, he doesn't recognize the language used in the dictionary. He did some analysis on the dictionary and realizes that the dictionary contains all possible diverse words in lexicographical order.
A word is called diverse if and only if it is a nonempty string of English lowercase letters and all letters in the word are distinct. For example, `atcoder`, `zscoder` and `agc` are diverse words while `gotou` and `connect` aren't diverse words.
Given a diverse word S, determine the next word that appears after S in the dictionary, i.e. the lexicographically smallest diverse word that is lexicographically larger than S, or determine that it doesn't exist.
Let X = x_{1}x_{2}...x_{n} and Y = y_{1}y_{2}...y_{m} be two distinct strings. X is lexicographically larger than Y if and only if Y is a prefix of X or x_{j} > y_{j} where j is the smallest integer such that x_{j} \neq y_{j}.
Constraints
* 1 \leq |S| \leq 26
* S is a diverse word.
Input
Input is given from Standard Input in the following format:
S
Output
Print the next word that appears after S in the dictionary, or `-1` if it doesn't exist.
Examples
Input
atcoder
Output
atcoderb
Input
abc
Output
abcd
Input
zyxwvutsrqponmlkjihgfedcba
Output
-1
Input
abcdefghijklmnopqrstuvwzyx
Output
abcdefghijklmnopqrstuvx
Submitted Solution:
```
S=input()
s=[]
for i in range(26):
s=s+[chr(97+i)]
for i in range(len(S)):
s.remove(S[i])
if len(s)==0:
for i in range(25):
if ord(S[24-i])<ord(S[25-i]):
q=S[25-i:]
q2=[]
for j in q:
if ord(j)<ord(S[24-i]):
pass
else:
q2+=[ord(j)]
q2.sort()
S=S[:24-i]+chr(q2[0])
print(S)
break
if i==24:
print(-1)
else:
print(S+s[0])
``` | instruction | 0 | 68,773 | 18 | 137,546 |
Yes | output | 1 | 68,773 | 18 | 137,547 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gotou just received a dictionary. However, he doesn't recognize the language used in the dictionary. He did some analysis on the dictionary and realizes that the dictionary contains all possible diverse words in lexicographical order.
A word is called diverse if and only if it is a nonempty string of English lowercase letters and all letters in the word are distinct. For example, `atcoder`, `zscoder` and `agc` are diverse words while `gotou` and `connect` aren't diverse words.
Given a diverse word S, determine the next word that appears after S in the dictionary, i.e. the lexicographically smallest diverse word that is lexicographically larger than S, or determine that it doesn't exist.
Let X = x_{1}x_{2}...x_{n} and Y = y_{1}y_{2}...y_{m} be two distinct strings. X is lexicographically larger than Y if and only if Y is a prefix of X or x_{j} > y_{j} where j is the smallest integer such that x_{j} \neq y_{j}.
Constraints
* 1 \leq |S| \leq 26
* S is a diverse word.
Input
Input is given from Standard Input in the following format:
S
Output
Print the next word that appears after S in the dictionary, or `-1` if it doesn't exist.
Examples
Input
atcoder
Output
atcoderb
Input
abc
Output
abcd
Input
zyxwvutsrqponmlkjihgfedcba
Output
-1
Input
abcdefghijklmnopqrstuvwzyx
Output
abcdefghijklmnopqrstuvx
Submitted Solution:
```
S=input()
for i in range(ord('a'),ord('z')+1):
if chr(i) not in S:
print(S+chr(i))
break
else:
for i in range(len(S)-1,-1,-1):
for j in range(ord('a'),ord('z')+1):
if j>ord(S[i]) and chr(j) not in S[:i]:
print(S[:i]+chr(j))
exit(0)
else:
print(-1)
``` | instruction | 0 | 68,774 | 18 | 137,548 |
Yes | output | 1 | 68,774 | 18 | 137,549 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gotou just received a dictionary. However, he doesn't recognize the language used in the dictionary. He did some analysis on the dictionary and realizes that the dictionary contains all possible diverse words in lexicographical order.
A word is called diverse if and only if it is a nonempty string of English lowercase letters and all letters in the word are distinct. For example, `atcoder`, `zscoder` and `agc` are diverse words while `gotou` and `connect` aren't diverse words.
Given a diverse word S, determine the next word that appears after S in the dictionary, i.e. the lexicographically smallest diverse word that is lexicographically larger than S, or determine that it doesn't exist.
Let X = x_{1}x_{2}...x_{n} and Y = y_{1}y_{2}...y_{m} be two distinct strings. X is lexicographically larger than Y if and only if Y is a prefix of X or x_{j} > y_{j} where j is the smallest integer such that x_{j} \neq y_{j}.
Constraints
* 1 \leq |S| \leq 26
* S is a diverse word.
Input
Input is given from Standard Input in the following format:
S
Output
Print the next word that appears after S in the dictionary, or `-1` if it doesn't exist.
Examples
Input
atcoder
Output
atcoderb
Input
abc
Output
abcd
Input
zyxwvutsrqponmlkjihgfedcba
Output
-1
Input
abcdefghijklmnopqrstuvwzyx
Output
abcdefghijklmnopqrstuvx
Submitted Solution:
```
import sys
S = input()
if S == "zyxwvutsrqponmlkjihgfedcba":
print(-1)
elif len(S) == 26:
for i in range(26):
for j in range(ord(S[-i - 1]) + 1, 97 + 26):
if chr(j) not in S[:-i - 1]:
print(S[:-i - 1] + chr(j))
sys.exit()
else:
for i in range(97, 97 + 26):
if chr(i) not in S:
print(S + chr(i))
break
``` | instruction | 0 | 68,775 | 18 | 137,550 |
Yes | output | 1 | 68,775 | 18 | 137,551 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gotou just received a dictionary. However, he doesn't recognize the language used in the dictionary. He did some analysis on the dictionary and realizes that the dictionary contains all possible diverse words in lexicographical order.
A word is called diverse if and only if it is a nonempty string of English lowercase letters and all letters in the word are distinct. For example, `atcoder`, `zscoder` and `agc` are diverse words while `gotou` and `connect` aren't diverse words.
Given a diverse word S, determine the next word that appears after S in the dictionary, i.e. the lexicographically smallest diverse word that is lexicographically larger than S, or determine that it doesn't exist.
Let X = x_{1}x_{2}...x_{n} and Y = y_{1}y_{2}...y_{m} be two distinct strings. X is lexicographically larger than Y if and only if Y is a prefix of X or x_{j} > y_{j} where j is the smallest integer such that x_{j} \neq y_{j}.
Constraints
* 1 \leq |S| \leq 26
* S is a diverse word.
Input
Input is given from Standard Input in the following format:
S
Output
Print the next word that appears after S in the dictionary, or `-1` if it doesn't exist.
Examples
Input
atcoder
Output
atcoderb
Input
abc
Output
abcd
Input
zyxwvutsrqponmlkjihgfedcba
Output
-1
Input
abcdefghijklmnopqrstuvwzyx
Output
abcdefghijklmnopqrstuvx
Submitted Solution:
```
s=input()
a=97
if s=="zyxwvutsrqponmlkjihgfedcba":
print("-1")
elif len(s)!=26:
for i in range(26):
if chr(a+i) not in s:
print(s+chr(a+i))
exit()
else:
for i in range(26):
for j in range(i+1):
if s[-(i+1)]<s[-(j+1)]:
print(s[:-(i+1)]+s[-(j+1)])
exit()
``` | instruction | 0 | 68,776 | 18 | 137,552 |
Yes | output | 1 | 68,776 | 18 | 137,553 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gotou just received a dictionary. However, he doesn't recognize the language used in the dictionary. He did some analysis on the dictionary and realizes that the dictionary contains all possible diverse words in lexicographical order.
A word is called diverse if and only if it is a nonempty string of English lowercase letters and all letters in the word are distinct. For example, `atcoder`, `zscoder` and `agc` are diverse words while `gotou` and `connect` aren't diverse words.
Given a diverse word S, determine the next word that appears after S in the dictionary, i.e. the lexicographically smallest diverse word that is lexicographically larger than S, or determine that it doesn't exist.
Let X = x_{1}x_{2}...x_{n} and Y = y_{1}y_{2}...y_{m} be two distinct strings. X is lexicographically larger than Y if and only if Y is a prefix of X or x_{j} > y_{j} where j is the smallest integer such that x_{j} \neq y_{j}.
Constraints
* 1 \leq |S| \leq 26
* S is a diverse word.
Input
Input is given from Standard Input in the following format:
S
Output
Print the next word that appears after S in the dictionary, or `-1` if it doesn't exist.
Examples
Input
atcoder
Output
atcoderb
Input
abc
Output
abcd
Input
zyxwvutsrqponmlkjihgfedcba
Output
-1
Input
abcdefghijklmnopqrstuvwzyx
Output
abcdefghijklmnopqrstuvx
Submitted Solution:
```
s = input(); a = []; b = [0]*27; t = ""
for i in range(len(s)): a.append(ord(s[i])); b[a[i]-97] = 1
if s == "zyxwvutsrqponmlkjihgfedcba": print(-1)
else:
for i in range(27):
if b[i] == 0: c = i; break
if c != 26:
a.append(c+97)
for i in range(len(a)): t += chr(a[i])
print(t)
else:
d = []
for _ in range(25):
if a[-2] > a[-1]: d.append(a.pop())
else:
d.append(a.pop()); a.pop(); d.sort()
for i in range(len(d)):
if a[-1] < d[i]: a.append(d[i]); break
break
for i in range(len(a)): t += chr(a[i])
print(t)
``` | instruction | 0 | 68,778 | 18 | 137,556 |
No | output | 1 | 68,778 | 18 | 137,557 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gotou just received a dictionary. However, he doesn't recognize the language used in the dictionary. He did some analysis on the dictionary and realizes that the dictionary contains all possible diverse words in lexicographical order.
A word is called diverse if and only if it is a nonempty string of English lowercase letters and all letters in the word are distinct. For example, `atcoder`, `zscoder` and `agc` are diverse words while `gotou` and `connect` aren't diverse words.
Given a diverse word S, determine the next word that appears after S in the dictionary, i.e. the lexicographically smallest diverse word that is lexicographically larger than S, or determine that it doesn't exist.
Let X = x_{1}x_{2}...x_{n} and Y = y_{1}y_{2}...y_{m} be two distinct strings. X is lexicographically larger than Y if and only if Y is a prefix of X or x_{j} > y_{j} where j is the smallest integer such that x_{j} \neq y_{j}.
Constraints
* 1 \leq |S| \leq 26
* S is a diverse word.
Input
Input is given from Standard Input in the following format:
S
Output
Print the next word that appears after S in the dictionary, or `-1` if it doesn't exist.
Examples
Input
atcoder
Output
atcoderb
Input
abc
Output
abcd
Input
zyxwvutsrqponmlkjihgfedcba
Output
-1
Input
abcdefghijklmnopqrstuvwzyx
Output
abcdefghijklmnopqrstuvx
Submitted Solution:
```
S=input()
A=[chr(i) for i in range(97, 97+26)]
N=len(S)
print(S)
if N<26:
d={}
for s in S:
d[s]=1
for a in A:
if a not in d:
S+=a
break
print(S)
else:
cond=True
d=[-1]
for i in range(N):
n=A.index(S[-i-1])
if n>max(d):
d.append(n)
else:
ans=S[:N-i-1]+A[n+1]
cond=False
break
if cond:
print(-1)
else:
print(ans)
``` | instruction | 0 | 68,779 | 18 | 137,558 |
No | output | 1 | 68,779 | 18 | 137,559 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gotou just received a dictionary. However, he doesn't recognize the language used in the dictionary. He did some analysis on the dictionary and realizes that the dictionary contains all possible diverse words in lexicographical order.
