text stringlengths 594 23.8k | conversation_id int64 97 109k | embedding list | cluster int64 0 0 |
|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting only of lowercase Latin letters.
You can rearrange all letters of this string as you wish. Your task is to obtain a good string by rearranging the letters of the given string or report that it is impossible to do it.
Let's call a string good if it is not a palindrome. Palindrome is a string which is read from left to right the same as from right to left. For example, strings "abacaba", "aa" and "z" are palindromes and strings "bba", "xd" are not.
You have to answer t independent queries.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — number of queries.
Each of the next t lines contains one string. The i-th line contains a string s_i consisting only of lowercase Latin letter. It is guaranteed that the length of s_i is from 1 to 1000 (inclusive).
Output
Print t lines. In the i-th line print the answer to the i-th query: -1 if it is impossible to obtain a good string by rearranging the letters of s_i and any good string which can be obtained from the given one (by rearranging the letters) otherwise.
Example
Input
3
aa
abacaba
xdd
Output
-1
abaacba
xdd
Note
In the first query we cannot rearrange letters to obtain a good string.
Other examples (not all) of correct answers to the second query: "ababaca", "abcabaa", "baacaba".
In the third query we can do nothing to obtain a good string.
Tags: constructive algorithms, greedy, sortings, strings
Correct Solution:
```
T=int(input())
for i in range(T):
s=input()
if(s!=s[::-1]):
print(s)
elif(s==s[::-1]):
if(s.count(s[-1])==len(s)):
print(-1)
else:
print(s[1:]+s[0])
```
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Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting only of lowercase Latin letters.
You can rearrange all letters of this string as you wish. Your task is to obtain a good string by rearranging the letters of the given string or report that it is impossible to do it.
Let's call a string good if it is not a palindrome. Palindrome is a string which is read from left to right the same as from right to left. For example, strings "abacaba", "aa" and "z" are palindromes and strings "bba", "xd" are not.
You have to answer t independent queries.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — number of queries.
Each of the next t lines contains one string. The i-th line contains a string s_i consisting only of lowercase Latin letter. It is guaranteed that the length of s_i is from 1 to 1000 (inclusive).
Output
Print t lines. In the i-th line print the answer to the i-th query: -1 if it is impossible to obtain a good string by rearranging the letters of s_i and any good string which can be obtained from the given one (by rearranging the letters) otherwise.
Example
Input
3
aa
abacaba
xdd
Output
-1
abaacba
xdd
Note
In the first query we cannot rearrange letters to obtain a good string.
Other examples (not all) of correct answers to the second query: "ababaca", "abcabaa", "baacaba".
In the third query we can do nothing to obtain a good string.
Submitted Solution:
```
n = int(input())
for i in range(n):
s = input()
if s.count(s[0]) == len(s):
print(-1)
continue
a = [i for i in s]
if a[0] != a[len(s) -1]:
print(s)
continue
i = 0
while a[0] == a[i]:
i += 1
a[i], a[0] = a[0], a[i]
for i in range(len(a)):
print(a[i], end='')
print()
```
Yes
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Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting only of lowercase Latin letters.
You can rearrange all letters of this string as you wish. Your task is to obtain a good string by rearranging the letters of the given string or report that it is impossible to do it.
Let's call a string good if it is not a palindrome. Palindrome is a string which is read from left to right the same as from right to left. For example, strings "abacaba", "aa" and "z" are palindromes and strings "bba", "xd" are not.
You have to answer t independent queries.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — number of queries.
Each of the next t lines contains one string. The i-th line contains a string s_i consisting only of lowercase Latin letter. It is guaranteed that the length of s_i is from 1 to 1000 (inclusive).
Output
Print t lines. In the i-th line print the answer to the i-th query: -1 if it is impossible to obtain a good string by rearranging the letters of s_i and any good string which can be obtained from the given one (by rearranging the letters) otherwise.
Example
Input
3
aa
abacaba
xdd
Output
-1
abaacba
xdd
Note
In the first query we cannot rearrange letters to obtain a good string.
Other examples (not all) of correct answers to the second query: "ababaca", "abcabaa", "baacaba".
In the third query we can do nothing to obtain a good string.
Submitted Solution:
```
import re
q=int(input())
for x in range(q):
s=''
s1=''
n=input()
c=n.count('a')
s=s+'a'*c
l=['a',]
for y in n:
if y in l:
continue
else:
c=n.count(y)
s=s+y*c
l.append(y)
for y in s:
s1=y+s1
if re.match(s,s1):
print("-1")
else:
print(s)
```
Yes
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-0... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting only of lowercase Latin letters.
You can rearrange all letters of this string as you wish. Your task is to obtain a good string by rearranging the letters of the given string or report that it is impossible to do it.
Let's call a string good if it is not a palindrome. Palindrome is a string which is read from left to right the same as from right to left. For example, strings "abacaba", "aa" and "z" are palindromes and strings "bba", "xd" are not.
You have to answer t independent queries.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — number of queries.
Each of the next t lines contains one string. The i-th line contains a string s_i consisting only of lowercase Latin letter. It is guaranteed that the length of s_i is from 1 to 1000 (inclusive).
Output
Print t lines. In the i-th line print the answer to the i-th query: -1 if it is impossible to obtain a good string by rearranging the letters of s_i and any good string which can be obtained from the given one (by rearranging the letters) otherwise.
Example
Input
3
aa
abacaba
xdd
Output
-1
abaacba
xdd
Note
In the first query we cannot rearrange letters to obtain a good string.
Other examples (not all) of correct answers to the second query: "ababaca", "abcabaa", "baacaba".
In the third query we can do nothing to obtain a good string.
Submitted Solution:
```
n = int(input())
for item in range(n):
l = input()
if len(set(l)) == 1:
print(-1)
else:
print("".join(sorted(l)))
```
Yes
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Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting only of lowercase Latin letters.
You can rearrange all letters of this string as you wish. Your task is to obtain a good string by rearranging the letters of the given string or report that it is impossible to do it.
Let's call a string good if it is not a palindrome. Palindrome is a string which is read from left to right the same as from right to left. For example, strings "abacaba", "aa" and "z" are palindromes and strings "bba", "xd" are not.
You have to answer t independent queries.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — number of queries.
Each of the next t lines contains one string. The i-th line contains a string s_i consisting only of lowercase Latin letter. It is guaranteed that the length of s_i is from 1 to 1000 (inclusive).
Output
Print t lines. In the i-th line print the answer to the i-th query: -1 if it is impossible to obtain a good string by rearranging the letters of s_i and any good string which can be obtained from the given one (by rearranging the letters) otherwise.
Example
Input
3
aa
abacaba
xdd
Output
-1
abaacba
xdd
Note
In the first query we cannot rearrange letters to obtain a good string.
Other examples (not all) of correct answers to the second query: "ababaca", "abcabaa", "baacaba".
In the third query we can do nothing to obtain a good string.
Submitted Solution:
```
for cas in range(int(input())):
s = input()
s = list(s)
ln = len(s)
s.sort()
if s[0] == s[-1]:
print(-1)
else:print("".join(s))
```
Yes
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Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting only of lowercase Latin letters.
You can rearrange all letters of this string as you wish. Your task is to obtain a good string by rearranging the letters of the given string or report that it is impossible to do it.
Let's call a string good if it is not a palindrome. Palindrome is a string which is read from left to right the same as from right to left. For example, strings "abacaba", "aa" and "z" are palindromes and strings "bba", "xd" are not.
You have to answer t independent queries.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — number of queries.
Each of the next t lines contains one string. The i-th line contains a string s_i consisting only of lowercase Latin letter. It is guaranteed that the length of s_i is from 1 to 1000 (inclusive).
Output
Print t lines. In the i-th line print the answer to the i-th query: -1 if it is impossible to obtain a good string by rearranging the letters of s_i and any good string which can be obtained from the given one (by rearranging the letters) otherwise.
Example
Input
3
aa
abacaba
xdd
Output
-1
abaacba
xdd
Note
In the first query we cannot rearrange letters to obtain a good string.
Other examples (not all) of correct answers to the second query: "ababaca", "abcabaa", "baacaba".
In the third query we can do nothing to obtain a good string.
Submitted Solution:
```
n = int(input())
for i in range(n):
t= input()
if len(t) == t.count(t[0]):
print(-1)
elif t == t[::-1]:
for j in range(len(t)-1,-1,-1):
if t[j] != t[0]:
t = t[0] + t[-1] + t[2:len(t)-1] + t[j]
break
print(t)
else:
print(t)
```
No
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Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting only of lowercase Latin letters.
You can rearrange all letters of this string as you wish. Your task is to obtain a good string by rearranging the letters of the given string or report that it is impossible to do it.
Let's call a string good if it is not a palindrome. Palindrome is a string which is read from left to right the same as from right to left. For example, strings "abacaba", "aa" and "z" are palindromes and strings "bba", "xd" are not.
You have to answer t independent queries.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — number of queries.
Each of the next t lines contains one string. The i-th line contains a string s_i consisting only of lowercase Latin letter. It is guaranteed that the length of s_i is from 1 to 1000 (inclusive).
Output
Print t lines. In the i-th line print the answer to the i-th query: -1 if it is impossible to obtain a good string by rearranging the letters of s_i and any good string which can be obtained from the given one (by rearranging the letters) otherwise.
Example
Input
3
aa
abacaba
xdd
Output
-1
abaacba
xdd
Note
In the first query we cannot rearrange letters to obtain a good string.
Other examples (not all) of correct answers to the second query: "ababaca", "abcabaa", "baacaba".
In the third query we can do nothing to obtain a good string.
Submitted Solution:
```
for _ in range(int(input())):
s = input().strip()
if s == s[::-1] and len(set(s)) == 1 :
print(-1)
elif s != s[::-1]:
print(s)
else:
print(s[:len(s)//2] + s[len(s)//2:][::-1])
```
No
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Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting only of lowercase Latin letters.
You can rearrange all letters of this string as you wish. Your task is to obtain a good string by rearranging the letters of the given string or report that it is impossible to do it.
Let's call a string good if it is not a palindrome. Palindrome is a string which is read from left to right the same as from right to left. For example, strings "abacaba", "aa" and "z" are palindromes and strings "bba", "xd" are not.
You have to answer t independent queries.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — number of queries.
Each of the next t lines contains one string. The i-th line contains a string s_i consisting only of lowercase Latin letter. It is guaranteed that the length of s_i is from 1 to 1000 (inclusive).
Output
Print t lines. In the i-th line print the answer to the i-th query: -1 if it is impossible to obtain a good string by rearranging the letters of s_i and any good string which can be obtained from the given one (by rearranging the letters) otherwise.
Example
Input
3
aa
abacaba
xdd
Output
-1
abaacba
xdd
Note
In the first query we cannot rearrange letters to obtain a good string.
Other examples (not all) of correct answers to the second query: "ababaca", "abcabaa", "baacaba".
In the third query we can do nothing to obtain a good string.
Submitted Solution:
```
t = int(input())
answers = []
for i in range(0,t):
s = input()
res = ' '
x = s[0]
y = s[len(s)-1]
if (x != y):
res = s
else:
for i in range(1,len(s)):
if (s[i] != x and s[i] != y):
res = s[i] + x + s[2:i] + s[i+1:]
break
if (res == ' '):
answers.append(-1)
else:
answers.append(res)
for i in range(0, len(answers)):
print(answers[i])
```
No
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Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting only of lowercase Latin letters.
You can rearrange all letters of this string as you wish. Your task is to obtain a good string by rearranging the letters of the given string or report that it is impossible to do it.
Let's call a string good if it is not a palindrome. Palindrome is a string which is read from left to right the same as from right to left. For example, strings "abacaba", "aa" and "z" are palindromes and strings "bba", "xd" are not.
You have to answer t independent queries.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — number of queries.
Each of the next t lines contains one string. The i-th line contains a string s_i consisting only of lowercase Latin letter. It is guaranteed that the length of s_i is from 1 to 1000 (inclusive).
Output
Print t lines. In the i-th line print the answer to the i-th query: -1 if it is impossible to obtain a good string by rearranging the letters of s_i and any good string which can be obtained from the given one (by rearranging the letters) otherwise.
Example
Input
3
aa
abacaba
xdd
Output
-1
abaacba
xdd
Note
In the first query we cannot rearrange letters to obtain a good string.
Other examples (not all) of correct answers to the second query: "ababaca", "abcabaa", "baacaba".
In the third query we can do nothing to obtain a good string.
Submitted Solution:
```
t=int(input())
def reverse(string):
string = string[::-1]
return string
def palindrome(str):
rts=reverse(str)
if rts==(str):
return 1
else:
return 0
def special(str):
length=len(str)
a=str[0:int(length/2)]
b=str[length-1:int(length/2)-1:-1]
if palindrome(a+b)==0:
return a+b
else:
return special(a+b)
for i in range(t):
word=input()
change=-1
letter=word[0]
for char in word:
if char!=letter:
change=0
if len(word)==1 or len(word)==2 or change==-1:
print ("-1")
else:
if palindrome(word)==0:
print (word)
else:
print (special(word))
```
No
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Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between the easy and hard versions is that the given string s in the easy version is initially a palindrome, this condition is not always true for the hard version.
A palindrome is a string that reads the same left to right and right to left. For example, "101101" is a palindrome, while "0101" is not.
Alice and Bob are playing a game on a string s of length n consisting of the characters '0' and '1'. Both players take alternate turns with Alice going first.
In each turn, the player can perform one of the following operations:
1. Choose any i (1 ≤ i ≤ n), where s[i] = '0' and change s[i] to '1'. Pay 1 dollar.
2. Reverse the whole string, pay 0 dollars. This operation is only allowed if the string is currently not a palindrome, and the last operation was not reverse. That is, if Alice reverses the string, then Bob can't reverse in the next move, and vice versa.
Reversing a string means reordering its letters from the last to the first. For example, "01001" becomes "10010" after reversing.
The game ends when every character of string becomes '1'. The player who spends minimum dollars till this point wins the game and it is a draw if both spend equal dollars. If both players play optimally, output whether Alice wins, Bob wins, or if it is a draw.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^3). Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^3).
The second line of each test case contains the string s of length n, consisting of the characters '0' and '1'. It is guaranteed that the string s contains at least one '0'.
Note that there is no limit on the sum of n over test cases.
Output
For each test case print a single word in a new line:
* "ALICE", if Alice will win the game,
* "BOB", if Bob will win the game,
* "DRAW", if the game ends in a draw.
Example
Input
3
3
110
2
00
4
1010
Output
ALICE
BOB
ALICE
Note
In the first test case of example,
* in the 1-st move, Alice will use the 2-nd operation to reverse the string, since doing the 1-st operation will result in her loss anyway. This also forces Bob to use the 1-st operation.
* in the 2-nd move, Bob has to perform the 1-st operation, since the 2-nd operation cannot be performed twice in a row. All characters of the string are '1', game over.
Alice spends 0 dollars while Bob spends 1 dollar. Hence, Alice wins.
In the second test case of example,
* in the 1-st move Alice has to perform the 1-st operation, since the string is currently a palindrome.
* in the 2-nd move Bob reverses the string.
* in the 3-rd move Alice again has to perform the 1-st operation. All characters of the string are '1', game over.
Alice spends 2 dollars while Bob spends 0 dollars. Hence, Bob wins.
Tags: constructive algorithms, games
Correct Solution:
```
def palindrome(s):
n=len(s)
for i in range ((n+1)//2):
if s[i]!=s[n-i-1]:
return 0
return 1
def one(s):
n=len(s)
c=0
for i in range ((n+1)//2):
if s[i]!=s[n-i-1]:
c+=1
return c
for _ in range (int(input())):
n=int(input())
#a=list(map(int,input().split()))
s=input()
c=s.count('0')
if n%2 and s[n//2]=='0':
if c>2:
print("ALICE")
elif c==2:
print("DRAW")
else:
print("BOB")
elif palindrome(s):
print("BOB")
else:
print("ALICE")
```
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Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between the easy and hard versions is that the given string s in the easy version is initially a palindrome, this condition is not always true for the hard version.
A palindrome is a string that reads the same left to right and right to left. For example, "101101" is a palindrome, while "0101" is not.
Alice and Bob are playing a game on a string s of length n consisting of the characters '0' and '1'. Both players take alternate turns with Alice going first.
In each turn, the player can perform one of the following operations:
1. Choose any i (1 ≤ i ≤ n), where s[i] = '0' and change s[i] to '1'. Pay 1 dollar.
2. Reverse the whole string, pay 0 dollars. This operation is only allowed if the string is currently not a palindrome, and the last operation was not reverse. That is, if Alice reverses the string, then Bob can't reverse in the next move, and vice versa.
Reversing a string means reordering its letters from the last to the first. For example, "01001" becomes "10010" after reversing.
The game ends when every character of string becomes '1'. The player who spends minimum dollars till this point wins the game and it is a draw if both spend equal dollars. If both players play optimally, output whether Alice wins, Bob wins, or if it is a draw.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^3). Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^3).
The second line of each test case contains the string s of length n, consisting of the characters '0' and '1'. It is guaranteed that the string s contains at least one '0'.
Note that there is no limit on the sum of n over test cases.
Output
For each test case print a single word in a new line:
* "ALICE", if Alice will win the game,
* "BOB", if Bob will win the game,
* "DRAW", if the game ends in a draw.
Example
Input
3
3
110
2
00
4
1010
Output
ALICE
BOB
ALICE
Note
In the first test case of example,
* in the 1-st move, Alice will use the 2-nd operation to reverse the string, since doing the 1-st operation will result in her loss anyway. This also forces Bob to use the 1-st operation.
* in the 2-nd move, Bob has to perform the 1-st operation, since the 2-nd operation cannot be performed twice in a row. All characters of the string are '1', game over.
Alice spends 0 dollars while Bob spends 1 dollar. Hence, Alice wins.
In the second test case of example,
* in the 1-st move Alice has to perform the 1-st operation, since the string is currently a palindrome.
* in the 2-nd move Bob reverses the string.
* in the 3-rd move Alice again has to perform the 1-st operation. All characters of the string are '1', game over.
Alice spends 2 dollars while Bob spends 0 dollars. Hence, Bob wins.
Tags: constructive algorithms, games
Correct Solution:
```
for i in range(int(input())):
n = int(input())
s = input()
if s == s[::-1]:
print('DRAW' if n == s.count('1') else('BOB' if s.count('0') % 2 == 0 or s.count('0') == 1 else 'ALICE'))
else:
print('DRAW' if s.count('0') == 2 and n % 2 == 1 and s[n//2] == '0' else 'ALICE')
```
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Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between the easy and hard versions is that the given string s in the easy version is initially a palindrome, this condition is not always true for the hard version.
A palindrome is a string that reads the same left to right and right to left. For example, "101101" is a palindrome, while "0101" is not.
Alice and Bob are playing a game on a string s of length n consisting of the characters '0' and '1'. Both players take alternate turns with Alice going first.
In each turn, the player can perform one of the following operations:
1. Choose any i (1 ≤ i ≤ n), where s[i] = '0' and change s[i] to '1'. Pay 1 dollar.
2. Reverse the whole string, pay 0 dollars. This operation is only allowed if the string is currently not a palindrome, and the last operation was not reverse. That is, if Alice reverses the string, then Bob can't reverse in the next move, and vice versa.
Reversing a string means reordering its letters from the last to the first. For example, "01001" becomes "10010" after reversing.
The game ends when every character of string becomes '1'. The player who spends minimum dollars till this point wins the game and it is a draw if both spend equal dollars. If both players play optimally, output whether Alice wins, Bob wins, or if it is a draw.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^3). Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^3).
The second line of each test case contains the string s of length n, consisting of the characters '0' and '1'. It is guaranteed that the string s contains at least one '0'.
Note that there is no limit on the sum of n over test cases.
Output
For each test case print a single word in a new line:
* "ALICE", if Alice will win the game,
* "BOB", if Bob will win the game,
* "DRAW", if the game ends in a draw.
Example
Input
3
3
110
2
00
4
1010
Output
ALICE
BOB
ALICE
Note
In the first test case of example,
* in the 1-st move, Alice will use the 2-nd operation to reverse the string, since doing the 1-st operation will result in her loss anyway. This also forces Bob to use the 1-st operation.
* in the 2-nd move, Bob has to perform the 1-st operation, since the 2-nd operation cannot be performed twice in a row. All characters of the string are '1', game over.
Alice spends 0 dollars while Bob spends 1 dollar. Hence, Alice wins.
In the second test case of example,
* in the 1-st move Alice has to perform the 1-st operation, since the string is currently a palindrome.
* in the 2-nd move Bob reverses the string.
* in the 3-rd move Alice again has to perform the 1-st operation. All characters of the string are '1', game over.
Alice spends 2 dollars while Bob spends 0 dollars. Hence, Bob wins.
Tags: constructive algorithms, games
Correct Solution:
```
import sys
input = sys.stdin.readline
for _ in range(int(input())):
n = int(input())
S = list(input())[: -1]
zero = S.count("0")
one = n - zero
diff = 0
for i in range(n): diff += S[i] != S[-1 - i]
if diff == 0:
if zero == 1:
print("BOB")
continue
if zero % 2 == 0:
print("BOB")
continue
else:
print("ALICE")
continue
if diff == 2:
if zero == 1:
print("ALICE")
continue
if zero == 2:
print("DRAW")
continue
if zero % 2 == 0:
print("ALICE")
continue
else:
print("ALICE")
continue
if diff > 2:
print("ALICE")
continue
```
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Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between the easy and hard versions is that the given string s in the easy version is initially a palindrome, this condition is not always true for the hard version.
A palindrome is a string that reads the same left to right and right to left. For example, "101101" is a palindrome, while "0101" is not.
Alice and Bob are playing a game on a string s of length n consisting of the characters '0' and '1'. Both players take alternate turns with Alice going first.
In each turn, the player can perform one of the following operations:
1. Choose any i (1 ≤ i ≤ n), where s[i] = '0' and change s[i] to '1'. Pay 1 dollar.
2. Reverse the whole string, pay 0 dollars. This operation is only allowed if the string is currently not a palindrome, and the last operation was not reverse. That is, if Alice reverses the string, then Bob can't reverse in the next move, and vice versa.
Reversing a string means reordering its letters from the last to the first. For example, "01001" becomes "10010" after reversing.
The game ends when every character of string becomes '1'. The player who spends minimum dollars till this point wins the game and it is a draw if both spend equal dollars. If both players play optimally, output whether Alice wins, Bob wins, or if it is a draw.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^3). Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^3).
The second line of each test case contains the string s of length n, consisting of the characters '0' and '1'. It is guaranteed that the string s contains at least one '0'.
Note that there is no limit on the sum of n over test cases.