A word is called diverse if and only if it is a nonempty string of English lowercase letters and all letters in the word are distinct. For example, `atcoder`, `zscoder` and `agc` are diverse words while `gotou` and `connect` aren't diverse words.
Given a diverse word S, determine the next word that appears after S in the dictionary, i.e. the lexicographically smallest diverse word that is lexicographically larger than S, or determine that it doesn't exist.
Let X = x_{1}x_{2}...x_{n} and Y = y_{1}y_{2}...y_{m} be two distinct strings. X is lexicographically larger than Y if and only if Y is a prefix of X or x_{j} > y_{j} where j is the smallest integer such that x_{j} \neq y_{j}.
Constraints
* 1 \leq |S| \leq 26
* S is a diverse word.
Input
Input is given from Standard Input in the following format:
S
Output
Print the next word that appears after S in the dictionary, or `-1` if it doesn't exist.
Examples
Input
atcoder
Output
atcoderb
Input
abc
Output
abcd
Input
zyxwvutsrqponmlkjihgfedcba
Output
-1
Input
abcdefghijklmnopqrstuvwzyx
Output
abcdefghijklmnopqrstuvx
Submitted Solution:
```
str = input()
sortstr=sorted(str)
alpha = [chr(ord('a') + i) for i in range(26)]
for i in str:
if i in alpha:
alpha.remove(i)
if len(str)!=len(set(str)) or str=="zyxwvutsrqponmlkjihgfedcba":
print(-1)
elif list(str)==sorted(str):#sorted(str) is list
print(str+alpha[0])
elif len(str)==26:
alpha2=[chr(ord('a') + i) for i in range(26)]
for i in range(len(str)):
if str[i]==sortstr[i]:
continue
alpha2.remove(str[i])
else:
print(str[i-1]+alpha[0])
break
else:
print(str+alpha[0])
``` | instruction | 0 | 68,780 | 18 | 137,560 |
No | output | 1 | 68,780 | 18 | 137,561 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given a string consisting of only numbers from 0 to 9, consider the operation of creating a new string from that string according to the following rules. Read the given string one character at a time from the left end. Go, if the same number a continues r, write the number r and the number a in this order without separating them with a space. Read to the right end of the given character string, and what is on the way to the end of the last writing Even if there are times of writing, all of them are counted as one operation. For the second and subsequent operations, the same operation is performed with the character string written out by the previous operation as the given character string. For example, "122244" If the character string "" is given, the character string obtained by one operation is "113224" and "44444444444" (11) because one 1, three 2, two 4s are given in order from the left end. In the case of 4), the obtained character string is “114”.
Create a program that outputs a string obtained by performing the above operation n times on a given string of 100 characters or less, where n ≤ 20.
The input data consists of two lines, the first line contains the number of operations n, and the second line contains the first character string.
Input example
---
Five
11
Output example
13112221
input
The input consists of multiple datasets. Input ends when n is 0. The number of datasets does not exceed 5.
output
For each data set, the character string that has been operated the specified number of times is output on one line.
Example
Input
5
11
5
11
0
Output
13112221
13112221
Submitted Solution:
```
while True:
n = int(input())
if n==0:
break
t = input()
for i in range(n):
s = list(t)
S = []
for i in range(len(s)):
S.append(int(s[i]))
i = 1
c = 0
r = []
a = [S[0]]
while True:
if S[i-1]==S[i]:
c += 1
else:
r.append(c+1)
a.append(S[i])
c = 0
i += 1
if i==len(S):
r.append(c+1)
break
S.clear()
for j in range(len(r)):
S.append(str(r[j]))
S.append(str(a[j]))
t = ''.join(S)
print(t)
``` | instruction | 0 | 68,861 | 18 | 137,722 |
Yes | output | 1 | 68,861 | 18 | 137,723 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given a string consisting of only numbers from 0 to 9, consider the operation of creating a new string from that string according to the following rules. Read the given string one character at a time from the left end. Go, if the same number a continues r, write the number r and the number a in this order without separating them with a space. Read to the right end of the given character string, and what is on the way to the end of the last writing Even if there are times of writing, all of them are counted as one operation. For the second and subsequent operations, the same operation is performed with the character string written out by the previous operation as the given character string. For example, "122244" If the character string "" is given, the character string obtained by one operation is "113224" and "44444444444" (11) because one 1, three 2, two 4s are given in order from the left end. In the case of 4), the obtained character string is “114”.
Create a program that outputs a string obtained by performing the above operation n times on a given string of 100 characters or less, where n ≤ 20.
The input data consists of two lines, the first line contains the number of operations n, and the second line contains the first character string.
Input example
---
Five
11
Output example
13112221
input
The input consists of multiple datasets. Input ends when n is 0. The number of datasets does not exceed 5.
output
For each data set, the character string that has been operated the specified number of times is output on one line.
Example
Input
5
11
5
11
0
Output
13112221
13112221
Submitted Solution:
```
# coding: utf-8
def solve(li):
cnt = 1
ans = []
for i in range(len(li)-1):
tar = int(li[i])
if tar==int(li[i+1]):
if cnt>=10:
ans.append(cnt//10)
cnt = cnt%10
cnt+=1
else:
if cnt>=10:
ans.append(cnt//10)
ans.append(cnt%10)
else:
ans.append(cnt)
ans.append(tar)
cnt = 1
tar = int(li[i+1])
ans.append(cnt)
ans.append(tar)
return ans
while True:
n = int(input())
if n ==0: break
a = input()
la = list(a)
i=0
while i<n:
la = solve(la)
i+=1
for x in la:
print(x, end='')
print()
``` | instruction | 0 | 68,862 | 18 | 137,724 |
Yes | output | 1 | 68,862 | 18 | 137,725 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given a string consisting of only numbers from 0 to 9, consider the operation of creating a new string from that string according to the following rules. Read the given string one character at a time from the left end. Go, if the same number a continues r, write the number r and the number a in this order without separating them with a space. Read to the right end of the given character string, and what is on the way to the end of the last writing Even if there are times of writing, all of them are counted as one operation. For the second and subsequent operations, the same operation is performed with the character string written out by the previous operation as the given character string. For example, "122244" If the character string "" is given, the character string obtained by one operation is "113224" and "44444444444" (11) because one 1, three 2, two 4s are given in order from the left end. In the case of 4), the obtained character string is “114”.
Create a program that outputs a string obtained by performing the above operation n times on a given string of 100 characters or less, where n ≤ 20.
The input data consists of two lines, the first line contains the number of operations n, and the second line contains the first character string.
Input example
---
Five
11
Output example
13112221
input
The input consists of multiple datasets. Input ends when n is 0. The number of datasets does not exceed 5.
output
For each data set, the character string that has been operated the specified number of times is output on one line.
Example
Input
5
11
5
11
0
Output
13112221
13112221
Submitted Solution:
```
def convert(s):
convertStr = ""
l = len(s)
i = 0
while i < l:
if i + 1 < l:
word = s[i]
sequenceNum = 1
j = i + 1
while j < l:
if(word == s[j]):
sequenceNum += 1
else:
break
j += 1
convertStr += str(sequenceNum) + word
i = i + sequenceNum - 1
else:
convertStr += "1" + str(s[i])
i += 1
return convertStr
while True:
try:
n = int(input())
s = input()
for i in range(0,n):
s = convert(s)
print(s)
except:
break
``` | instruction | 0 | 68,863 | 18 | 137,726 |
Yes | output | 1 | 68,863 | 18 | 137,727 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given a string consisting of only numbers from 0 to 9, consider the operation of creating a new string from that string according to the following rules. Read the given string one character at a time from the left end. Go, if the same number a continues r, write the number r and the number a in this order without separating them with a space. Read to the right end of the given character string, and what is on the way to the end of the last writing Even if there are times of writing, all of them are counted as one operation. For the second and subsequent operations, the same operation is performed with the character string written out by the previous operation as the given character string. For example, "122244" If the character string "" is given, the character string obtained by one operation is "113224" and "44444444444" (11) because one 1, three 2, two 4s are given in order from the left end. In the case of 4), the obtained character string is “114”.
Create a program that outputs a string obtained by performing the above operation n times on a given string of 100 characters or less, where n ≤ 20.
The input data consists of two lines, the first line contains the number of operations n, and the second line contains the first character string.
Input example
---
Five
11
Output example
13112221
input
The input consists of multiple datasets. Input ends when n is 0. The number of datasets does not exceed 5.
output
For each data set, the character string that has been operated the specified number of times is output on one line.
Example
Input
5
11
5
11
0
Output
13112221
13112221
Submitted Solution:
```
def change(x) :
i = 0
ans = []
while True :
if i >= len(x) : break
cnt = 1
j = 1
while True :
if i+j >= len(x) or x[i+j] != x[i] :
ans = ans + list(str(cnt))
ans.append(x[i])
i += cnt
break
else :
cnt += 1
j += 1
return ans
while True :
n = int(input())
if n == 0 :
break
lst = list(input())
for i in range(n) :
lst = change(lst)
print(*lst, sep='')
``` | instruction | 0 | 68,864 | 18 | 137,728 |
Yes | output | 1 | 68,864 | 18 | 137,729 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given a string consisting of only numbers from 0 to 9, consider the operation of creating a new string from that string according to the following rules. Read the given string one character at a time from the left end. Go, if the same number a continues r, write the number r and the number a in this order without separating them with a space. Read to the right end of the given character string, and what is on the way to the end of the last writing Even if there are times of writing, all of them are counted as one operation. For the second and subsequent operations, the same operation is performed with the character string written out by the previous operation as the given character string. For example, "122244" If the character string "" is given, the character string obtained by one operation is "113224" and "44444444444" (11) because one 1, three 2, two 4s are given in order from the left end. In the case of 4), the obtained character string is “114”.
Create a program that outputs a string obtained by performing the above operation n times on a given string of 100 characters or less, where n ≤ 20.
The input data consists of two lines, the first line contains the number of operations n, and the second line contains the first character string.
Input example
---
Five
11
Output example
13112221
input
The input consists of multiple datasets. Input ends when n is 0. The number of datasets does not exceed 5.
output
For each data set, the character string that has been operated the specified number of times is output on one line.
Example
Input
5
11
5
11
0
Output
13112221
13112221
Submitted Solution:
```
while True:
n = int(input())
if not n:
break
ss = [s for s in input()[::-1]]
for i in range(n):
new = []
app = new.append
last = ss.pop()
count = 1
while ss:
a = ss.pop()
if a == last:
count += 1
else:
app(str(count))
app(last)
last = a
count = 1
app(str(count))
app(last)
ss = [s for s in new[::-1]]
print("".join(new))
``` | instruction | 0 | 68,865 | 18 | 137,730 |
No | output | 1 | 68,865 | 18 | 137,731 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given a string consisting of only numbers from 0 to 9, consider the operation of creating a new string from that string according to the following rules. Read the given string one character at a time from the left end. Go, if the same number a continues r, write the number r and the number a in this order without separating them with a space. Read to the right end of the given character string, and what is on the way to the end of the last writing Even if there are times of writing, all of them are counted as one operation. For the second and subsequent operations, the same operation is performed with the character string written out by the previous operation as the given character string. For example, "122244" If the character string "" is given, the character string obtained by one operation is "113224" and "44444444444" (11) because one 1, three 2, two 4s are given in order from the left end. In the case of 4), the obtained character string is “114”.