Output
For each test case print a single word in a new line:
* "ALICE", if Alice will win the game,
* "BOB", if Bob will win the game,
* "DRAW", if the game ends in a draw.
Example
Input
3
3
110
2
00
4
1010
Output
ALICE
BOB
ALICE
Note
In the first test case of example,
* in the 1-st move, Alice will use the 2-nd operation to reverse the string, since doing the 1-st operation will result in her loss anyway. This also forces Bob to use the 1-st operation.
* in the 2-nd move, Bob has to perform the 1-st operation, since the 2-nd operation cannot be performed twice in a row. All characters of the string are '1', game over.
Alice spends 0 dollars while Bob spends 1 dollar. Hence, Alice wins.
In the second test case of example,
* in the 1-st move Alice has to perform the 1-st operation, since the string is currently a palindrome.
* in the 2-nd move Bob reverses the string.
* in the 3-rd move Alice again has to perform the 1-st operation. All characters of the string are '1', game over.
Alice spends 2 dollars while Bob spends 0 dollars. Hence, Bob wins.
Tags: constructive algorithms, games
Correct Solution:
```
def getResult(n,s):
count = 0
for c in s:
if c == "0":
count +=1
v = check(s)
if v == 0:
if count%2==0 or count == 1:
return 'BOB'
else:
return "ALICE"
elif v == 1:
count -= 1
if count==0:
return 'ALICE'
elif count == 1:
return "DRAW"
elif count%2 == 0:
return "ALICE"
else:
return "ALICE"
else:
return "ALICE"
def check(s):
l = len(s)
v = 0
for i in range(l//2):
if s[i]!=s[l-i-1]:
v += 1
return v
t = int(input())
for _ in range(t):
n = int(input())
s = input()
print(getResult(n,s))
```
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Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between the easy and hard versions is that the given string s in the easy version is initially a palindrome, this condition is not always true for the hard version.
A palindrome is a string that reads the same left to right and right to left. For example, "101101" is a palindrome, while "0101" is not.
Alice and Bob are playing a game on a string s of length n consisting of the characters '0' and '1'. Both players take alternate turns with Alice going first.
In each turn, the player can perform one of the following operations:
1. Choose any i (1 ≤ i ≤ n), where s[i] = '0' and change s[i] to '1'. Pay 1 dollar.
2. Reverse the whole string, pay 0 dollars. This operation is only allowed if the string is currently not a palindrome, and the last operation was not reverse. That is, if Alice reverses the string, then Bob can't reverse in the next move, and vice versa.
Reversing a string means reordering its letters from the last to the first. For example, "01001" becomes "10010" after reversing.
The game ends when every character of string becomes '1'. The player who spends minimum dollars till this point wins the game and it is a draw if both spend equal dollars. If both players play optimally, output whether Alice wins, Bob wins, or if it is a draw.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^3). Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^3).
The second line of each test case contains the string s of length n, consisting of the characters '0' and '1'. It is guaranteed that the string s contains at least one '0'.
Note that there is no limit on the sum of n over test cases.
Output
For each test case print a single word in a new line:
* "ALICE", if Alice will win the game,
* "BOB", if Bob will win the game,
* "DRAW", if the game ends in a draw.
Example
Input
3
3
110
2
00
4
1010
Output
ALICE
BOB
ALICE
Note
In the first test case of example,
* in the 1-st move, Alice will use the 2-nd operation to reverse the string, since doing the 1-st operation will result in her loss anyway. This also forces Bob to use the 1-st operation.
* in the 2-nd move, Bob has to perform the 1-st operation, since the 2-nd operation cannot be performed twice in a row. All characters of the string are '1', game over.
Alice spends 0 dollars while Bob spends 1 dollar. Hence, Alice wins.
In the second test case of example,
* in the 1-st move Alice has to perform the 1-st operation, since the string is currently a palindrome.
* in the 2-nd move Bob reverses the string.
* in the 3-rd move Alice again has to perform the 1-st operation. All characters of the string are '1', game over.
Alice spends 2 dollars while Bob spends 0 dollars. Hence, Bob wins.
Tags: constructive algorithms, games
Correct Solution:
```
def divisors(M):
d=[]
i=1
while M>=i**2:
if M%i==0:
d.append(i)
if i**2!=M:
d.append(M//i)
i=i+1
return d
def popcount(x):
x = x - ((x >> 1) & 0x55555555)
x = (x & 0x33333333) + ((x >> 2) & 0x33333333)
x = (x + (x >> 4)) & 0x0f0f0f0f
x = x + (x >> 8)
x = x + (x >> 16)
return x & 0x0000007f
def eratosthenes(n):
res=[0 for i in range(n+1)]
prime=set([])
for i in range(2,n+1):
if not res[i]:
prime.add(i)
for j in range(1,n//i+1):
res[i*j]=1
return prime
def factorization(n):
res=[]
for p in prime:
if n%p==0:
while n%p==0:
n//=p
res.append(p)
if n!=1:
res.append(n)
return res
def euler_phi(n):
res = n
for x in range(2,n+1):
if x ** 2 > n:
break
if n%x==0:
res = res//x * (x-1)
while n%x==0:
n //= x
if n!=1:
res = res//n * (n-1)
return res
def ind(b,n):
res=0
while n%b==0:
res+=1
n//=b
return res
def isPrimeMR(n):
d = n - 1
d = d // (d & -d)
L = [2, 3, 5, 7, 11, 13, 17]
for a in L:
t = d
y = pow(a, t, n)
if y == 1: continue
while y != n - 1:
y = (y * y) % n
if y == 1 or t == n - 1: return 0
t <<= 1
return 1
def findFactorRho(n):
from math import gcd
m = 1 << n.bit_length() // 8
for c in range(1, 99):
f = lambda x: (x * x + c) % n
y, r, q, g = 2, 1, 1, 1
while g == 1:
x = y
for i in range(r):
y = f(y)
k = 0
while k < r and g == 1:
ys = y
for i in range(min(m, r - k)):
y = f(y)
q = q * abs(x - y) % n
g = gcd(q, n)
k += m
r <<= 1
if g == n:
g = 1
while g == 1:
ys = f(ys)
g = gcd(abs(x - ys), n)
if g < n:
if isPrimeMR(g): return g
elif isPrimeMR(n // g): return n // g
return findFactorRho(g)
def primeFactor(n):
i = 2
ret = {}
rhoFlg = 0
while i*i <= n:
k = 0
while n % i == 0:
n //= i
k += 1
if k: ret[i] = k
i += 1 + i % 2
if i == 101 and n >= 2 ** 20:
while n > 1:
if isPrimeMR(n):
ret[n], n = 1, 1
else:
rhoFlg = 1
j = findFactorRho(n)
k = 0
while n % j == 0:
n //= j
k += 1
ret[j] = k
if n > 1: ret[n] = 1
if rhoFlg: ret = {x: ret[x] for x in sorted(ret)}
return ret
def divisors(n):
res = [1]
prime = primeFactor(n)
for p in prime:
newres = []
for d in res:
for j in range(prime[p]+1):
newres.append(d*p**j)
res = newres
res.sort()
return res
def xorfactorial(num):
if num==0:
return 0
elif num==1:
return 1
elif num==2:
return 3
elif num==3:
return 0
else:
x=baseorder(num)
return (2**x)*((num-2**x+1)%2)+function(num-2**x)
def xorconv(n,X,Y):
if n==0:
res=[(X[0]*Y[0])%mod]
return res
x=[X[i]+X[i+2**(n-1)] for i in range(2**(n-1))]
y=[Y[i]+Y[i+2**(n-1)] for i in range(2**(n-1))]
z=[X[i]-X[i+2**(n-1)] for i in range(2**(n-1))]
w=[Y[i]-Y[i+2**(n-1)] for i in range(2**(n-1))]
res1=xorconv(n-1,x,y)
res2=xorconv(n-1,z,w)
former=[(res1[i]+res2[i])*inv for i in range(2**(n-1))]
latter=[(res1[i]-res2[i])*inv for i in range(2**(n-1))]
former=list(map(lambda x:x%mod,former))
latter=list(map(lambda x:x%mod,latter))
return former+latter
def merge_sort(A,B):
pos_A,pos_B = 0,0
n,m = len(A),len(B)
res = []
while pos_A < n and pos_B < m:
a,b = A[pos_A],B[pos_B]
if a < b:
res.append(a)
pos_A += 1
else:
res.append(b)
pos_B += 1
res += A[pos_A:]
res += B[pos_B:]
return res
class UnionFindVerSize():
def __init__(self, N):
self._parent = [n for n in range(0, N)]
self._size = [1] * N
self.group = N
def find_root(self, x):
if self._parent[x] == x: return x
self._parent[x] = self.find_root(self._parent[x])
stack = [x]
while self._parent[stack[-1]]!=stack[-1]:
stack.append(self._parent[stack[-1]])
for v in stack:
self._parent[v] = stack[-1]
return self._parent[x]
def unite(self, x, y):
gx = self.find_root(x)
gy = self.find_root(y)
if gx == gy: return
self.group -= 1
if self._size[gx] < self._size[gy]:
self._parent[gx] = gy
self._size[gy] += self._size[gx]
else:
self._parent[gy] = gx
self._size[gx] += self._size[gy]
def get_size(self, x):
return self._size[self.find_root(x)]
def is_same_group(self, x, y):
return self.find_root(x) == self.find_root(y)
class WeightedUnionFind():
def __init__(self,N):
self.parent = [i for i in range(N)]
self.size = [1 for i in range(N)]
self.val = [0 for i in range(N)]
self.flag = True
self.edge = [[] for i in range(N)]
def dfs(self,v,pv):
stack = [(v,pv)]
new_parent = self.parent[pv]
while stack:
v,pv = stack.pop()
self.parent[v] = new_parent
for nv,w in self.edge[v]:
if nv!=pv:
self.val[nv] = self.val[v] + w
stack.append((nv,v))
def unite(self,x,y,w):
if not self.flag:
return
if self.parent[x]==self.parent[y]:
self.flag = (self.val[x] - self.val[y] == w)
return
if self.size[self.parent[x]]>self.size[self.parent[y]]:
self.edge[x].append((y,-w))
self.edge[y].append((x,w))
self.size[x] += self.size[y]
self.val[y] = self.val[x] - w
self.dfs(y,x)
else:
self.edge[x].append((y,-w))
self.edge[y].append((x,w))
self.size[y] += self.size[x]
self.val[x] = self.val[y] + w
self.dfs(x,y)
class Dijkstra():
class Edge():
def __init__(self, _to, _cost):
self.to = _to
self.cost = _cost
def __init__(self, V):
self.G = [[] for i in range(V)]
self._E = 0
self._V = V
@property
def E(self):
return self._E
@property
def V(self):
return self._V
def add_edge(self, _from, _to, _cost):
self.G[_from].append(self.Edge(_to, _cost))
self._E += 1
def shortest_path(self, s):
import heapq
que = []
d = [10**15] * self.V
d[s] = 0
heapq.heappush(que, (0, s))
while len(que) != 0:
cost, v = heapq.heappop(que)
if d[v] < cost: continue
for i in range(len(self.G[v])):
e = self.G[v][i]
if d[e.to] > d[v] + e.cost:
d[e.to] = d[v] + e.cost
heapq.heappush(que, (d[e.to], e.to))
return d
#Z[i]:length of the longest list starting from S[i] which is also a prefix of S
#O(|S|)
def Z_algorithm(s):
N = len(s)
Z_alg = [0]*N
Z_alg[0] = N
i = 1
j = 0
while i < N:
while i+j < N and s[j] == s[i+j]:
j += 1
Z_alg[i] = j
if j == 0:
i += 1
continue
k = 1
while i+k < N and k + Z_alg[k]<j:
Z_alg[i+k] = Z_alg[k]
k += 1
i += k
j -= k
return Z_alg
class BIT():
def __init__(self,n,mod=0):
self.BIT = [0]*(n+1)
self.num = n
self.mod = mod
def query(self,idx):
res_sum = 0
mod = self.mod
while idx > 0:
res_sum += self.BIT[idx]
if mod:
res_sum %= mod
idx -= idx&(-idx)
return res_sum
#Ai += x O(logN)
def update(self,idx,x):
mod = self.mod
while idx <= self.num:
self.BIT[idx] += x
if mod:
self.BIT[idx] %= mod
idx += idx&(-idx)
return
class dancinglink():
def __init__(self,n,debug=False):
self.n = n
self.debug = debug
self._left = [i-1 for i in range(n)]
self._right = [i+1 for i in range(n)]
self.exist = [True for i in range(n)]
def pop(self,k):
if self.debug:
assert self.exist[k]
L = self._left[k]
R = self._right[k]
if L!=-1:
if R!=self.n:
self._right[L],self._left[R] = R,L
else:
self._right[L] = self.n
elif R!=self.n:
self._left[R] = -1
self.exist[k] = False
def left(self,idx,k=1):
if self.debug:
assert self.exist[idx]
res = idx
while k:
res = self._left[res]
if res==-1:
break
k -= 1
return res
def right(self,idx,k=1):
if self.debug:
assert self.exist[idx]
res = idx
while k:
res = self._right[res]
if res==self.n:
break
k -= 1
return res
class SparseTable():
def __init__(self,A,merge_func,ide_ele):
N=len(A)
n=N.bit_length()
self.table=[[ide_ele for i in range(n)] for i in range(N)]
self.merge_func=merge_func
for i in range(N):
self.table[i][0]=A[i]
for j in range(1,n):
for i in range(0,N-2**j+1):
f=self.table[i][j-1]
s=self.table[i+2**(j-1)][j-1]
self.table[i][j]=self.merge_func(f,s)
def query(self,s,t):
b=t-s+1
m=b.bit_length()-1
return self.merge_func(self.table[s][m],self.table[t-2**m+1][m])
class BinaryTrie:
class node:
def __init__(self,val):
self.left = None
self.right = None
self.max = val
def __init__(self):
self.root = self.node(-10**15)
def append(self,key,val):
pos = self.root
for i in range(29,-1,-1):
pos.max = max(pos.max,val)
if key>>i & 1:
if pos.right is None:
pos.right = self.node(val)
pos = pos.right
else:
pos = pos.right
else:
if pos.left is None:
pos.left = self.node(val)
pos = pos.left
else:
pos = pos.left
pos.max = max(pos.max,val)
def search(self,M,xor):
res = -10**15
pos = self.root
for i in range(29,-1,-1):
if pos is None:
break
if M>>i & 1:
if xor>>i & 1:
if pos.right:
res = max(res,pos.right.max)
pos = pos.left
else:
if pos.left:
res = max(res,pos.left.max)
pos = pos.right
else:
if xor>>i & 1:
pos = pos.right
else:
pos = pos.left
if pos:
res = max(res,pos.max)
return res
def solveequation(edge,ans,n,m):
#edge=[[to,dire,id]...]