Create a program that outputs a string obtained by performing the above operation n times on a given string of 100 characters or less, where n ≤ 20.
The input data consists of two lines, the first line contains the number of operations n, and the second line contains the first character string.
Input example
---
Five
11
Output example
13112221
input
The input consists of multiple datasets. Input ends when n is 0. The number of datasets does not exceed 5.
output
For each data set, the character string that has been operated the specified number of times is output on one line.
Example
Input
5
11
5
11
0
Output
13112221
13112221
Submitted Solution:
```
# coding: utf-8
def solve(li):
cnt = 1
ans = []
for i in range(len(li)-1):
tar = int(li[i])
if tar==int(li[i+1]):
cnt+=1
else:
ans.append(cnt)
ans.append(tar)
cnt = 1
tar = int(li[i+1])
ans.append(cnt)
ans.append(tar)
return ans
while True:
n = int(input())
if n ==0: break
a = input()
la = list(a)
i=0
while i<n:
la = solve(la)
i+=1
for x in la:
print(x, end='')
print()
print()
``` | instruction | 0 | 68,866 | 18 | 137,732 |
No | output | 1 | 68,866 | 18 | 137,733 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given a string consisting of only numbers from 0 to 9, consider the operation of creating a new string from that string according to the following rules. Read the given string one character at a time from the left end. Go, if the same number a continues r, write the number r and the number a in this order without separating them with a space. Read to the right end of the given character string, and what is on the way to the end of the last writing Even if there are times of writing, all of them are counted as one operation. For the second and subsequent operations, the same operation is performed with the character string written out by the previous operation as the given character string. For example, "122244" If the character string "" is given, the character string obtained by one operation is "113224" and "44444444444" (11) because one 1, three 2, two 4s are given in order from the left end. In the case of 4), the obtained character string is “114”.
Create a program that outputs a string obtained by performing the above operation n times on a given string of 100 characters or less, where n ≤ 20.
The input data consists of two lines, the first line contains the number of operations n, and the second line contains the first character string.
Input example
---
Five
11
Output example
13112221
input
The input consists of multiple datasets. Input ends when n is 0. The number of datasets does not exceed 5.
output
For each data set, the character string that has been operated the specified number of times is output on one line.
Example
Input
5
11
5
11
0
Output
13112221
13112221
Submitted Solution:
```
# coding: utf-8
def solve(li):
cnt = 1
ans = []
for i in range(len(li)-1):
tar = int(li[i])
if tar==int(li[i+1]):
if cnt>=10:
ans.append(cnt//10)
cnt = cnt%10
cnt+=1
else:
if cnt>=10:
print('a')
ans.append(cnt//10)
ans.append(cnt%10)
else:
print('b')
ans.append(cnt)
ans.append(tar)
cnt = 1
tar = int(li[i+1])
ans.append(cnt)
ans.append(tar)
return ans
while True:
n = int(input())
if n ==0: break
a = input()
la = list(a)
i=0
while i<n:
la = solve(la)
i+=1
for x in la:
print(x, end='')
print()
``` | instruction | 0 | 68,868 | 18 | 137,736 |
No | output | 1 | 68,868 | 18 | 137,737 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Limak is a little polar bear. Polar bears hate long strings and thus they like to compress them. You should also know that Limak is so young that he knows only first six letters of the English alphabet: 'a', 'b', 'c', 'd', 'e' and 'f'.
You are given a set of q possible operations. Limak can perform them in any order, any operation may be applied any number of times. The i-th operation is described by a string ai of length two and a string bi of length one. No two of q possible operations have the same string ai.
When Limak has a string s he can perform the i-th operation on s if the first two letters of s match a two-letter string ai. Performing the i-th operation removes first two letters of s and inserts there a string bi. See the notes section for further clarification.
You may note that performing an operation decreases the length of a string s exactly by 1. Also, for some sets of operations there may be a string that cannot be compressed any further, because the first two letters don't match any ai.
Limak wants to start with a string of length n and perform n - 1 operations to finally get a one-letter string "a". In how many ways can he choose the starting string to be able to get "a"? Remember that Limak can use only letters he knows.
Input
The first line contains two integers n and q (2 ≤ n ≤ 6, 1 ≤ q ≤ 36) — the length of the initial string and the number of available operations.
The next q lines describe the possible operations. The i-th of them contains two strings ai and bi (|ai| = 2, |bi| = 1). It's guaranteed that ai ≠ aj for i ≠ j and that all ai and bi consist of only first six lowercase English letters.
Output
Print the number of strings of length n that Limak will be able to transform to string "a" by applying only operations given in the input.
Examples
Input
3 5
ab a
cc c
ca a
ee c
ff d
Output
4
Input
2 8
af e
dc d
cc f
bc b
da b
eb a
bb b
ff c
Output
1
Input
6 2
bb a
ba a
Output
0
Note
In the first sample, we count initial strings of length 3 from which Limak can get a required string "a". There are 4 such strings: "abb", "cab", "cca", "eea". The first one Limak can compress using operation 1 two times (changing "ab" to a single "a"). The first operation would change "abb" to "ab" and the second operation would change "ab" to "a".
Other three strings may be compressed as follows:
* "cab" <image> "ab" <image> "a"
* "cca" <image> "ca" <image> "a"
* "eea" <image> "ca" <image> "a"
In the second sample, the only correct initial string is "eb" because it can be immediately compressed to "a".
Submitted Solution:
```
from collections import defaultdict
from itertools import product
def generate_strings(n):
for s in product("abcdef", repeat = n):
yield ''.join(s)
def compress(s, d):
while len(s) > 1:
p = s[:2]
if p not in d:
break
s = d[p] + s[2:]
return s
def solve(n, d):
return sum(compress(s, d) == "a" for s in generate_strings(n))
n, q = map(int, input().split())
d = defaultdict(str)
for i in range(q):
s1, s2 = input().split()
d[s1] = s2
print(solve(n, d))
``` | instruction | 0 | 70,300 | 18 | 140,600 |
Yes | output | 1 | 70,300 | 18 | 140,601 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Limak is a little polar bear. Polar bears hate long strings and thus they like to compress them. You should also know that Limak is so young that he knows only first six letters of the English alphabet: 'a', 'b', 'c', 'd', 'e' and 'f'.
You are given a set of q possible operations. Limak can perform them in any order, any operation may be applied any number of times. The i-th operation is described by a string ai of length two and a string bi of length one. No two of q possible operations have the same string ai.
When Limak has a string s he can perform the i-th operation on s if the first two letters of s match a two-letter string ai. Performing the i-th operation removes first two letters of s and inserts there a string bi. See the notes section for further clarification.
You may note that performing an operation decreases the length of a string s exactly by 1. Also, for some sets of operations there may be a string that cannot be compressed any further, because the first two letters don't match any ai.
Limak wants to start with a string of length n and perform n - 1 operations to finally get a one-letter string "a". In how many ways can he choose the starting string to be able to get "a"? Remember that Limak can use only letters he knows.
Input
The first line contains two integers n and q (2 ≤ n ≤ 6, 1 ≤ q ≤ 36) — the length of the initial string and the number of available operations.
The next q lines describe the possible operations. The i-th of them contains two strings ai and bi (|ai| = 2, |bi| = 1). It's guaranteed that ai ≠ aj for i ≠ j and that all ai and bi consist of only first six lowercase English letters.
Output
Print the number of strings of length n that Limak will be able to transform to string "a" by applying only operations given in the input.
Examples
Input
3 5
ab a
cc c
ca a
ee c
ff d
Output
4
Input
2 8
af e
dc d
cc f
bc b
da b
eb a
bb b
ff c
Output
1
Input
6 2
bb a
ba a
Output
0
Note
In the first sample, we count initial strings of length 3 from which Limak can get a required string "a". There are 4 such strings: "abb", "cab", "cca", "eea". The first one Limak can compress using operation 1 two times (changing "ab" to a single "a"). The first operation would change "abb" to "ab" and the second operation would change "ab" to "a".
Other three strings may be compressed as follows:
* "cab" <image> "ab" <image> "a"
* "cca" <image> "ca" <image> "a"
* "eea" <image> "ca" <image> "a"
In the second sample, the only correct initial string is "eb" because it can be immediately compressed to "a".
Submitted Solution:
```
n, m = map(int, input().split())
d = {}
c ={}
def calc(ch, cnt):
if cnt == n:
if ch not in d:
return 0
return d[ch]
ans = 0;
if ch not in c:
return 0
for i in c[ch]:
ans += calc(i, cnt + 1)
return ans
for i in range(0,m):
a,b = input().split()
if b in d:
d[b]+=1
else:
d[b] = 1
if b in c:
c[b].append(a[0])
else:
c[b] = [a[0]]
print(calc('a',2))
``` | instruction | 0 | 70,301 | 18 | 140,602 |
Yes | output | 1 | 70,301 | 18 | 140,603 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Limak is a little polar bear. Polar bears hate long strings and thus they like to compress them. You should also know that Limak is so young that he knows only first six letters of the English alphabet: 'a', 'b', 'c', 'd', 'e' and 'f'.
You are given a set of q possible operations. Limak can perform them in any order, any operation may be applied any number of times. The i-th operation is described by a string ai of length two and a string bi of length one. No two of q possible operations have the same string ai.
When Limak has a string s he can perform the i-th operation on s if the first two letters of s match a two-letter string ai. Performing the i-th operation removes first two letters of s and inserts there a string bi. See the notes section for further clarification.
You may note that performing an operation decreases the length of a string s exactly by 1. Also, for some sets of operations there may be a string that cannot be compressed any further, because the first two letters don't match any ai.
Limak wants to start with a string of length n and perform n - 1 operations to finally get a one-letter string "a". In how many ways can he choose the starting string to be able to get "a"? Remember that Limak can use only letters he knows.
Input
The first line contains two integers n and q (2 ≤ n ≤ 6, 1 ≤ q ≤ 36) — the length of the initial string and the number of available operations.
The next q lines describe the possible operations. The i-th of them contains two strings ai and bi (|ai| = 2, |bi| = 1). It's guaranteed that ai ≠ aj for i ≠ j and that all ai and bi consist of only first six lowercase English letters.
Output
Print the number of strings of length n that Limak will be able to transform to string "a" by applying only operations given in the input.
Examples
Input
3 5
ab a
cc c
ca a
ee c
ff d
Output
4
Input
2 8
af e
dc d
cc f
bc b
da b
eb a
bb b
ff c
Output
1
Input
6 2
bb a
ba a
Output
0
Note
In the first sample, we count initial strings of length 3 from which Limak can get a required string "a". There are 4 such strings: "abb", "cab", "cca", "eea". The first one Limak can compress using operation 1 two times (changing "ab" to a single "a"). The first operation would change "abb" to "ab" and the second operation would change "ab" to "a".
Other three strings may be compressed as follows:
* "cab" <image> "ab" <image> "a"
* "cca" <image> "ca" <image> "a"
* "eea" <image> "ca" <image> "a"
In the second sample, the only correct initial string is "eb" because it can be immediately compressed to "a".
Submitted Solution:
```
n, q = map(int, input().split())
lexicon = {}
for _ in range(q):
ai, bi = input().split()
if bi in lexicon:
lexicon[bi].append(ai)
else:
lexicon[bi] = [ai]
memo = {}
def solve(s):
if s in memo:
return memo[s]
else:
count = 0
if len(s) == n:
count = 1
else:
if s[0] in lexicon:
for ss in lexicon[s[0]]:
count += solve(ss + s[1:])
memo[s] = count
return count
print(solve('a'))
``` | instruction | 0 | 70,302 | 18 | 140,604 |
Yes | output | 1 | 70,302 | 18 | 140,605 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Limak is a little polar bear. Polar bears hate long strings and thus they like to compress them. You should also know that Limak is so young that he knows only first six letters of the English alphabet: 'a', 'b', 'c', 'd', 'e' and 'f'.
You are given a set of q possible operations. Limak can perform them in any order, any operation may be applied any number of times. The i-th operation is described by a string ai of length two and a string bi of length one. No two of q possible operations have the same string ai.
When Limak has a string s he can perform the i-th operation on s if the first two letters of s match a two-letter string ai. Performing the i-th operation removes first two letters of s and inserts there a string bi. See the notes section for further clarification.
You may note that performing an operation decreases the length of a string s exactly by 1. Also, for some sets of operations there may be a string that cannot be compressed any further, because the first two letters don't match any ai.
Limak wants to start with a string of length n and perform n - 1 operations to finally get a one-letter string "a". In how many ways can he choose the starting string to be able to get "a"? Remember that Limak can use only letters he knows.
Input
The first line contains two integers n and q (2 ≤ n ≤ 6, 1 ≤ q ≤ 36) — the length of the initial string and the number of available operations.
The next q lines describe the possible operations. The i-th of them contains two strings ai and bi (|ai| = 2, |bi| = 1). It's guaranteed that ai ≠ aj for i ≠ j and that all ai and bi consist of only first six lowercase English letters.
Output
Print the number of strings of length n that Limak will be able to transform to string "a" by applying only operations given in the input.
Examples
Input
3 5
ab a
cc c
ca a
ee c
ff d
Output
4
Input
2 8
af e
dc d
cc f
bc b
da b
eb a
bb b
ff c
Output
1
Input
6 2
bb a
ba a
Output
0
Note
In the first sample, we count initial strings of length 3 from which Limak can get a required string "a". There are 4 such strings: "abb", "cab", "cca", "eea". The first one Limak can compress using operation 1 two times (changing "ab" to a single "a"). The first operation would change "abb" to "ab" and the second operation would change "ab" to "a".
Other three strings may be compressed as follows:
* "cab" <image> "ab" <image> "a"
* "cca" <image> "ca" <image> "a"
* "eea" <image> "ca" <image> "a"
In the second sample, the only correct initial string is "eb" because it can be immediately compressed to "a".
Submitted Solution:
```
links=[[],[],[],[],[],[]]
n,q=map(int,input().split())
for _ in range(q):
a,b=input().split()
links[ord(b)-ord('a')].append(a[0])
def func(b,depth):
if depth==n:
return 1
count=0
for a in links[ord(b)-ord('a')]:
count+=func(a,depth+1)
return count
print(func('a',1))
``` | instruction | 0 | 70,303 | 18 | 140,606 |
Yes | output | 1 | 70,303 | 18 | 140,607 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Limak is a little polar bear. Polar bears hate long strings and thus they like to compress them. You should also know that Limak is so young that he knows only first six letters of the English alphabet: 'a', 'b', 'c', 'd', 'e' and 'f'.
You are given a set of q possible operations. Limak can perform them in any order, any operation may be applied any number of times. The i-th operation is described by a string ai of length two and a string bi of length one. No two of q possible operations have the same string ai.
When Limak has a string s he can perform the i-th operation on s if the first two letters of s match a two-letter string ai. Performing the i-th operation removes first two letters of s and inserts there a string bi. See the notes section for further clarification.
You may note that performing an operation decreases the length of a string s exactly by 1. Also, for some sets of operations there may be a string that cannot be compressed any further, because the first two letters don't match any ai.
Limak wants to start with a string of length n and perform n - 1 operations to finally get a one-letter string "a". In how many ways can he choose the starting string to be able to get "a"? Remember that Limak can use only letters he knows.
Input
The first line contains two integers n and q (2 ≤ n ≤ 6, 1 ≤ q ≤ 36) — the length of the initial string and the number of available operations.
The next q lines describe the possible operations. The i-th of them contains two strings ai and bi (|ai| = 2, |bi| = 1). It's guaranteed that ai ≠ aj for i ≠ j and that all ai and bi consist of only first six lowercase English letters.
Output
Print the number of strings of length n that Limak will be able to transform to string "a" by applying only operations given in the input.
Examples
Input
3 5
ab a
cc c
ca a
ee c
ff d
Output
4
Input
2 8
af e
dc d
cc f
bc b
da b
eb a
bb b
ff c
Output
1
Input
6 2
bb a
ba a
Output
0
Note
In the first sample, we count initial strings of length 3 from which Limak can get a required string "a". There are 4 such strings: "abb", "cab", "cca", "eea". The first one Limak can compress using operation 1 two times (changing "ab" to a single "a"). The first operation would change "abb" to "ab" and the second operation would change "ab" to "a".
Other three strings may be compressed as follows:
* "cab" <image> "ab" <image> "a"
* "cca" <image> "ca" <image> "a"
* "eea" <image> "ca" <image> "a"
In the second sample, the only correct initial string is "eb" because it can be immediately compressed to "a".
Submitted Solution:
```
from collections import defaultdict
n, q = map(int, input().split())
es = defaultdict(list)
for _ in range(q):
a, b = input().split()
es[b].append(a)
ss = set('a')
for i in range(n):
ssp = set()
for s in ss:
for e in es[s[0]]:
ssp.add(e + s[1:])
ss = ssp
if not ss:
break
print(len(ss))
``` | instruction | 0 | 70,304 | 18 | 140,608 |
No | output | 1 | 70,304 | 18 | 140,609 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Limak is a little polar bear. Polar bears hate long strings and thus they like to compress them. You should also know that Limak is so young that he knows only first six letters of the English alphabet: 'a', 'b', 'c', 'd', 'e' and 'f'.
You are given a set of q possible operations. Limak can perform them in any order, any operation may be applied any number of times. The i-th operation is described by a string ai of length two and a string bi of length one. No two of q possible operations have the same string ai.
When Limak has a string s he can perform the i-th operation on s if the first two letters of s match a two-letter string ai. Performing the i-th operation removes first two letters of s and inserts there a string bi. See the notes section for further clarification.
You may note that performing an operation decreases the length of a string s exactly by 1. Also, for some sets of operations there may be a string that cannot be compressed any further, because the first two letters don't match any ai.
Limak wants to start with a string of length n and perform n - 1 operations to finally get a one-letter string "a". In how many ways can he choose the starting string to be able to get "a"? Remember that Limak can use only letters he knows.
Input
The first line contains two integers n and q (2 ≤ n ≤ 6, 1 ≤ q ≤ 36) — the length of the initial string and the number of available operations.
The next q lines describe the possible operations. The i-th of them contains two strings ai and bi (|ai| = 2, |bi| = 1). It's guaranteed that ai ≠ aj for i ≠ j and that all ai and bi consist of only first six lowercase English letters.
Output
Print the number of strings of length n that Limak will be able to transform to string "a" by applying only operations given in the input.
Examples
Input
3 5
ab a
cc c
ca a
ee c
ff d
Output
4
Input
2 8
af e
dc d
cc f
bc b
da b
eb a
bb b
ff c
Output
1
Input
6 2
bb a
ba a
Output
0
Note
In the first sample, we count initial strings of length 3 from which Limak can get a required string "a". There are 4 such strings: "abb", "cab", "cca", "eea". The first one Limak can compress using operation 1 two times (changing "ab" to a single "a"). The first operation would change "abb" to "ab" and the second operation would change "ab" to "a".
Other three strings may be compressed as follows:
* "cab" <image> "ab" <image> "a"
* "cca" <image> "ca" <image> "a"
* "eea" <image> "ca" <image> "a"
In the second sample, the only correct initial string is "eb" because it can be immediately compressed to "a".
Submitted Solution:
```
n,k = map(int,input().split())
l = {}
cnt = 0
for i in range(k):
a,b = input().split()
l[a] = b;
if n==2:
for i in l:
if i[1]=='a':
cnt+=1
print(cnt)
exit(0)
def allcombs(s,prefix,l):
global cnt
if l==0:
if (check(prefix)):
cnt+=1
return
for i in s:
allcombs(s,prefix+i,l-1)
def check(s):
while len(s)>1:
yes = False
if s[:2] in l:
s = l[s[:2]] + s[2:]
yes = True
if not yes:
return False
if s=='a':
return True
else:
return False
allcombs(['a','b','c','d','e','f'],"",n)
print(cnt)
``` | instruction | 0 | 70,305 | 18 | 140,610 |
No | output | 1 | 70,305 | 18 | 140,611 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Limak is a little polar bear. Polar bears hate long strings and thus they like to compress them. You should also know that Limak is so young that he knows only first six letters of the English alphabet: 'a', 'b', 'c', 'd', 'e' and 'f'.
You are given a set of q possible operations. Limak can perform them in any order, any operation may be applied any number of times. The i-th operation is described by a string ai of length two and a string bi of length one. No two of q possible operations have the same string ai.
When Limak has a string s he can perform the i-th operation on s if the first two letters of s match a two-letter string ai. Performing the i-th operation removes first two letters of s and inserts there a string bi. See the notes section for further clarification.
You may note that performing an operation decreases the length of a string s exactly by 1. Also, for some sets of operations there may be a string that cannot be compressed any further, because the first two letters don't match any ai.
Limak wants to start with a string of length n and perform n - 1 operations to finally get a one-letter string "a". In how many ways can he choose the starting string to be able to get "a"? Remember that Limak can use only letters he knows.
Input
The first line contains two integers n and q (2 ≤ n ≤ 6, 1 ≤ q ≤ 36) — the length of the initial string and the number of available operations.
The next q lines describe the possible operations. The i-th of them contains two strings ai and bi (|ai| = 2, |bi| = 1). It's guaranteed that ai ≠ aj for i ≠ j and that all ai and bi consist of only first six lowercase English letters.
Output
Print the number of strings of length n that Limak will be able to transform to string "a" by applying only operations given in the input.
Examples
Input
3 5
ab a
cc c
ca a
ee c
ff d
Output
4
Input
2 8
af e
dc d
cc f
bc b
da b
eb a
bb b
ff c
Output
1
Input
6 2
bb a
ba a
Output
0
Note
In the first sample, we count initial strings of length 3 from which Limak can get a required string "a". There are 4 such strings: "abb", "cab", "cca", "eea". The first one Limak can compress using operation 1 two times (changing "ab" to a single "a"). The first operation would change "abb" to "ab" and the second operation would change "ab" to "a".
Other three strings may be compressed as follows:
* "cab" <image> "ab" <image> "a"
* "cca" <image> "ca" <image> "a"
* "eea" <image> "ca" <image> "a"
In the second sample, the only correct initial string is "eb" because it can be immediately compressed to "a".