x=[0]*m
used=[False]*n
for v in range(n):
if used[v]:
continue
y = dfs(v)
if y!=0:
return False
return x
def dfs(v):
used[v]=True
r=ans[v]
for to,dire,id in edge[v]:
if used[to]:
continue
y=dfs(to)
if dire==-1:
x[id]=y
else:
x[id]=-y
r+=y
return r
class SegmentTree:
def __init__(self, init_val, segfunc, ide_ele):
n = len(init_val)
self.segfunc = segfunc
self.ide_ele = ide_ele
self.num = 1 << (n - 1).bit_length()
self.tree = [ide_ele] * 2 * self.num
self.size = n
for i in range(n):
self.tree[self.num + i] = init_val[i]
for i in range(self.num - 1, 0, -1):
self.tree[i] = self.segfunc(self.tree[2 * i], self.tree[2 * i + 1])
def update(self, k, x):
k += self.num
self.tree[k] = x
while k > 1:
self.tree[k >> 1] = self.segfunc(self.tree[k], self.tree[k ^ 1])
k >>= 1
def query(self, l, r):
if r==self.size:
r = self.num
res = self.ide_ele
l += self.num
r += self.num
while l < r:
if l & 1:
res = self.segfunc(res, self.tree[l])
l += 1
if r & 1:
res = self.segfunc(res, self.tree[r - 1])
l >>= 1
r >>= 1
return res
def bisect_l(self,l,r,x):
l += self.num
r += self.num
Lmin = -1
Rmin = -1
while l<r:
if l & 1:
if self.tree[l] <= x and Lmin==-1:
Lmin = l
l += 1
if r & 1:
if self.tree[r-1] <=x:
Rmin = r-1
l >>= 1
r >>= 1
if Lmin != -1:
pos = Lmin
while pos<self.num:
if self.tree[2 * pos] <=x:
pos = 2 * pos
else:
pos = 2 * pos +1
return pos-self.num
elif Rmin != -1:
pos = Rmin
while pos<self.num:
if self.tree[2 * pos] <=x:
pos = 2 * pos
else:
pos = 2 * pos +1
return pos-self.num
else:
return -1
import sys,random,bisect
from collections import deque,defaultdict
from heapq import heapify,heappop,heappush
from itertools import permutations
from math import gcd,log
input = lambda :sys.stdin.readline().rstrip()
mi = lambda :map(int,input().split())
li = lambda :list(mi())
for _ in range(int(input())):
n = int(input())
s = input()
if n%2==0:
a,b = 0,0
for i in range(n//2):
if s[i]=="0" and s[-i-1]=="0":
a += 1
elif (s[i],s[-i-1])!=("1","1"):
b += 1
if a!=0:
if b!=0:
print("ALICE")
else:
print("BOB")
else:
print("ALICE")
else:
if s[n//2]=="1":
a,b = 0,0
for i in range(n//2):
if s[i]=="0" and s[-i-1]=="0":
a += 1
elif (s[i],s[-i-1])!=("1","1"):
b += 1
if a!=0:
if b!=0:
print("ALICE")
else:
print("BOB")
else:
print("ALICE")
else:
a,b = 0,0
for i in range(n//2):
if s[i]=="0" and s[-i-1]=="0":
a += 1
elif (s[i],s[-i-1])!=("1","1"):
b += 1
if a==0:
if b==0:
print("BOB")
elif b==1:
print("DRAW")
else:
print("ALICE")
else:
print("ALICE")
```
| 37,067 | [
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0.92822265625,
1.0068359375,
0.0689697265625,
0.332763671875,
-0.560546875,
-0.73193359375,
-0.2451171875,
-0.91064453125,
... | 0 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between the easy and hard versions is that the given string s in the easy version is initially a palindrome, this condition is not always true for the hard version.
A palindrome is a string that reads the same left to right and right to left. For example, "101101" is a palindrome, while "0101" is not.
Alice and Bob are playing a game on a string s of length n consisting of the characters '0' and '1'. Both players take alternate turns with Alice going first.
In each turn, the player can perform one of the following operations:
1. Choose any i (1 ≤ i ≤ n), where s[i] = '0' and change s[i] to '1'. Pay 1 dollar.
2. Reverse the whole string, pay 0 dollars. This operation is only allowed if the string is currently not a palindrome, and the last operation was not reverse. That is, if Alice reverses the string, then Bob can't reverse in the next move, and vice versa.
Reversing a string means reordering its letters from the last to the first. For example, "01001" becomes "10010" after reversing.
The game ends when every character of string becomes '1'. The player who spends minimum dollars till this point wins the game and it is a draw if both spend equal dollars. If both players play optimally, output whether Alice wins, Bob wins, or if it is a draw.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^3). Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^3).
The second line of each test case contains the string s of length n, consisting of the characters '0' and '1'. It is guaranteed that the string s contains at least one '0'.
Note that there is no limit on the sum of n over test cases.
Output
For each test case print a single word in a new line:
* "ALICE", if Alice will win the game,
* "BOB", if Bob will win the game,
* "DRAW", if the game ends in a draw.
Example
Input
3
3
110
2
00
4
1010
Output
ALICE
BOB
ALICE
Note
In the first test case of example,
* in the 1-st move, Alice will use the 2-nd operation to reverse the string, since doing the 1-st operation will result in her loss anyway. This also forces Bob to use the 1-st operation.
* in the 2-nd move, Bob has to perform the 1-st operation, since the 2-nd operation cannot be performed twice in a row. All characters of the string are '1', game over.
Alice spends 0 dollars while Bob spends 1 dollar. Hence, Alice wins.
In the second test case of example,
* in the 1-st move Alice has to perform the 1-st operation, since the string is currently a palindrome.
* in the 2-nd move Bob reverses the string.
* in the 3-rd move Alice again has to perform the 1-st operation. All characters of the string are '1', game over.
Alice spends 2 dollars while Bob spends 0 dollars. Hence, Bob wins.
Tags: constructive algorithms, games
Correct Solution:
```
# Author : nitish420 --------------------------------------------------------------------
import os
import sys
from io import BufferedIOBase, BytesIO, IOBase
def ispal(arr,n):
l=0
r=n-1
count=0
while l<r:
if arr[l]==arr[r]:
pass
else:
arr[l]=1
arr[r]=1
count+=1
l+=1
r-=1
return count
def main():
for _ in range(int(input())):
n=int(input())
arr=list(map(int,input()))
count=ispal(arr,n)
if count:
bob=count
zeroes=arr.count(0)
if n%2 and arr[n//2]==0:
if bob==1 and zeroes==1:
print("DRAW")
else:
print("ALICE")
else:
print("ALICE")
else:
zeroes=arr.count(0)
if n%2 and arr[n//2]==0:
zeroes-=1
if zeroes:
print("ALICE")
else:
print("BOB")
else:
print("BOB")
#----------------------------------------------------------------------------------------
def nouse0():
# This is to save my code from plag due to use of FAST IO template in it.
a=420
b=420
print(f'i am nitish{(a+b)//2}')
def nouse1():
# This is to save my code from plag due to use of FAST IO template in it.
a=420
b=420
print(f'i am nitish{(a+b)//2}')
def nouse2():
# This is to save my code from plag due to use of FAST IO template in it.
a=420
b=420
print(f'i am nitish{(a+b)//2}')
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = 'x' in file.mode or 'r' not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b'\n') + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode('ascii'))
self.read = lambda: self.buffer.read().decode('ascii')
self.readline = lambda: self.buffer.readline().decode('ascii')
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
def nouse3():
# This is to save my code from plag due to use of FAST IO template in it.
a=420
b=420
print(f'i am nitish{(a+b)//2}')
def nouse4():
# This is to save my code from plag due to use of FAST IO template in it.
a=420
b=420
print(f'i am nitish{(a+b)//2}')
def nouse5():
# This is to save my code from plag due to use of FAST IO template in it.
a=420
b=420
print(f'i am nitish{(a+b)//2}')
# endregion
if __name__ == '__main__':
main()
```
| 37,068 | [
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-0.07318115234375,
0.92822265625,
1.0068359375,
0.0689697265625,
0.332763671875,
-0.560546875,
-0.73193359375,
-0.2451171875,
-0.91064453125,
... | 0 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between the easy and hard versions is that the given string s in the easy version is initially a palindrome, this condition is not always true for the hard version.