Submitted Solution:
```
def deep(ch, curn):
s = 0
for i in d[ch]:
if i[0] in 'abcdef' and i[1] in 'abcdef':
if curn and i[0] in d:
s += deep(i[0], curn-1)
elif i[0] in d:
s += 1
return s
n, m = map(int, input().split())
d = {}
for i in range(m):
dec, en = input().split()
if en in d:
d[en].append(dec)
else:
d[en] = [dec]
if n == 1:
print(1)
elif 'a' in d:
print(deep('a', n-1))
else:
print(0)
``` | instruction | 0 | 70,306 | 18 | 140,612 |
No | output | 1 | 70,306 | 18 | 140,613 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Limak is a little polar bear. Polar bears hate long strings and thus they like to compress them. You should also know that Limak is so young that he knows only first six letters of the English alphabet: 'a', 'b', 'c', 'd', 'e' and 'f'.
You are given a set of q possible operations. Limak can perform them in any order, any operation may be applied any number of times. The i-th operation is described by a string ai of length two and a string bi of length one. No two of q possible operations have the same string ai.
When Limak has a string s he can perform the i-th operation on s if the first two letters of s match a two-letter string ai. Performing the i-th operation removes first two letters of s and inserts there a string bi. See the notes section for further clarification.
You may note that performing an operation decreases the length of a string s exactly by 1. Also, for some sets of operations there may be a string that cannot be compressed any further, because the first two letters don't match any ai.
Limak wants to start with a string of length n and perform n - 1 operations to finally get a one-letter string "a". In how many ways can he choose the starting string to be able to get "a"? Remember that Limak can use only letters he knows.
Input
The first line contains two integers n and q (2 ≤ n ≤ 6, 1 ≤ q ≤ 36) — the length of the initial string and the number of available operations.
The next q lines describe the possible operations. The i-th of them contains two strings ai and bi (|ai| = 2, |bi| = 1). It's guaranteed that ai ≠ aj for i ≠ j and that all ai and bi consist of only first six lowercase English letters.
Output
Print the number of strings of length n that Limak will be able to transform to string "a" by applying only operations given in the input.
Examples
Input
3 5
ab a
cc c
ca a
ee c
ff d
Output
4
Input
2 8
af e
dc d
cc f
bc b
da b
eb a
bb b
ff c
Output
1
Input
6 2
bb a
ba a
Output
0
Note
In the first sample, we count initial strings of length 3 from which Limak can get a required string "a". There are 4 such strings: "abb", "cab", "cca", "eea". The first one Limak can compress using operation 1 two times (changing "ab" to a single "a"). The first operation would change "abb" to "ab" and the second operation would change "ab" to "a".
Other three strings may be compressed as follows:
* "cab" <image> "ab" <image> "a"
* "cca" <image> "ca" <image> "a"
* "eea" <image> "ca" <image> "a"
In the second sample, the only correct initial string is "eb" because it can be immediately compressed to "a".
Submitted Solution:
```
temp = input().split(" ")
maxLength, initialCount = [int(x) for x in temp]
resultSet = {1: [], 2: []}
dataSet = dict()
for i in range(initialCount):
temp = input().split(" ")
a, b = [x for x in temp]
dataSet[a] = b
resultSet[1] = b
resultSet[2] = a
def populate(toValue):
fromValues = []
if len(toValue) == 1:
for key, value in dataSet.items():
if value == toValue and not key in fromValues:
fromValues.append(key)
else:
process = toValue[:1]
tail = toValue[1:]
firstPart = populate(process)
if len(firstPart) > 0:
fromValues = fromValues + [x + tail for x in firstPart if (not x + tail in fromValues and (not x + tail in dataSet.keys()))]
return fromValues
count = 0
currents = ["a"]
currentCount = 1
while currentCount < maxLength:
buffer = []
print(currents)
for current in currents:
result = populate(current)
for key in result:
dataSet[key] = current
buffer = buffer + result
if len(buffer) > 0:
currentCount += 1
currents = buffer
if currentCount == maxLength:
count = len(buffer)
else:
break
print(count)
``` | instruction | 0 | 70,307 | 18 | 140,614 |
No | output | 1 | 70,307 | 18 | 140,615 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Modern text editors usually show some information regarding the document being edited. For example, the number of words, the number of pages, or the number of characters.
In this problem you should implement the similar functionality.
You are given a string which only consists of:
* uppercase and lowercase English letters,
* underscore symbols (they are used as separators),
* parentheses (both opening and closing).
It is guaranteed that each opening parenthesis has a succeeding closing parenthesis. Similarly, each closing parentheses has a preceding opening parentheses matching it. For each pair of matching parentheses there are no other parenthesis between them. In other words, each parenthesis in the string belongs to a matching "opening-closing" pair, and such pairs can't be nested.
For example, the following string is valid: "_Hello_Vasya(and_Petya)__bye_(and_OK)".
Word is a maximal sequence of consecutive letters, i.e. such sequence that the first character to the left and the first character to the right of it is an underscore, a parenthesis, or it just does not exist. For example, the string above consists of seven words: "Hello", "Vasya", "and", "Petya", "bye", "and" and "OK". Write a program that finds:
* the length of the longest word outside the parentheses (print 0, if there is no word outside the parentheses),
* the number of words inside the parentheses (print 0, if there is no word inside the parentheses).
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 255) — the length of the given string. The second line contains the string consisting of only lowercase and uppercase English letters, parentheses and underscore symbols.
Output
Print two space-separated integers:
* the length of the longest word outside the parentheses (print 0, if there is no word outside the parentheses),
* the number of words inside the parentheses (print 0, if there is no word inside the parentheses).
Examples
Input
37
_Hello_Vasya(and_Petya)__bye_(and_OK)
Output
5 4
Input
37
_a_(_b___c)__de_f(g_)__h__i(j_k_l)m__
Output
2 6
Input
27
(LoooonG)__shOrt__(LoooonG)
Output
5 2
Input
5
(___)
Output
0 0
Note
In the first sample, the words "Hello", "Vasya" and "bye" are outside any of the parentheses, and the words "and", "Petya", "and" and "OK" are inside. Note, that the word "and" is given twice and you should count it twice in the answer.
Submitted Solution:
```
_, s, x = input(), list(input()), 0
while '(' in s:
x += len([i for i in ''.join(s[s.index('(') + 1 : s.index(')')]).split('_') if i])
s[s.index('('): s.index(')') + 1] = '_'
print(max([len(i) if i else 0 for i in ''.join(s).split('_')]), x)
``` | instruction | 0 | 70,336 | 18 | 140,672 |
Yes | output | 1 | 70,336 | 18 | 140,673 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Modern text editors usually show some information regarding the document being edited. For example, the number of words, the number of pages, or the number of characters.
In this problem you should implement the similar functionality.
You are given a string which only consists of:
* uppercase and lowercase English letters,
* underscore symbols (they are used as separators),
* parentheses (both opening and closing).
It is guaranteed that each opening parenthesis has a succeeding closing parenthesis. Similarly, each closing parentheses has a preceding opening parentheses matching it. For each pair of matching parentheses there are no other parenthesis between them. In other words, each parenthesis in the string belongs to a matching "opening-closing" pair, and such pairs can't be nested.
For example, the following string is valid: "_Hello_Vasya(and_Petya)__bye_(and_OK)".
Word is a maximal sequence of consecutive letters, i.e. such sequence that the first character to the left and the first character to the right of it is an underscore, a parenthesis, or it just does not exist. For example, the string above consists of seven words: "Hello", "Vasya", "and", "Petya", "bye", "and" and "OK". Write a program that finds:
* the length of the longest word outside the parentheses (print 0, if there is no word outside the parentheses),
* the number of words inside the parentheses (print 0, if there is no word inside the parentheses).
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 255) — the length of the given string. The second line contains the string consisting of only lowercase and uppercase English letters, parentheses and underscore symbols.
Output
Print two space-separated integers:
* the length of the longest word outside the parentheses (print 0, if there is no word outside the parentheses),
* the number of words inside the parentheses (print 0, if there is no word inside the parentheses).
Examples
Input
37
_Hello_Vasya(and_Petya)__bye_(and_OK)
Output
5 4
Input
37
_a_(_b___c)__de_f(g_)__h__i(j_k_l)m__
Output
2 6
Input
27
(LoooonG)__shOrt__(LoooonG)
Output
5 2
Input
5
(___)
Output
0 0
Note
In the first sample, the words "Hello", "Vasya" and "bye" are outside any of the parentheses, and the words "and", "Petya", "and" and "OK" are inside. Note, that the word "and" is given twice and you should count it twice in the answer.
Submitted Solution:
```
n = int(input())
s = input()
b = 0
d = 0
max_d = 0
m = False
k = 0
for i in range(n):
if s[i] == '(':
b += 1
elif s[i] == ')':
b -= 1
if b == 0:
if s[i] == '_' or s[i] == ')' or s[i] == '(':
if d > max_d:
max_d = d
d = 0
else:
d += 1
else:
if s[i] == '_' or s[i] == ')' or s[i] == '(':
m = True
else:
if m:
k += 1
m = False
if d > max_d:
max_d = d
print(max_d, k)
``` | instruction | 0 | 70,337 | 18 | 140,674 |
Yes | output | 1 | 70,337 | 18 | 140,675 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Modern text editors usually show some information regarding the document being edited. For example, the number of words, the number of pages, or the number of characters.
In this problem you should implement the similar functionality.
You are given a string which only consists of:
* uppercase and lowercase English letters,
* underscore symbols (they are used as separators),
* parentheses (both opening and closing).
It is guaranteed that each opening parenthesis has a succeeding closing parenthesis. Similarly, each closing parentheses has a preceding opening parentheses matching it. For each pair of matching parentheses there are no other parenthesis between them. In other words, each parenthesis in the string belongs to a matching "opening-closing" pair, and such pairs can't be nested.
For example, the following string is valid: "_Hello_Vasya(and_Petya)__bye_(and_OK)".
Word is a maximal sequence of consecutive letters, i.e. such sequence that the first character to the left and the first character to the right of it is an underscore, a parenthesis, or it just does not exist. For example, the string above consists of seven words: "Hello", "Vasya", "and", "Petya", "bye", "and" and "OK". Write a program that finds:
* the length of the longest word outside the parentheses (print 0, if there is no word outside the parentheses),
* the number of words inside the parentheses (print 0, if there is no word inside the parentheses).
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 255) — the length of the given string. The second line contains the string consisting of only lowercase and uppercase English letters, parentheses and underscore symbols.
Output
Print two space-separated integers:
* the length of the longest word outside the parentheses (print 0, if there is no word outside the parentheses),
* the number of words inside the parentheses (print 0, if there is no word inside the parentheses).
Examples
Input
37
_Hello_Vasya(and_Petya)__bye_(and_OK)
Output
5 4
Input
37
_a_(_b___c)__de_f(g_)__h__i(j_k_l)m__
Output
2 6
Input
27
(LoooonG)__shOrt__(LoooonG)
Output
5 2
Input
5
(___)
Output
0 0
Note
In the first sample, the words "Hello", "Vasya" and "bye" are outside any of the parentheses, and the words "and", "Petya", "and" and "OK" are inside. Note, that the word "and" is given twice and you should count it twice in the answer.
Submitted Solution:
```
lens = int(input())
ins = input()
p, t, c, m = 0, 0, 0, 0
for i in range(lens):
if ins[i] == "_":
if p and t: c += 1
if not p and t>m: m = t
t = 0
elif ins[i] == "(":
p = 1
if t > m: m = t
t = 0
elif ins[i] == ")":
p = 0
if t: c += 1
t = 0
else:
t += 1
if t > m : m = t
print(m, c)
``` | instruction | 0 | 70,338 | 18 | 140,676 |
Yes | output | 1 | 70,338 | 18 | 140,677 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Modern text editors usually show some information regarding the document being edited. For example, the number of words, the number of pages, or the number of characters.
In this problem you should implement the similar functionality.
You are given a string which only consists of:
* uppercase and lowercase English letters,
* underscore symbols (they are used as separators),
* parentheses (both opening and closing).
It is guaranteed that each opening parenthesis has a succeeding closing parenthesis. Similarly, each closing parentheses has a preceding opening parentheses matching it. For each pair of matching parentheses there are no other parenthesis between them. In other words, each parenthesis in the string belongs to a matching "opening-closing" pair, and such pairs can't be nested.
For example, the following string is valid: "_Hello_Vasya(and_Petya)__bye_(and_OK)".
Word is a maximal sequence of consecutive letters, i.e. such sequence that the first character to the left and the first character to the right of it is an underscore, a parenthesis, or it just does not exist. For example, the string above consists of seven words: "Hello", "Vasya", "and", "Petya", "bye", "and" and "OK". Write a program that finds:
* the length of the longest word outside the parentheses (print 0, if there is no word outside the parentheses),
* the number of words inside the parentheses (print 0, if there is no word inside the parentheses).
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 255) — the length of the given string. The second line contains the string consisting of only lowercase and uppercase English letters, parentheses and underscore symbols.
Output
Print two space-separated integers:
* the length of the longest word outside the parentheses (print 0, if there is no word outside the parentheses),
* the number of words inside the parentheses (print 0, if there is no word inside the parentheses).
Examples
Input
37
_Hello_Vasya(and_Petya)__bye_(and_OK)
Output
5 4
Input
37
_a_(_b___c)__de_f(g_)__h__i(j_k_l)m__
Output
2 6
Input
27
(LoooonG)__shOrt__(LoooonG)
Output
5 2
Input
5
(___)
Output
0 0
Note
In the first sample, the words "Hello", "Vasya" and "bye" are outside any of the parentheses, and the words "and", "Petya", "and" and "OK" are inside. Note, that the word "and" is given twice and you should count it twice in the answer.
Submitted Solution:
```
a=input()
s=list(input())
i=0
ss=sc=max1=0
fal=0
for i in range(len(s)):
if s[i]=='_':
sc=0
if s[i]=='(':
fal+=1
sc=0
if s[i+1]!=')' and s[i+1]!='_':
ss+=1
if (fal==0) and s[i]!='_' and s[i]!=')'and s[i]!='(':
sc=sc+1
if sc>max1:
max1=sc
if s[i]==')':
fal-=1
sc=0
if fal>=1:
if (s[i]=='_'):
ss+=1
if (s[i+1]=='_' or s[i+1]==')' or s[i-1]==')' ):
ss-=1
print(max1, (ss))
``` | instruction | 0 | 70,339 | 18 | 140,678 |
Yes | output | 1 | 70,339 | 18 | 140,679 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Modern text editors usually show some information regarding the document being edited. For example, the number of words, the number of pages, or the number of characters.
In this problem you should implement the similar functionality.
You are given a string which only consists of:
* uppercase and lowercase English letters,
* underscore symbols (they are used as separators),
* parentheses (both opening and closing).
It is guaranteed that each opening parenthesis has a succeeding closing parenthesis. Similarly, each closing parentheses has a preceding opening parentheses matching it. For each pair of matching parentheses there are no other parenthesis between them. In other words, each parenthesis in the string belongs to a matching "opening-closing" pair, and such pairs can't be nested.
For example, the following string is valid: "_Hello_Vasya(and_Petya)__bye_(and_OK)".
Word is a maximal sequence of consecutive letters, i.e. such sequence that the first character to the left and the first character to the right of it is an underscore, a parenthesis, or it just does not exist. For example, the string above consists of seven words: "Hello", "Vasya", "and", "Petya", "bye", "and" and "OK". Write a program that finds:
* the length of the longest word outside the parentheses (print 0, if there is no word outside the parentheses),
* the number of words inside the parentheses (print 0, if there is no word inside the parentheses).
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 255) — the length of the given string. The second line contains the string consisting of only lowercase and uppercase English letters, parentheses and underscore symbols.
Output
Print two space-separated integers:
* the length of the longest word outside the parentheses (print 0, if there is no word outside the parentheses),
* the number of words inside the parentheses (print 0, if there is no word inside the parentheses).
Examples
Input
37
_Hello_Vasya(and_Petya)__bye_(and_OK)
Output
5 4
Input
37
_a_(_b___c)__de_f(g_)__h__i(j_k_l)m__
Output
2 6
Input
27
(LoooonG)__shOrt__(LoooonG)
Output
5 2
Input
5
(___)
Output
0 0
Note
In the first sample, the words "Hello", "Vasya" and "bye" are outside any of the parentheses, and the words "and", "Petya", "and" and "OK" are inside. Note, that the word "and" is given twice and you should count it twice in the answer.
Submitted Solution:
```
input()
k = input().replace('()','+')
l = k.replace('(',' ').replace(')',' ').split()
mx = 0
n = 0
for i in range(len(l)):
s = l[i].replace('_', ' ').replace('+',' ').split()
if k[0]!='(':
if i%2==0:
for j in s:
mx = max(mx, len(j))
else:
n += len(s)
else:
if i%2==0:
n += len(s)
else:
for j in s:
mx = max(mx,len(j))
print(mx, n)
``` | instruction | 0 | 70,340 | 18 | 140,680 |
No | output | 1 | 70,340 | 18 | 140,681 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Modern text editors usually show some information regarding the document being edited. For example, the number of words, the number of pages, or the number of characters.
In this problem you should implement the similar functionality.
You are given a string which only consists of:
* uppercase and lowercase English letters,
* underscore symbols (they are used as separators),
* parentheses (both opening and closing).
It is guaranteed that each opening parenthesis has a succeeding closing parenthesis. Similarly, each closing parentheses has a preceding opening parentheses matching it. For each pair of matching parentheses there are no other parenthesis between them. In other words, each parenthesis in the string belongs to a matching "opening-closing" pair, and such pairs can't be nested.
For example, the following string is valid: "_Hello_Vasya(and_Petya)__bye_(and_OK)".
Word is a maximal sequence of consecutive letters, i.e. such sequence that the first character to the left and the first character to the right of it is an underscore, a parenthesis, or it just does not exist. For example, the string above consists of seven words: "Hello", "Vasya", "and", "Petya", "bye", "and" and "OK". Write a program that finds:
* the length of the longest word outside the parentheses (print 0, if there is no word outside the parentheses),
* the number of words inside the parentheses (print 0, if there is no word inside the parentheses).
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 255) — the length of the given string. The second line contains the string consisting of only lowercase and uppercase English letters, parentheses and underscore symbols.
Output
Print two space-separated integers:
* the length of the longest word outside the parentheses (print 0, if there is no word outside the parentheses),
* the number of words inside the parentheses (print 0, if there is no word inside the parentheses).
Examples
Input
37
_Hello_Vasya(and_Petya)__bye_(and_OK)
Output
5 4
Input
37
_a_(_b___c)__de_f(g_)__h__i(j_k_l)m__
Output
2 6
Input
27
(LoooonG)__shOrt__(LoooonG)
Output
5 2
Input
5
(___)
Output
0 0
Note
In the first sample, the words "Hello", "Vasya" and "bye" are outside any of the parentheses, and the words "and", "Petya", "and" and "OK" are inside. Note, that the word "and" is given twice and you should count it twice in the answer.
Submitted Solution:
```
n = int(input())
a = input() + ' '
count = 0
counts = a.count("(")
for x in range(counts):
try:
start = a.index("(")
stop = a.index(")")
b = a[start:stop + 1].strip("()").split("_")
for x in b:
if len(x) > 0:
count += 1
a = a[:start]+"_" + a[stop + 1:]
except:
pass
a = a.strip("_ ").split("_")
print(a)
c = [len(x) for x in a]
print(max(c), count)
``` | instruction | 0 | 70,341 | 18 | 140,682 |
No | output | 1 | 70,341 | 18 | 140,683 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Modern text editors usually show some information regarding the document being edited. For example, the number of words, the number of pages, or the number of characters.
In this problem you should implement the similar functionality.
You are given a string which only consists of:
* uppercase and lowercase English letters,
* underscore symbols (they are used as separators),
* parentheses (both opening and closing).
It is guaranteed that each opening parenthesis has a succeeding closing parenthesis. Similarly, each closing parentheses has a preceding opening parentheses matching it. For each pair of matching parentheses there are no other parenthesis between them. In other words, each parenthesis in the string belongs to a matching "opening-closing" pair, and such pairs can't be nested.
For example, the following string is valid: "_Hello_Vasya(and_Petya)__bye_(and_OK)".
Word is a maximal sequence of consecutive letters, i.e. such sequence that the first character to the left and the first character to the right of it is an underscore, a parenthesis, or it just does not exist. For example, the string above consists of seven words: "Hello", "Vasya", "and", "Petya", "bye", "and" and "OK". Write a program that finds:
* the length of the longest word outside the parentheses (print 0, if there is no word outside the parentheses),
* the number of words inside the parentheses (print 0, if there is no word inside the parentheses).
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 255) — the length of the given string. The second line contains the string consisting of only lowercase and uppercase English letters, parentheses and underscore symbols.
Output
Print two space-separated integers:
* the length of the longest word outside the parentheses (print 0, if there is no word outside the parentheses),
* the number of words inside the parentheses (print 0, if there is no word inside the parentheses).
Examples
Input
37
_Hello_Vasya(and_Petya)__bye_(and_OK)
Output
5 4
Input
37
_a_(_b___c)__de_f(g_)__h__i(j_k_l)m__
Output
2 6
Input
27
(LoooonG)__shOrt__(LoooonG)
Output
5 2
Input
5
(___)
Output
0 0
Note
In the first sample, the words "Hello", "Vasya" and "bye" are outside any of the parentheses, and the words "and", "Petya", "and" and "OK" are inside. Note, that the word "and" is given twice and you should count it twice in the answer.
Submitted Solution:
```
n = int(input())
lst = input()
flag = 0; len = 0
num = 0; max = 0
for i in range(0,n):
if lst[i]=='_'or lst[i]=='('or lst[i]==')'or lst[i]=='\0':
if flag == 0 and max < len:
max = len
else:
if flag == 1 and len != 0:
num += 1
len = 0
if lst[i]=='(':
flag = 1
else:
if lst[i]==')':
flag = 0
else:
len += 1
print(max,num)
``` | instruction | 0 | 70,342 | 18 | 140,684 |
No | output | 1 | 70,342 | 18 | 140,685 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Modern text editors usually show some information regarding the document being edited. For example, the number of words, the number of pages, or the number of characters.
In this problem you should implement the similar functionality.
You are given a string which only consists of:
* uppercase and lowercase English letters,
* underscore symbols (they are used as separators),
* parentheses (both opening and closing).