A palindrome is a string that reads the same left to right and right to left. For example, "101101" is a palindrome, while "0101" is not.
Alice and Bob are playing a game on a string s of length n consisting of the characters '0' and '1'. Both players take alternate turns with Alice going first.
In each turn, the player can perform one of the following operations:
1. Choose any i (1 ≤ i ≤ n), where s[i] = '0' and change s[i] to '1'. Pay 1 dollar.
2. Reverse the whole string, pay 0 dollars. This operation is only allowed if the string is currently not a palindrome, and the last operation was not reverse. That is, if Alice reverses the string, then Bob can't reverse in the next move, and vice versa.
Reversing a string means reordering its letters from the last to the first. For example, "01001" becomes "10010" after reversing.
The game ends when every character of string becomes '1'. The player who spends minimum dollars till this point wins the game and it is a draw if both spend equal dollars. If both players play optimally, output whether Alice wins, Bob wins, or if it is a draw.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^3). Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^3).
The second line of each test case contains the string s of length n, consisting of the characters '0' and '1'. It is guaranteed that the string s contains at least one '0'.
Note that there is no limit on the sum of n over test cases.
Output
For each test case print a single word in a new line:
* "ALICE", if Alice will win the game,
* "BOB", if Bob will win the game,
* "DRAW", if the game ends in a draw.
Example
Input
3
3
110
2
00
4
1010
Output
ALICE
BOB
ALICE
Note
In the first test case of example,
* in the 1-st move, Alice will use the 2-nd operation to reverse the string, since doing the 1-st operation will result in her loss anyway. This also forces Bob to use the 1-st operation.
* in the 2-nd move, Bob has to perform the 1-st operation, since the 2-nd operation cannot be performed twice in a row. All characters of the string are '1', game over.
Alice spends 0 dollars while Bob spends 1 dollar. Hence, Alice wins.
In the second test case of example,
* in the 1-st move Alice has to perform the 1-st operation, since the string is currently a palindrome.
* in the 2-nd move Bob reverses the string.
* in the 3-rd move Alice again has to perform the 1-st operation. All characters of the string are '1', game over.
Alice spends 2 dollars while Bob spends 0 dollars. Hence, Bob wins.
Tags: constructive algorithms, games
Correct Solution:
```
#------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
# from fractions import *
from heapq import *
from bisect import *
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
ALPHA='abcde'
M = 10**9 + 7
EPS = 1e-6
def Ceil(a,b): return a//b+int(a%b>0)
def INT():return int(input())
def STR():return input()
def INTs():return tuple(map(int,input().split()))
def ARRINT():return [int(i) for i in input().split()]
def ARRSTR():return [i for i in input().split()]
#-------------------------code---------------------------#
def solve(S):
n = len(S)
mid=0
X=0
Y=0
if n%2 and S[n//2]=="0":
mid=1
for i in range(n//2):
if S[i]=="0" and S[n-1-i]=="0":
X+=1
for i in range(n):
if S[i]=="0" and S[n-1-i]=="1":
Y+=1
if Y==0:
if X==0:
if mid==0:
print("DRAW")
else:
print("BOB")
else:
if mid==0:
print("BOB")
else:
print("ALICE")
elif Y==1:
if X==0 and mid==1:
print("DRAW")
elif mid==0:
print("ALICE")
else:
print("ALICE")
else:
print("ALICE")
for _ in range(INT()):
_ = INT()
solve(STR())
```
| 37,069 | [
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0.92822265625,
1.0068359375,
0.0689697265625,
0.332763671875,
-0.560546875,
-0.73193359375,
-0.2451171875,
-0.91064453125,
... | 0 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between the easy and hard versions is that the given string s in the easy version is initially a palindrome, this condition is not always true for the hard version.
A palindrome is a string that reads the same left to right and right to left. For example, "101101" is a palindrome, while "0101" is not.
Alice and Bob are playing a game on a string s of length n consisting of the characters '0' and '1'. Both players take alternate turns with Alice going first.
In each turn, the player can perform one of the following operations:
1. Choose any i (1 ≤ i ≤ n), where s[i] = '0' and change s[i] to '1'. Pay 1 dollar.
2. Reverse the whole string, pay 0 dollars. This operation is only allowed if the string is currently not a palindrome, and the last operation was not reverse. That is, if Alice reverses the string, then Bob can't reverse in the next move, and vice versa.
Reversing a string means reordering its letters from the last to the first. For example, "01001" becomes "10010" after reversing.
The game ends when every character of string becomes '1'. The player who spends minimum dollars till this point wins the game and it is a draw if both spend equal dollars. If both players play optimally, output whether Alice wins, Bob wins, or if it is a draw.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^3). Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^3).
The second line of each test case contains the string s of length n, consisting of the characters '0' and '1'. It is guaranteed that the string s contains at least one '0'.
Note that there is no limit on the sum of n over test cases.
Output
For each test case print a single word in a new line:
* "ALICE", if Alice will win the game,
* "BOB", if Bob will win the game,
* "DRAW", if the game ends in a draw.
Example
Input
3
3
110
2
00
4
1010
Output
ALICE
BOB
ALICE
Note
In the first test case of example,
* in the 1-st move, Alice will use the 2-nd operation to reverse the string, since doing the 1-st operation will result in her loss anyway. This also forces Bob to use the 1-st operation.
* in the 2-nd move, Bob has to perform the 1-st operation, since the 2-nd operation cannot be performed twice in a row. All characters of the string are '1', game over.
Alice spends 0 dollars while Bob spends 1 dollar. Hence, Alice wins.
In the second test case of example,
* in the 1-st move Alice has to perform the 1-st operation, since the string is currently a palindrome.
* in the 2-nd move Bob reverses the string.
* in the 3-rd move Alice again has to perform the 1-st operation. All characters of the string are '1', game over.
Alice spends 2 dollars while Bob spends 0 dollars. Hence, Bob wins.
Tags: constructive algorithms, games
Correct Solution:
```
#b1
for _ in range(int(input())):
input()
inp=str(input())
zz=0
zo=0
for i in range(len(inp)):
if inp[i]==inp[-i-1] and inp[i]=="0":
zz+=1
if inp[i]!=inp[-i-1]:
zo+=1
win=0
zo/=2
if zo==0:
if (zz%2==0 and zz>=2) or zz==1:
win=-1
if zz%2==1 and zz>1:
win=1
else:
if not(zz==zo and zz==1):
win=1
if win==1:
print("ALICE")
if win==-1:
print("BOB")
if win==0:
print("DRAW")
```
| 37,070 | [
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-0.07318115234375,
0.92822265625,
1.0068359375,
0.0689697265625,
0.332763671875,
-0.560546875,
-0.73193359375,
-0.2451171875,
-0.91064453125,
... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between the easy and hard versions is that the given string s in the easy version is initially a palindrome, this condition is not always true for the hard version.
A palindrome is a string that reads the same left to right and right to left. For example, "101101" is a palindrome, while "0101" is not.
Alice and Bob are playing a game on a string s of length n consisting of the characters '0' and '1'. Both players take alternate turns with Alice going first.
In each turn, the player can perform one of the following operations:
1. Choose any i (1 ≤ i ≤ n), where s[i] = '0' and change s[i] to '1'. Pay 1 dollar.
2. Reverse the whole string, pay 0 dollars. This operation is only allowed if the string is currently not a palindrome, and the last operation was not reverse. That is, if Alice reverses the string, then Bob can't reverse in the next move, and vice versa.
Reversing a string means reordering its letters from the last to the first. For example, "01001" becomes "10010" after reversing.
The game ends when every character of string becomes '1'. The player who spends minimum dollars till this point wins the game and it is a draw if both spend equal dollars. If both players play optimally, output whether Alice wins, Bob wins, or if it is a draw.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^3). Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^3).
The second line of each test case contains the string s of length n, consisting of the characters '0' and '1'. It is guaranteed that the string s contains at least one '0'.
Note that there is no limit on the sum of n over test cases.
Output
For each test case print a single word in a new line:
* "ALICE", if Alice will win the game,
* "BOB", if Bob will win the game,
* "DRAW", if the game ends in a draw.
Example
Input
3
3
110
2
00
4
1010
Output
ALICE
BOB
ALICE
Note
In the first test case of example,
* in the 1-st move, Alice will use the 2-nd operation to reverse the string, since doing the 1-st operation will result in her loss anyway. This also forces Bob to use the 1-st operation.
* in the 2-nd move, Bob has to perform the 1-st operation, since the 2-nd operation cannot be performed twice in a row. All characters of the string are '1', game over.
Alice spends 0 dollars while Bob spends 1 dollar. Hence, Alice wins.
In the second test case of example,
* in the 1-st move Alice has to perform the 1-st operation, since the string is currently a palindrome.
* in the 2-nd move Bob reverses the string.
* in the 3-rd move Alice again has to perform the 1-st operation. All characters of the string are '1', game over.
Alice spends 2 dollars while Bob spends 0 dollars. Hence, Bob wins.
Submitted Solution:
```
import sys
#input = sys.stdin.readline
for _ in range(int(input())):
n=int(input())
s=input()
p=0
for i in range(n//2):
if s[i]!=s[-i-1]:
p+=1
if p==0:
o=s.count('0')
if o%2:
#o-=1
if o==1:
print('BOB')
else:
print('ALICE')
else:
if (o//2)%2==0:
print('BOB')
else:
print('BOB')
else:
o=s.count('0')
if o==2 and n%2 and s[n//2]=='0':
print('DRAW')
else:
print('ALICE')
```
Yes
| 37,071 | [
-0.053924560546875,
0.01129913330078125,
-0.314697265625,
-0.08673095703125,
-0.51123046875,
-0.537109375,
-0.12005615234375,
-0.239990234375,
-0.09130859375,
0.9541015625,
0.8740234375,
0.0711669921875,
0.34375,
-0.56640625,
-0.66259765625,
-0.257080078125,
-0.8740234375,
-0.76416... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between the easy and hard versions is that the given string s in the easy version is initially a palindrome, this condition is not always true for the hard version.
A palindrome is a string that reads the same left to right and right to left. For example, "101101" is a palindrome, while "0101" is not.
Alice and Bob are playing a game on a string s of length n consisting of the characters '0' and '1'. Both players take alternate turns with Alice going first.
In each turn, the player can perform one of the following operations:
1. Choose any i (1 ≤ i ≤ n), where s[i] = '0' and change s[i] to '1'. Pay 1 dollar.
2. Reverse the whole string, pay 0 dollars. This operation is only allowed if the string is currently not a palindrome, and the last operation was not reverse. That is, if Alice reverses the string, then Bob can't reverse in the next move, and vice versa.
Reversing a string means reordering its letters from the last to the first. For example, "01001" becomes "10010" after reversing.
The game ends when every character of string becomes '1'. The player who spends minimum dollars till this point wins the game and it is a draw if both spend equal dollars. If both players play optimally, output whether Alice wins, Bob wins, or if it is a draw.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^3). Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^3).
The second line of each test case contains the string s of length n, consisting of the characters '0' and '1'. It is guaranteed that the string s contains at least one '0'.
Note that there is no limit on the sum of n over test cases.
Output
For each test case print a single word in a new line:
* "ALICE", if Alice will win the game,
* "BOB", if Bob will win the game,
* "DRAW", if the game ends in a draw.