It is guaranteed that each opening parenthesis has a succeeding closing parenthesis. Similarly, each closing parentheses has a preceding opening parentheses matching it. For each pair of matching parentheses there are no other parenthesis between them. In other words, each parenthesis in the string belongs to a matching "opening-closing" pair, and such pairs can't be nested.
For example, the following string is valid: "_Hello_Vasya(and_Petya)__bye_(and_OK)".
Word is a maximal sequence of consecutive letters, i.e. such sequence that the first character to the left and the first character to the right of it is an underscore, a parenthesis, or it just does not exist. For example, the string above consists of seven words: "Hello", "Vasya", "and", "Petya", "bye", "and" and "OK". Write a program that finds:
* the length of the longest word outside the parentheses (print 0, if there is no word outside the parentheses),
* the number of words inside the parentheses (print 0, if there is no word inside the parentheses).
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 255) — the length of the given string. The second line contains the string consisting of only lowercase and uppercase English letters, parentheses and underscore symbols.
Output
Print two space-separated integers:
* the length of the longest word outside the parentheses (print 0, if there is no word outside the parentheses),
* the number of words inside the parentheses (print 0, if there is no word inside the parentheses).
Examples
Input
37
_Hello_Vasya(and_Petya)__bye_(and_OK)
Output
5 4
Input
37
_a_(_b___c)__de_f(g_)__h__i(j_k_l)m__
Output
2 6
Input
27
(LoooonG)__shOrt__(LoooonG)
Output
5 2
Input
5
(___)
Output
0 0
Note
In the first sample, the words "Hello", "Vasya" and "bye" are outside any of the parentheses, and the words "and", "Petya", "and" and "OK" are inside. Note, that the word "and" is given twice and you should count it twice in the answer.
Submitted Solution:
```
n = int(input())
s = input()
ans1 = 0
ans2 = 0
para = False
us = False
c = 0
for i in s:
if not para and (i!='(' and i!='_' and i!=')'):
c += 1
elif para:
if i!='(' and i!='_' and i!=')':
c += 1
us = True
else:
if us:
ans2 += 1
c = 0
us = False
if i==')':
para = False
else:
ans1 = max(ans1,c)
c = 0
if i=='(':
para = True
print(ans1,ans2)
``` | instruction | 0 | 70,343 | 18 | 140,686 |
No | output | 1 | 70,343 | 18 | 140,687 |
Provide a correct Python 3 solution for this coding contest problem.
It is important to use strong passwords to make the Internet more secure. At the same time, it is very important not to reuse the same password. No matter how strong your password is, if the plaintext is leaked in one place, it will be very easy to break. Of course, if all applications hash passwords and use SALT properly, such damage will not occur even if leaked, but there are still applications that store passwords in clear text.
Well, the contest is about to begin. I have to register a team account ...
problem
A string representing the password is given. Check if the character string meets all of the following conditions, and if it meets all of the following conditions, output "VALID", and if there is an item that does not meet even one, output "INVALID".
* String length is 6 or more
* Contains one or more numbers
* Contains one or more uppercase letters
* Contains one or more lowercase letters
input
A string representing the password is given on one line.
output
Print "VALID" if the password string meets all the conditions in question, and "INVALID" if there are conditions that do not.
Constraint
* The password string must be between 1 and 20 characters.
* Password string contains only uppercase letters, lowercase letters, and numbers
Input / output example
Input 1
password
Output 1
INVALID
It is a password that is widely used all over the world.
Input 2
AizuCamp2013
Output 2
VALID
Great password.
Input 3
1234
Output 3
INVALID
Widely used as a PIN.
Input 4
NaBiO3
Output 4
VALID
It is barely 6 characters or more.
Example
Input
password
Output
INVALID | instruction | 0 | 70,697 | 18 | 141,394 |
"Correct Solution:
```
A=0
n=str(input())
if len(n)>=6:
A=A+1
B=n.count('0')+n.count('1')+n.count('2')+n.count('3')+n.count('4')+n.count('5')+n.count('6')+n.count('7')+n.count('8')+n.count('9')
if B>0:
A=A+1
C=n.count('a')+n.count('b')+n.count('c')+n.count('d')+n.count('e')+n.count('f')+n.count('g')+n.count('h')+n.count('i')+n.count('j')+n.count('k')+n.count('l')+n.count('m')+n.count('n')+n.count('o')+n.count('p')+n.count('q')+n.count('r')+n.count('s')+n.count('t')+n.count('u')+n.count('v')+n.count('w')++n.count('x')+n.count('y')+n.count('z')
if C>0:
A=A+1
D=n.count('A')+n.count('B')+n.count('C')+n.count('D')+n.count('E')+n.count('F')+n.count('G')+n.count('H')+n.count('I')+n.count('J')+n.count('K')+n.count('L')+n.count('M')+n.count('N')+n.count('O')+n.count('P')+n.count('Q')+n.count('R')+n.count('S')+n.count('T')+n.count('U')+n.count('V')+n.count('W')++n.count('X')+n.count('Y')+n.count('Z')
if D>0:
A=A+1
if A==4:
print('VALID')
else:
print('INVALID')
``` | output | 1 | 70,697 | 18 | 141,395 |
Provide a correct Python 3 solution for this coding contest problem.
It is important to use strong passwords to make the Internet more secure. At the same time, it is very important not to reuse the same password. No matter how strong your password is, if the plaintext is leaked in one place, it will be very easy to break. Of course, if all applications hash passwords and use SALT properly, such damage will not occur even if leaked, but there are still applications that store passwords in clear text.
Well, the contest is about to begin. I have to register a team account ...
problem
A string representing the password is given. Check if the character string meets all of the following conditions, and if it meets all of the following conditions, output "VALID", and if there is an item that does not meet even one, output "INVALID".
* String length is 6 or more
* Contains one or more numbers
* Contains one or more uppercase letters
* Contains one or more lowercase letters
input
A string representing the password is given on one line.
output
Print "VALID" if the password string meets all the conditions in question, and "INVALID" if there are conditions that do not.
Constraint
* The password string must be between 1 and 20 characters.
* Password string contains only uppercase letters, lowercase letters, and numbers
Input / output example
Input 1
password
Output 1
INVALID
It is a password that is widely used all over the world.
Input 2
AizuCamp2013
Output 2
VALID
Great password.
Input 3
1234
Output 3
INVALID
Widely used as a PIN.
Input 4
NaBiO3
Output 4
VALID
It is barely 6 characters or more.
Example
Input
password
Output
INVALID | instruction | 0 | 70,698 | 18 | 141,396 |
"Correct Solution:
```
x=input()
if len(x)<6 or x.isalpha() or x.islower() or x.isupper() or x.isdigit():
print("INVALID")
else:
print("VALID")
``` | output | 1 | 70,698 | 18 | 141,397 |
Provide a correct Python 3 solution for this coding contest problem.
It is important to use strong passwords to make the Internet more secure. At the same time, it is very important not to reuse the same password. No matter how strong your password is, if the plaintext is leaked in one place, it will be very easy to break. Of course, if all applications hash passwords and use SALT properly, such damage will not occur even if leaked, but there are still applications that store passwords in clear text.
Well, the contest is about to begin. I have to register a team account ...
problem
A string representing the password is given. Check if the character string meets all of the following conditions, and if it meets all of the following conditions, output "VALID", and if there is an item that does not meet even one, output "INVALID".
* String length is 6 or more
* Contains one or more numbers
* Contains one or more uppercase letters
* Contains one or more lowercase letters
input
A string representing the password is given on one line.
output
Print "VALID" if the password string meets all the conditions in question, and "INVALID" if there are conditions that do not.
Constraint
* The password string must be between 1 and 20 characters.
* Password string contains only uppercase letters, lowercase letters, and numbers
Input / output example
Input 1
password
Output 1
INVALID
It is a password that is widely used all over the world.
Input 2
AizuCamp2013
Output 2
VALID
Great password.
Input 3
1234
Output 3
INVALID
Widely used as a PIN.
Input 4
NaBiO3
Output 4
VALID
It is barely 6 characters or more.
Example
Input
password
Output
INVALID | instruction | 0 | 70,699 | 18 | 141,398 |
"Correct Solution:
```
p = input()
if len(p) < 6 or p.isalpha() or p.isdigit() or p.islower() or p.isupper():
print("INVALID")
else:
print("VALID")
``` | output | 1 | 70,699 | 18 | 141,399 |
Provide a correct Python 3 solution for this coding contest problem.
It is important to use strong passwords to make the Internet more secure. At the same time, it is very important not to reuse the same password. No matter how strong your password is, if the plaintext is leaked in one place, it will be very easy to break. Of course, if all applications hash passwords and use SALT properly, such damage will not occur even if leaked, but there are still applications that store passwords in clear text.
Well, the contest is about to begin. I have to register a team account ...
problem
A string representing the password is given. Check if the character string meets all of the following conditions, and if it meets all of the following conditions, output "VALID", and if there is an item that does not meet even one, output "INVALID".
* String length is 6 or more
* Contains one or more numbers
* Contains one or more uppercase letters
* Contains one or more lowercase letters
input
A string representing the password is given on one line.
output
Print "VALID" if the password string meets all the conditions in question, and "INVALID" if there are conditions that do not.
Constraint
* The password string must be between 1 and 20 characters.
* Password string contains only uppercase letters, lowercase letters, and numbers
Input / output example
Input 1
password
Output 1
INVALID
It is a password that is widely used all over the world.
Input 2
AizuCamp2013
Output 2
VALID
Great password.
Input 3
1234
Output 3
INVALID
Widely used as a PIN.
Input 4
NaBiO3
Output 4
VALID
It is barely 6 characters or more.
Example
Input
password
Output
INVALID | instruction | 0 | 70,700 | 18 | 141,400 |
"Correct Solution:
```
s=input()
print('INVALID' if s.isalpha() or s.isdigit() or s.islower() or s.isupper() or len(s)<6 else 'VALID')
``` | output | 1 | 70,700 | 18 | 141,401 |
Provide a correct Python 3 solution for this coding contest problem.
It is important to use strong passwords to make the Internet more secure. At the same time, it is very important not to reuse the same password. No matter how strong your password is, if the plaintext is leaked in one place, it will be very easy to break. Of course, if all applications hash passwords and use SALT properly, such damage will not occur even if leaked, but there are still applications that store passwords in clear text.
Well, the contest is about to begin. I have to register a team account ...
problem
A string representing the password is given. Check if the character string meets all of the following conditions, and if it meets all of the following conditions, output "VALID", and if there is an item that does not meet even one, output "INVALID".
* String length is 6 or more
* Contains one or more numbers
* Contains one or more uppercase letters
* Contains one or more lowercase letters
input
A string representing the password is given on one line.
output
Print "VALID" if the password string meets all the conditions in question, and "INVALID" if there are conditions that do not.
Constraint
* The password string must be between 1 and 20 characters.
* Password string contains only uppercase letters, lowercase letters, and numbers
Input / output example
Input 1
password
Output 1
INVALID
It is a password that is widely used all over the world.
Input 2
AizuCamp2013
Output 2
VALID
Great password.
Input 3
1234
Output 3
INVALID
Widely used as a PIN.
Input 4
NaBiO3
Output 4
VALID
It is barely 6 characters or more.