Example
Input
3
3
110
2
00
4
1010
Output
ALICE
BOB
ALICE
Note
In the first test case of example,
* in the 1-st move, Alice will use the 2-nd operation to reverse the string, since doing the 1-st operation will result in her loss anyway. This also forces Bob to use the 1-st operation.
* in the 2-nd move, Bob has to perform the 1-st operation, since the 2-nd operation cannot be performed twice in a row. All characters of the string are '1', game over.
Alice spends 0 dollars while Bob spends 1 dollar. Hence, Alice wins.
In the second test case of example,
* in the 1-st move Alice has to perform the 1-st operation, since the string is currently a palindrome.
* in the 2-nd move Bob reverses the string.
* in the 3-rd move Alice again has to perform the 1-st operation. All characters of the string are '1', game over.
Alice spends 2 dollars while Bob spends 0 dollars. Hence, Bob wins.
Submitted Solution:
```
from collections import defaultdict
for _ in range(int(input())):
n=int(input())
s=list(input())
if s.count('0')==0:
print('DRAW')
elif n==1:
if s[0]=='0':
print('BOB')
else:
print('DRAW')
elif s==s[::-1]:
if s.count('0')%2==0 or s.count('0')==1:
print('BOB')
else:
print('ALICE')
else:
l=0
r=n-1
c=0
zero=0
if n%2:
if s[n//2]=='0':
zero+=1
while(l<r):
if s[l]!=s[r]:
c+=1
else:
if s[l]=='0':
zero+=2
l+=1
r-=1
if zero==1 and c==1:
print('DRAW')
else:
print('ALICE')
```
Yes
| 37,072 | [
-0.053924560546875,
0.01129913330078125,
-0.314697265625,
-0.08673095703125,
-0.51123046875,
-0.537109375,
-0.12005615234375,
-0.239990234375,
-0.09130859375,
0.9541015625,
0.8740234375,
0.0711669921875,
0.34375,
-0.56640625,
-0.66259765625,
-0.257080078125,
-0.8740234375,
-0.76416... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between the easy and hard versions is that the given string s in the easy version is initially a palindrome, this condition is not always true for the hard version.
A palindrome is a string that reads the same left to right and right to left. For example, "101101" is a palindrome, while "0101" is not.
Alice and Bob are playing a game on a string s of length n consisting of the characters '0' and '1'. Both players take alternate turns with Alice going first.
In each turn, the player can perform one of the following operations:
1. Choose any i (1 ≤ i ≤ n), where s[i] = '0' and change s[i] to '1'. Pay 1 dollar.
2. Reverse the whole string, pay 0 dollars. This operation is only allowed if the string is currently not a palindrome, and the last operation was not reverse. That is, if Alice reverses the string, then Bob can't reverse in the next move, and vice versa.
Reversing a string means reordering its letters from the last to the first. For example, "01001" becomes "10010" after reversing.
The game ends when every character of string becomes '1'. The player who spends minimum dollars till this point wins the game and it is a draw if both spend equal dollars. If both players play optimally, output whether Alice wins, Bob wins, or if it is a draw.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^3). Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^3).
The second line of each test case contains the string s of length n, consisting of the characters '0' and '1'. It is guaranteed that the string s contains at least one '0'.
Note that there is no limit on the sum of n over test cases.
Output
For each test case print a single word in a new line:
* "ALICE", if Alice will win the game,
* "BOB", if Bob will win the game,
* "DRAW", if the game ends in a draw.
Example
Input
3
3
110
2
00
4
1010
Output
ALICE
BOB
ALICE
Note
In the first test case of example,
* in the 1-st move, Alice will use the 2-nd operation to reverse the string, since doing the 1-st operation will result in her loss anyway. This also forces Bob to use the 1-st operation.
* in the 2-nd move, Bob has to perform the 1-st operation, since the 2-nd operation cannot be performed twice in a row. All characters of the string are '1', game over.
Alice spends 0 dollars while Bob spends 1 dollar. Hence, Alice wins.
In the second test case of example,
* in the 1-st move Alice has to perform the 1-st operation, since the string is currently a palindrome.
* in the 2-nd move Bob reverses the string.
* in the 3-rd move Alice again has to perform the 1-st operation. All characters of the string are '1', game over.
Alice spends 2 dollars while Bob spends 0 dollars. Hence, Bob wins.
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
s = input()
zeros = s.count('0')
if s == s[::-1]:
if zeros == 1:
print('BOB')
continue
elif zeros % 2 == 1:
print('ALICE')
continue
print('BOB')
else:
if n % 2 == 1 and s[n // 2] == '0' and zeros == 2:
print('DRAW')
else:
print('ALICE')
```
Yes
| 37,073 | [
-0.053924560546875,
0.01129913330078125,
-0.314697265625,
-0.08673095703125,
-0.51123046875,
-0.537109375,
-0.12005615234375,
-0.239990234375,
-0.09130859375,
0.9541015625,
0.8740234375,
0.0711669921875,
0.34375,
-0.56640625,
-0.66259765625,
-0.257080078125,
-0.8740234375,
-0.76416... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between the easy and hard versions is that the given string s in the easy version is initially a palindrome, this condition is not always true for the hard version.
A palindrome is a string that reads the same left to right and right to left. For example, "101101" is a palindrome, while "0101" is not.
Alice and Bob are playing a game on a string s of length n consisting of the characters '0' and '1'. Both players take alternate turns with Alice going first.
In each turn, the player can perform one of the following operations:
1. Choose any i (1 ≤ i ≤ n), where s[i] = '0' and change s[i] to '1'. Pay 1 dollar.
2. Reverse the whole string, pay 0 dollars. This operation is only allowed if the string is currently not a palindrome, and the last operation was not reverse. That is, if Alice reverses the string, then Bob can't reverse in the next move, and vice versa.
Reversing a string means reordering its letters from the last to the first. For example, "01001" becomes "10010" after reversing.
The game ends when every character of string becomes '1'. The player who spends minimum dollars till this point wins the game and it is a draw if both spend equal dollars. If both players play optimally, output whether Alice wins, Bob wins, or if it is a draw.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^3). Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^3).
The second line of each test case contains the string s of length n, consisting of the characters '0' and '1'. It is guaranteed that the string s contains at least one '0'.
Note that there is no limit on the sum of n over test cases.
Output
For each test case print a single word in a new line:
* "ALICE", if Alice will win the game,
* "BOB", if Bob will win the game,
* "DRAW", if the game ends in a draw.
Example
Input
3
3
110
2
00
4
1010
Output
ALICE
BOB
ALICE
Note
In the first test case of example,
* in the 1-st move, Alice will use the 2-nd operation to reverse the string, since doing the 1-st operation will result in her loss anyway. This also forces Bob to use the 1-st operation.
* in the 2-nd move, Bob has to perform the 1-st operation, since the 2-nd operation cannot be performed twice in a row. All characters of the string are '1', game over.
Alice spends 0 dollars while Bob spends 1 dollar. Hence, Alice wins.
In the second test case of example,
* in the 1-st move Alice has to perform the 1-st operation, since the string is currently a palindrome.
* in the 2-nd move Bob reverses the string.
* in the 3-rd move Alice again has to perform the 1-st operation. All characters of the string are '1', game over.
Alice spends 2 dollars while Bob spends 0 dollars. Hence, Bob wins.
Submitted Solution:
```
for _ in range(int(input())):
n=int(input())
s=input()
cnt=0
for i in s:
if i=='0':
cnt+=1
if s==s[::-1]:
if cnt>1 and s[n//2]=='0' and n%2!=0:
print("ALICE")
else:
print("BOB")
else:
if cnt==2 and s[n//2]=='0' and n%2!=0:
print("DRAW")
else:
print("ALICE")
```
Yes
| 37,074 | [
-0.053924560546875,
0.01129913330078125,
-0.314697265625,
-0.08673095703125,
-0.51123046875,
-0.537109375,
-0.12005615234375,
-0.239990234375,
-0.09130859375,
0.9541015625,
0.8740234375,
0.0711669921875,
0.34375,
-0.56640625,
-0.66259765625,
-0.257080078125,
-0.8740234375,
-0.76416... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between the easy and hard versions is that the given string s in the easy version is initially a palindrome, this condition is not always true for the hard version.
A palindrome is a string that reads the same left to right and right to left. For example, "101101" is a palindrome, while "0101" is not.
Alice and Bob are playing a game on a string s of length n consisting of the characters '0' and '1'. Both players take alternate turns with Alice going first.
In each turn, the player can perform one of the following operations:
1. Choose any i (1 ≤ i ≤ n), where s[i] = '0' and change s[i] to '1'. Pay 1 dollar.
2. Reverse the whole string, pay 0 dollars. This operation is only allowed if the string is currently not a palindrome, and the last operation was not reverse. That is, if Alice reverses the string, then Bob can't reverse in the next move, and vice versa.
Reversing a string means reordering its letters from the last to the first. For example, "01001" becomes "10010" after reversing.
The game ends when every character of string becomes '1'. The player who spends minimum dollars till this point wins the game and it is a draw if both spend equal dollars. If both players play optimally, output whether Alice wins, Bob wins, or if it is a draw.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^3). Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^3).
The second line of each test case contains the string s of length n, consisting of the characters '0' and '1'. It is guaranteed that the string s contains at least one '0'.
Note that there is no limit on the sum of n over test cases.
Output
For each test case print a single word in a new line:
* "ALICE", if Alice will win the game,
* "BOB", if Bob will win the game,
* "DRAW", if the game ends in a draw.
Example
Input
3
3
110
2
00
4
1010
Output
ALICE
BOB
ALICE
Note
In the first test case of example,
* in the 1-st move, Alice will use the 2-nd operation to reverse the string, since doing the 1-st operation will result in her loss anyway. This also forces Bob to use the 1-st operation.
* in the 2-nd move, Bob has to perform the 1-st operation, since the 2-nd operation cannot be performed twice in a row. All characters of the string are '1', game over.
Alice spends 0 dollars while Bob spends 1 dollar. Hence, Alice wins.
In the second test case of example,
* in the 1-st move Alice has to perform the 1-st operation, since the string is currently a palindrome.
* in the 2-nd move Bob reverses the string.
* in the 3-rd move Alice again has to perform the 1-st operation. All characters of the string are '1', game over.
Alice spends 2 dollars while Bob spends 0 dollars. Hence, Bob wins.
Submitted Solution:
```
def isPalindrome(s):
for i in range(len(s) // 2):
if s[i] != s[len(s)-i-1]: return False
return True
for _ in range(int(input())):
player = 0
reverse = False
n = int(input())
s = list(input())
count = s.count('1')
score = [0, 0]
# isPalindrome = True
palindex = 0
try:
while True:
# print(''.join(s), player, palindex)
if not reverse and not isPalindrome(s):
reverse = True
elif len(s) % 2 and s[len(s) // 2] == '0':
s[len(s) // 2] = '1'
palindex = len(s) // 2
score[player] += 1
else:
if s[palindex] == '1':
palindex = s.index('0')
s[palindex] = '1'
# isPalindrome = False
else:
palindex = n - palindex - 1
s[palindex] = '1'
# isPalindrome = True
count += 1
score[player] += 1
player = (player + 1) % 2
# print(''.join(s), palindex)
# print(score)
# print()
except: pass
if score[0] == score[1]:
print('DRAW')
elif score[0] < score[1]:
print('ALICE')
else:
print('BOB')
```
No
| 37,075 | [
-0.053924560546875,
0.01129913330078125,
-0.314697265625,
-0.08673095703125,
-0.51123046875,
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-0.09130859375,
0.9541015625,
0.8740234375,
0.0711669921875,
0.34375,
-0.56640625,
-0.66259765625,
-0.257080078125,
-0.8740234375,
-0.76416... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between the easy and hard versions is that the given string s in the easy version is initially a palindrome, this condition is not always true for the hard version.
A palindrome is a string that reads the same left to right and right to left. For example, "101101" is a palindrome, while "0101" is not.
Alice and Bob are playing a game on a string s of length n consisting of the characters '0' and '1'. Both players take alternate turns with Alice going first.
In each turn, the player can perform one of the following operations:
1. Choose any i (1 ≤ i ≤ n), where s[i] = '0' and change s[i] to '1'. Pay 1 dollar.
2. Reverse the whole string, pay 0 dollars. This operation is only allowed if the string is currently not a palindrome, and the last operation was not reverse. That is, if Alice reverses the string, then Bob can't reverse in the next move, and vice versa.
Reversing a string means reordering its letters from the last to the first. For example, "01001" becomes "10010" after reversing.
The game ends when every character of string becomes '1'. The player who spends minimum dollars till this point wins the game and it is a draw if both spend equal dollars. If both players play optimally, output whether Alice wins, Bob wins, or if it is a draw.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^3). Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^3).
The second line of each test case contains the string s of length n, consisting of the characters '0' and '1'. It is guaranteed that the string s contains at least one '0'.
Note that there is no limit on the sum of n over test cases.
Output
For each test case print a single word in a new line:
* "ALICE", if Alice will win the game,
* "BOB", if Bob will win the game,
* "DRAW", if the game ends in a draw.
Example
Input
3
3
110
2
00
4
1010
Output
ALICE
BOB
ALICE
Note
In the first test case of example,
* in the 1-st move, Alice will use the 2-nd operation to reverse the string, since doing the 1-st operation will result in her loss anyway. This also forces Bob to use the 1-st operation.
* in the 2-nd move, Bob has to perform the 1-st operation, since the 2-nd operation cannot be performed twice in a row. All characters of the string are '1', game over.
Alice spends 0 dollars while Bob spends 1 dollar. Hence, Alice wins.
In the second test case of example,
* in the 1-st move Alice has to perform the 1-st operation, since the string is currently a palindrome.
* in the 2-nd move Bob reverses the string.
* in the 3-rd move Alice again has to perform the 1-st operation. All characters of the string are '1', game over.
Alice spends 2 dollars while Bob spends 0 dollars. Hence, Bob wins.
Submitted Solution:
```
t = int(input())
while t:
n = int(input())
s = input()
n = len(s)
cnt1 = cnt2 = 0
mid = int((n-1)/2)
for i in range(0, int((n+1)/2)):
if(s[i] != s[n-1-i]):
cnt1+=1
else:
if(s[i] == '0' and s[n-1-i] == '0'):
cnt2+=1
if ((n%2)==0) or (n%2!=0 and s[mid]=='1'):
if(cnt1==0):
if(cnt2==0):
print("DRAW")
else:
print("BOB")
else:
print("ALICE")
else:
if(cnt1==0):
if(cnt2==0):
print("BOB")
else:
print("ALICE")
else:
if(cnt2==0):
print("DRAW")
else:
if(cnt2>1):
print("ALICE")
else:
print("BOB")
t-=1
```
No
| 37,076 | [
-0.053924560546875,
0.01129913330078125,
-0.314697265625,
-0.08673095703125,
-0.51123046875,
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-0.239990234375,
-0.09130859375,
0.9541015625,
0.8740234375,
0.0711669921875,
0.34375,
-0.56640625,
-0.66259765625,
-0.257080078125,
-0.8740234375,
-0.76416... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between the easy and hard versions is that the given string s in the easy version is initially a palindrome, this condition is not always true for the hard version.
A palindrome is a string that reads the same left to right and right to left. For example, "101101" is a palindrome, while "0101" is not.
Alice and Bob are playing a game on a string s of length n consisting of the characters '0' and '1'. Both players take alternate turns with Alice going first.
In each turn, the player can perform one of the following operations:
1. Choose any i (1 ≤ i ≤ n), where s[i] = '0' and change s[i] to '1'. Pay 1 dollar.
2. Reverse the whole string, pay 0 dollars. This operation is only allowed if the string is currently not a palindrome, and the last operation was not reverse. That is, if Alice reverses the string, then Bob can't reverse in the next move, and vice versa.
Reversing a string means reordering its letters from the last to the first. For example, "01001" becomes "10010" after reversing.
The game ends when every character of string becomes '1'. The player who spends minimum dollars till this point wins the game and it is a draw if both spend equal dollars. If both players play optimally, output whether Alice wins, Bob wins, or if it is a draw.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^3). Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^3).
The second line of each test case contains the string s of length n, consisting of the characters '0' and '1'. It is guaranteed that the string s contains at least one '0'.
Note that there is no limit on the sum of n over test cases.
Output
For each test case print a single word in a new line:
* "ALICE", if Alice will win the game,
* "BOB", if Bob will win the game,
* "DRAW", if the game ends in a draw.
Example
Input
3
3
110
2
00
4
1010
Output
ALICE
BOB
ALICE
Note
In the first test case of example,
* in the 1-st move, Alice will use the 2-nd operation to reverse the string, since doing the 1-st operation will result in her loss anyway. This also forces Bob to use the 1-st operation.
* in the 2-nd move, Bob has to perform the 1-st operation, since the 2-nd operation cannot be performed twice in a row. All characters of the string are '1', game over.
Alice spends 0 dollars while Bob spends 1 dollar. Hence, Alice wins.
In the second test case of example,
* in the 1-st move Alice has to perform the 1-st operation, since the string is currently a palindrome.
* in the 2-nd move Bob reverses the string.
* in the 3-rd move Alice again has to perform the 1-st operation. All characters of the string are '1', game over.
Alice spends 2 dollars while Bob spends 0 dollars. Hence, Bob wins.
Submitted Solution:
```
from bisect import bisect,bisect_left
from collections import *
from heapq import *
from math import gcd,ceil,sqrt,floor,inf
from itertools import *
from operator import add,mul,sub,xor,truediv,floordiv
from functools import *
#------------------------------------------------------------------------
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#------------------------------------------------------------------------
def RL(): return map(int, sys.stdin.readline().split())
def RLL(): return list(map(int, sys.stdin.readline().split()))
def N(): return int(input())
def A(n):return [0]*n
def AI(n,x): return [x]*n
def A2(n,m): return [[0]*m for i in range(n)]
def G(n): return [[] for i in range(n)]
def GP(it): return [[ch,len(list(g))] for ch,g in groupby(it)]
#------------------------------------------------------------------------
from types import GeneratorType
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
mod=10**9+7
farr=[1]
ifa=[]
def fact(x,mod=0):
if mod:
while x>=len(farr):
farr.append(farr[-1]*len(farr)%mod)
else:
while x>=len(farr):
farr.append(farr[-1]*len(farr))
return farr[x]
def ifact(x,mod):
global ifa
fact(x,mod)
ifa.append(pow(farr[-1],mod-2,mod))
for i in range(x,0,-1):
ifa.append(ifa[-1]*i%mod)
ifa.reverse()
def per(i,j,mod=0):
if i<j: return 0
if not mod:
return fact(i)//fact(i-j)
return farr[i]*ifa[i-j]%mod
def com(i,j,mod=0):
if i<j: return 0
if not mod:
return per(i,j)//fact(j)
return per(i,j,mod)*ifa[j]%mod
def catalan(n):
return com(2*n,n)//(n+1)
def isprime(n):
for i in range(2,int(n**0.5)+1):
if n%i==0:
return False
return True
def floorsum(a,b,c,n):#sum((a*i+b)//c for i in range(n+1))
if a==0:return b//c*(n+1)
if a>=c or b>=c: return floorsum(a%c,b%c,c,n)+b//c*(n+1)+a//c*n*(n+1)//2
m=(a*n+b)//c
return n*m-floorsum(c,c-b-1,a,m-1)
def inverse(a,m):
a%=m
if a<=1: return a
return ((1-inverse(m,a)*m)//a)%m
def lowbit(n):
return n&-n
class BIT:
def __init__(self,arr):
self.arr=arr
self.n=len(arr)-1
def update(self,x,v):
while x<=self.n:
self.arr[x]+=v
x+=x&-x
def query(self,x):
ans=0
while x:
ans+=self.arr[x]
x&=x-1
return ans
class ST:
def __init__(self,arr):#n!=0
n=len(arr)
mx=n.bit_length()#取不到
self.st=[[0]*mx for i in range(n)]
for i in range(n):
self.st[i][0]=arr[i]
for j in range(1,mx):
for i in range(n-(1<<j)+1):
self.st[i][j]=max(self.st[i][j-1],self.st[i+(1<<j-1)][j-1])
def query(self,l,r):
if l>r:return -inf
s=(r+1-l).bit_length()-1
return max(self.st[l][s],self.st[r-(1<<s)+1][s])
class DSU:#容量+路径压缩
def __init__(self,n):
self.c=[-1]*n
def same(self,x,y):
return self.find(x)==self.find(y)
def find(self,x):
if self.c[x]<0:
return x
self.c[x]=self.find(self.c[x])
return self.c[x]
def union(self,u,v):
u,v=self.find(u),self.find(v)
if u==v:
return False
if self.c[u]>self.c[v]:
u,v=v,u
self.c[u]+=self.c[v]
self.c[v]=u
return True
def size(self,x): return -self.c[self.find(x)]
class UFS:#秩+路径
def __init__(self,n):
self.parent=[i for i in range(n)]
self.ranks=[0]*n
def find(self,x):
if x!=self.parent[x]:
self.parent[x]=self.find(self.parent[x])
return self.parent[x]
def union(self,u,v):
pu,pv=self.find(u),self.find(v)
if pu==pv:
return False
if self.ranks[pu]>=self.ranks[pv]:
self.parent[pv]=pu
if self.ranks[pv]==self.ranks[pu]:
self.ranks[pu]+=1
else:
self.parent[pu]=pv
class UF:#秩+路径+容量，边数
def __init__(self,n):
self.parent=[i for i in range(n)]
self.ranks=[0]*n
self.size=AI(n,1)
self.edge=A(n)
def find(self,x):
if x!=self.parent[x]:
self.parent[x]=self.find(self.parent[x])
return self.parent[x]
def union(self,u,v):
pu,pv=self.find(u),self.find(v)
if pu==pv:
self.edge[pu]+=1
return False
if self.ranks[pu]>=self.ranks[pv]:
self.parent[pv]=pu
self.edge[pu]+=self.edge[pv]+1
self.size[pu]+=self.size[pv]
if self.ranks[pv]==self.ranks[pu]:
self.ranks[pu]+=1
else:
self.parent[pu]=pv
self.edge[pv]+=self.edge[pu]+1
self.size[pv]+=self.size[pu]
def Prime(n):
c=0
prime=[]
flag=[0]*(n+1)
for i in range(2,n+1):
if not flag[i]:
prime.append(i)
c+=1
for j in range(c):
if i*prime[j]>n: break
flag[i*prime[j]]=prime[j]
if i%prime[j]==0: break
return prime
def dij(s,graph):