Example
Input
password
Output
INVALID | instruction | 0 | 70,701 | 18 | 141,402 |
"Correct Solution:
```
s = input()
if len(s) >= 6 and\
any([(c in s) for c in "1234567890"]) and\
any([(c in s) for c in "ABCDEFGHIJKLMNOPQRSTUVWXYZ"]) and\
any([(c in s) for c in "abcdefghijklmnopqrstuvwxyz"]):
print("VALID")
else:
print("INVALID")
``` | output | 1 | 70,701 | 18 | 141,403 |
Provide a correct Python 3 solution for this coding contest problem.
It is important to use strong passwords to make the Internet more secure. At the same time, it is very important not to reuse the same password. No matter how strong your password is, if the plaintext is leaked in one place, it will be very easy to break. Of course, if all applications hash passwords and use SALT properly, such damage will not occur even if leaked, but there are still applications that store passwords in clear text.
Well, the contest is about to begin. I have to register a team account ...
problem
A string representing the password is given. Check if the character string meets all of the following conditions, and if it meets all of the following conditions, output "VALID", and if there is an item that does not meet even one, output "INVALID".
* String length is 6 or more
* Contains one or more numbers
* Contains one or more uppercase letters
* Contains one or more lowercase letters
input
A string representing the password is given on one line.
output
Print "VALID" if the password string meets all the conditions in question, and "INVALID" if there are conditions that do not.
Constraint
* The password string must be between 1 and 20 characters.
* Password string contains only uppercase letters, lowercase letters, and numbers
Input / output example
Input 1
password
Output 1
INVALID
It is a password that is widely used all over the world.
Input 2
AizuCamp2013
Output 2
VALID
Great password.
Input 3
1234
Output 3
INVALID
Widely used as a PIN.
Input 4
NaBiO3
Output 4
VALID
It is barely 6 characters or more.
Example
Input
password
Output
INVALID | instruction | 0 | 70,702 | 18 | 141,404 |
"Correct Solution:
```
x=input()
a=0
b=0
c=0
if len(x)<6:
print("INVALID")
else:
for i in range(len(x)):
X=ord(x[i])
if 48<=X and X<=57:
a+=1
if 97<=X and X<=122:
b+=1
if 65<=X and X<=90:
c+=1
if a==0 or b==0 or c==0:
print("INVALID")
else:
print("VALID")
``` | output | 1 | 70,702 | 18 | 141,405 |
Provide a correct Python 3 solution for this coding contest problem.
It is important to use strong passwords to make the Internet more secure. At the same time, it is very important not to reuse the same password. No matter how strong your password is, if the plaintext is leaked in one place, it will be very easy to break. Of course, if all applications hash passwords and use SALT properly, such damage will not occur even if leaked, but there are still applications that store passwords in clear text.
Well, the contest is about to begin. I have to register a team account ...
problem
A string representing the password is given. Check if the character string meets all of the following conditions, and if it meets all of the following conditions, output "VALID", and if there is an item that does not meet even one, output "INVALID".
* String length is 6 or more
* Contains one or more numbers
* Contains one or more uppercase letters
* Contains one or more lowercase letters
input
A string representing the password is given on one line.
output
Print "VALID" if the password string meets all the conditions in question, and "INVALID" if there are conditions that do not.
Constraint
* The password string must be between 1 and 20 characters.
* Password string contains only uppercase letters, lowercase letters, and numbers
Input / output example
Input 1
password
Output 1
INVALID
It is a password that is widely used all over the world.
Input 2
AizuCamp2013
Output 2
VALID
Great password.
Input 3
1234
Output 3
INVALID
Widely used as a PIN.
Input 4
NaBiO3
Output 4
VALID
It is barely 6 characters or more.
Example
Input
password
Output
INVALID | instruction | 0 | 70,703 | 18 | 141,406 |
"Correct Solution:
```
pas = input()
ans1,ans2,ans3 = 0,0,0
if len(pas)<6:
print("INVALID")
else:
for i in range(len(pas)):
if pas[i].isupper(): ans1 += 1
if pas[i].islower(): ans2 += 1
if pas[i].isdecimal(): ans3 += 1
if ans1>0 and ans2>0 and ans3>0: print("VALID")
else: print("INVALID")
``` | output | 1 | 70,703 | 18 | 141,407 |
Provide a correct Python 3 solution for this coding contest problem.
It is important to use strong passwords to make the Internet more secure. At the same time, it is very important not to reuse the same password. No matter how strong your password is, if the plaintext is leaked in one place, it will be very easy to break. Of course, if all applications hash passwords and use SALT properly, such damage will not occur even if leaked, but there are still applications that store passwords in clear text.
Well, the contest is about to begin. I have to register a team account ...
problem
A string representing the password is given. Check if the character string meets all of the following conditions, and if it meets all of the following conditions, output "VALID", and if there is an item that does not meet even one, output "INVALID".
* String length is 6 or more
* Contains one or more numbers
* Contains one or more uppercase letters
* Contains one or more lowercase letters
input
A string representing the password is given on one line.
output
Print "VALID" if the password string meets all the conditions in question, and "INVALID" if there are conditions that do not.
Constraint
* The password string must be between 1 and 20 characters.
* Password string contains only uppercase letters, lowercase letters, and numbers
Input / output example
Input 1
password
Output 1
INVALID
It is a password that is widely used all over the world.
Input 2
AizuCamp2013
Output 2
VALID
Great password.
Input 3
1234
Output 3
INVALID
Widely used as a PIN.
Input 4
NaBiO3
Output 4
VALID
It is barely 6 characters or more.
Example
Input
password
Output
INVALID | instruction | 0 | 70,704 | 18 | 141,408 |
"Correct Solution:
```
import re
s=input()
if len(s)>=6 and re.findall("[0-9]", s) and re.findall("[A-Z]", s) and re.findall("[a-z]", s):
print("VALID")
else:
print("INVALID")
``` | output | 1 | 70,704 | 18 | 141,409 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is important to use strong passwords to make the Internet more secure. At the same time, it is very important not to reuse the same password. No matter how strong your password is, if the plaintext is leaked in one place, it will be very easy to break. Of course, if all applications hash passwords and use SALT properly, such damage will not occur even if leaked, but there are still applications that store passwords in clear text.
Well, the contest is about to begin. I have to register a team account ...
problem
A string representing the password is given. Check if the character string meets all of the following conditions, and if it meets all of the following conditions, output "VALID", and if there is an item that does not meet even one, output "INVALID".
* String length is 6 or more
* Contains one or more numbers
* Contains one or more uppercase letters
* Contains one or more lowercase letters
input
A string representing the password is given on one line.
output
Print "VALID" if the password string meets all the conditions in question, and "INVALID" if there are conditions that do not.
Constraint
* The password string must be between 1 and 20 characters.
* Password string contains only uppercase letters, lowercase letters, and numbers
Input / output example
Input 1
password
Output 1
INVALID
It is a password that is widely used all over the world.
Input 2
AizuCamp2013
Output 2
VALID
Great password.
Input 3
1234
Output 3
INVALID
Widely used as a PIN.
Input 4
NaBiO3
Output 4
VALID
It is barely 6 characters or more.
Example
Input
password
Output
INVALID
Submitted Solution:
```
#2522
l = []
word = input()
for i in word:
l.append(i)
a = 0
b = 0
c = 0
d = 0
if len(l) >= 6:
a += 1
for i in l:
if 48 <= ord(i) <= 57:
b += 1
elif 97 <= ord(i) <= 122:
c += 1
elif 65 <= ord(i) <= 90:
d += 1
else:
c += 0
if a >= 1 and b >= 1 and c >= 1 and d >= 1:
print('VALID')
else:
print('INVALID')
``` | instruction | 0 | 70,705 | 18 | 141,410 |
Yes | output | 1 | 70,705 | 18 | 141,411 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is important to use strong passwords to make the Internet more secure. At the same time, it is very important not to reuse the same password. No matter how strong your password is, if the plaintext is leaked in one place, it will be very easy to break. Of course, if all applications hash passwords and use SALT properly, such damage will not occur even if leaked, but there are still applications that store passwords in clear text.
Well, the contest is about to begin. I have to register a team account ...
problem
A string representing the password is given. Check if the character string meets all of the following conditions, and if it meets all of the following conditions, output "VALID", and if there is an item that does not meet even one, output "INVALID".
* String length is 6 or more
* Contains one or more numbers
* Contains one or more uppercase letters
* Contains one or more lowercase letters
input
A string representing the password is given on one line.
output
Print "VALID" if the password string meets all the conditions in question, and "INVALID" if there are conditions that do not.
Constraint
* The password string must be between 1 and 20 characters.
* Password string contains only uppercase letters, lowercase letters, and numbers
Input / output example
Input 1
password
Output 1
INVALID
It is a password that is widely used all over the world.
Input 2
AizuCamp2013
Output 2
VALID
Great password.
Input 3
1234
Output 3
INVALID
Widely used as a PIN.
Input 4
NaBiO3
Output 4
VALID
It is barely 6 characters or more.
Example
Input
password
Output
INVALID
Submitted Solution:
```
pas = input()
if len(pas) < 6 :
print("INVALID")
else :
num = 0
Lstr = 0
Sstr = 0
for i in range(len(pas)) :
if ord(pas[i]) >= 48 and ord(pas[i]) <= 57 :
num += 1
if ord(pas[i]) >= 97 and ord(pas[i]) <= 122 :
Lstr += 1
if ord(pas[i]) >= 65 and ord(pas[i]) <= 90 :
Sstr += 1
if num == 0 or Lstr == 0 or Sstr == 0 :
print("INVALID")
else :
print("VALID")
``` | instruction | 0 | 70,706 | 18 | 141,412 |
Yes | output | 1 | 70,706 | 18 | 141,413 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is important to use strong passwords to make the Internet more secure. At the same time, it is very important not to reuse the same password. No matter how strong your password is, if the plaintext is leaked in one place, it will be very easy to break. Of course, if all applications hash passwords and use SALT properly, such damage will not occur even if leaked, but there are still applications that store passwords in clear text.
Well, the contest is about to begin. I have to register a team account ...
problem
A string representing the password is given. Check if the character string meets all of the following conditions, and if it meets all of the following conditions, output "VALID", and if there is an item that does not meet even one, output "INVALID".
* String length is 6 or more
* Contains one or more numbers
* Contains one or more uppercase letters
* Contains one or more lowercase letters
input
A string representing the password is given on one line.
output
Print "VALID" if the password string meets all the conditions in question, and "INVALID" if there are conditions that do not.
Constraint
* The password string must be between 1 and 20 characters.
* Password string contains only uppercase letters, lowercase letters, and numbers
Input / output example
Input 1
password
Output 1
INVALID
It is a password that is widely used all over the world.
Input 2
AizuCamp2013
Output 2
VALID
Great password.
Input 3
1234
Output 3
INVALID
Widely used as a PIN.
Input 4
NaBiO3
Output 4
VALID
It is barely 6 characters or more.
Example
Input
password
Output
INVALID
Submitted Solution:
```
import re
lst = input()
if len(lst) >= 6:
if re.findall("[0-9]",lst):
if re.findall("[A-Z]",lst):
if re.findall("[a-z]",lst):
print("VALID")
else: print("INVALID")
else: print("INVALID")
else: print("INVALID")
else: print("INVALID")
``` | instruction | 0 | 70,707 | 18 | 141,414 |
Yes | output | 1 | 70,707 | 18 | 141,415 |
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