d=AI(n,inf)
d[s]=0
heap=[(0,s)]
vis=A(n)
while heap:
dis,u=heappop(heap)
if vis[u]:
continue
vis[u]=1
for v,w in graph[u]:
if d[v]>d[u]+w:
d[v]=d[u]+w
heappush(heap,(d[v],v))
return d
def bell(s,g):#bellman-Ford
dis=AI(n,inf)
dis[s]=0
for i in range(n-1):
for u,v,w in edge:
if dis[v]>dis[u]+w:
dis[v]=dis[u]+w
change=A(n)
for i in range(n):
for u,v,w in edge:
if dis[v]>dis[u]+w:
dis[v]=dis[u]+w
change[v]=1
return dis
def lcm(a,b): return a*b//gcd(a,b)
def lis(nums):
res=[]
for k in nums:
i=bisect.bisect_left(res,k)
if i==len(res):
res.append(k)
else:
res[i]=k
return len(res)
def RP(nums):#逆序对
n = len(nums)
s=set(nums)
d={}
for i,k in enumerate(sorted(s),1):
d[k]=i
bi=BIT([0]*(len(s)+1))
ans=0
for i in range(n-1,-1,-1):
ans+=bi.query(d[nums[i]]-1)
bi.update(d[nums[i]],1)
return ans
class DLN:
def __init__(self,val):
self.val=val
self.pre=None
self.next=None
def nb(i,j,n,m):
for ni,nj in [[i+1,j],[i-1,j],[i,j-1],[i,j+1]]:
if 0<=ni<n and 0<=nj<m:
yield ni,nj
def topo(n):
q=deque()
res=[]
for i in range(1,n+1):
if ind[i]==0:
q.append(i)
res.append(i)
while q:
u=q.popleft()
for v in g[u]:
ind[v]-=1
if ind[v]==0:
q.append(v)
res.append(v)
return res
@bootstrap
def gdfs(r,p):
for ch in g[r]:
if ch!=p:
yield gdfs(ch,r)
yield None
#from random import randint
a=1001
dp=[[[inf]*2 for i in range(a)] for i in range(a)]
dp[0][0][0]=dp[0][0][1]=0
for i in range(a):
for j in range(a-i*2):
if i>1:
dp[i][j][0]=min(1-dp[i-2][j+1][1],dp[i][j][0])
if i&1:
dp[i][j][0]=min(1-dp[i-1][j][1],dp[i][j][0])
if j:
dp[i][j][0]=min(dp[i][j][0],1-dp[i][j-1][1])
if not j:
dp[i][j][1]=dp[i][j][0]
else:
dp[i][j][1]=min(-dp[i][j][0],dp[i][j][0])
'''
@lru_cache(None)
def check(s,t):
if s=='1'*n:
return 0
if s==s[::-1]:
t=0
res=inf
if t==0:
s=list(s)
for i in range(n):
if s[i]=='0':
s[i]='1'
tmp=''.join(s)
res=min(1-check(tmp,1),res)
s[i]='0'
return res
else:
return min(check(s,0),-check(s,0))'''
t=N()
#n=10
#t=1<<n
for i in range(t):
n=N()
'''
s=['0']*n
for j in range(n):
if i&1<<j:
s[j]='1'
s=''.join(s)'''
s=input()
d=0
cp=cs=0
l=0
r=n-1
while l<r:
if s[l]==s[r]:
if s[l]=='0':
cp+=2
else:
cs+=1
l+=1
r-=1
if l==r and s[l]=='0':
cp+=1
d=dp[cp][cs][1]
#ans=check(s,1)
#if d!=ans:
# print(s,d,ans)
if d<0:
ans="ALICE"
elif d==0:
ans='DRAW'
else:
ans="BOB"
print(ans)
'''
sys.setrecursionlimit(200000)
import threading
threading.stack_size(10**8)
t=threading.Thr
ead(target=main)
t.start()
t.join()
'''
```
No
| 37,077 | [
-0.053924560546875,
0.01129913330078125,
-0.314697265625,
-0.08673095703125,
-0.51123046875,
-0.537109375,
-0.12005615234375,
-0.239990234375,
-0.09130859375,
0.9541015625,
0.8740234375,
0.0711669921875,
0.34375,
-0.56640625,
-0.66259765625,
-0.257080078125,
-0.8740234375,
-0.76416... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between the easy and hard versions is that the given string s in the easy version is initially a palindrome, this condition is not always true for the hard version.
A palindrome is a string that reads the same left to right and right to left. For example, "101101" is a palindrome, while "0101" is not.
Alice and Bob are playing a game on a string s of length n consisting of the characters '0' and '1'. Both players take alternate turns with Alice going first.
In each turn, the player can perform one of the following operations:
1. Choose any i (1 ≤ i ≤ n), where s[i] = '0' and change s[i] to '1'. Pay 1 dollar.
2. Reverse the whole string, pay 0 dollars. This operation is only allowed if the string is currently not a palindrome, and the last operation was not reverse. That is, if Alice reverses the string, then Bob can't reverse in the next move, and vice versa.
Reversing a string means reordering its letters from the last to the first. For example, "01001" becomes "10010" after reversing.
The game ends when every character of string becomes '1'. The player who spends minimum dollars till this point wins the game and it is a draw if both spend equal dollars. If both players play optimally, output whether Alice wins, Bob wins, or if it is a draw.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^3). Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^3).
The second line of each test case contains the string s of length n, consisting of the characters '0' and '1'. It is guaranteed that the string s contains at least one '0'.
Note that there is no limit on the sum of n over test cases.
Output
For each test case print a single word in a new line:
* "ALICE", if Alice will win the game,
* "BOB", if Bob will win the game,
* "DRAW", if the game ends in a draw.
Example
Input
3
3
110
2
00
4
1010
Output
ALICE
BOB
ALICE
Note
In the first test case of example,
* in the 1-st move, Alice will use the 2-nd operation to reverse the string, since doing the 1-st operation will result in her loss anyway. This also forces Bob to use the 1-st operation.
* in the 2-nd move, Bob has to perform the 1-st operation, since the 2-nd operation cannot be performed twice in a row. All characters of the string are '1', game over.
Alice spends 0 dollars while Bob spends 1 dollar. Hence, Alice wins.
In the second test case of example,
* in the 1-st move Alice has to perform the 1-st operation, since the string is currently a palindrome.
* in the 2-nd move Bob reverses the string.
* in the 3-rd move Alice again has to perform the 1-st operation. All characters of the string are '1', game over.
Alice spends 2 dollars while Bob spends 0 dollars. Hence, Bob wins.
Submitted Solution:
```
#!/usr/bin/env python
from __future__ import division, print_function
import os
import sys
from io import BytesIO, IOBase
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
def main():
t = int(input())
for _ in range(t):
n = int(input())
s = input()
z = s.count('0')
a = sum(s[i] != s[-i - 1] for i in range(n // 2))
f = ((z - a) // 2) + 1
s = ((z - a) // 2) - 1
if n % 2:
f, s = s + 1, f
if (z - a) == 0:
f, s = 0, 0
if a:
# skip, change
sc = f + 1 < (a - 1) + s
scd = f + 1 == (a - 1) + s
# skip, no change
sn = s < a + s
snd = s == a + s
# no skip, change
nc = (a - 1) + s < f + 1
ncd = (a - 1) + s == f + 1
# no skip, no change
nn = a + f < s
nnd = a + f == s
if sc or sn or (nc and nn):
print("ALICE")
elif scd or snd or (ncd and nnd):
print("DRAW")
else:
print("BOB")
else:
print("ALICE" if f < s else "BOB")
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
```
No
| 37,078 | [
-0.053924560546875,
0.01129913330078125,
-0.314697265625,
-0.08673095703125,
-0.51123046875,
-0.537109375,
-0.12005615234375,
-0.239990234375,
-0.09130859375,
0.9541015625,
0.8740234375,
0.0711669921875,
0.34375,
-0.56640625,
-0.66259765625,
-0.257080078125,
-0.8740234375,
-0.76416... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given P, a permutation of (1,\ 2,\ ...\ N).
A string S of length N consisting of `0` and `1` is a good string when it meets the following criterion:
* The sequences X and Y are constructed as follows:
* First, let X and Y be empty sequences.
* For each i=1,\ 2,\ ...\ N, in this order, append P_i to the end of X if S_i= `0`, and append it to the end of Y if S_i= `1`.
* If X and Y have the same number of high elements, S is a good string. Here, the i-th element of a sequence is called high when that element is the largest among the elements from the 1-st to i-th element in the sequence.
Determine if there exists a good string. If it exists, find the lexicographically smallest such string.
Constraints
* 1 \leq N \leq 2 \times 10^5
* 1 \leq P_i \leq N
* P_1,\ P_2,\ ...\ P_N are all distinct.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
P_1 P_2 ... P_N
Output
If a good string does not exist, print `-1`. If it exists, print the lexicographically smallest such string.
Examples
Input
6
3 1 4 6 2 5
Output
001001
Input
5
1 2 3 4 5
Output
-1
Input
7
1 3 2 5 6 4 7
Output
0001101
Input
30
1 2 6 3 5 7 9 8 11 12 10 13 16 23 15 18 14 24 22 26 19 21 28 17 4 27 29 25 20 30
Output
000000000001100101010010011101
Submitted Solution:
```
import itertools
N = int(input())
P = [int(x) for x in input().split()]
for S in itertools.product("01", repeat=N):
X_max, X_high = 0, 0
Y_max, Y_high = 0, 0
for i, s in enumerate(S):
t = P[i]
if s == "0" and t >= X_max:
X_max = t
X_high += 1
elif s == "1" and t >= Y_max:
Y_max = t
Y_high += 1
if X_high == Y_high:
print("".join(S))
exit()
else:
print(-1)
```
No
| 37,571 | [
0.197998046875,
-0.07159423828125,
0.32177734375,
0.345458984375,
-0.56982421875,
-0.482177734375,
-0.2225341796875,
0.231201171875,
0.3212890625,
0.87060546875,
0.84375,
-0.0894775390625,
0.058837890625,
-1.080078125,
-0.70947265625,
-0.032257080078125,
-0.58056640625,
-0.82910156... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Some programming website is establishing a secure communication protocol. For security reasons, they want to choose several more or less random strings.
Initially, they have a string s consisting of lowercase English letters. Now they want to choose q strings using the following steps, and you are to help them.
1. A string x consisting of lowercase English letters and integers l and r (1 ≤ l ≤ r ≤ |s|) are chosen.
2. Consider all non-empty distinct substrings of the s_l s_{l + 1} … s_r, that is all distinct strings s_i s_{i+1} … s_{j} where l ≤ i ≤ j ≤ r. Among all of them choose all strings that are lexicographically greater than x.
3. If there are no such strings, you should print -1. Otherwise print the lexicographically smallest among them.
String a is lexicographically less than string b, if either a is a prefix of b and a ≠ b, or there exists such a position i (1 ≤ i ≤ min(|a|, |b|)), such that a_i < b_i and for all j (1 ≤ j < i) a_j = b_j. Here |a| denotes the length of the string a.
Input
The first line of input contains a non-empty string s (1 ≤ |s| ≤ 10^{5}) consisting of lowercase English letters.
The second line contains an integer q (1 ≤ q ≤ 2 ⋅ 10^5) — the number of strings to select.
Each of the next q lines contains two integers l, r (1 ≤ l ≤ r ≤ |s|) and a non-empty string x consisting of lowercase English letters. The total length of strings x for all queries does not exceed 2 ⋅ 10^{5}.
Output
Output q lines, each of them should contain the desired string or -1, if there is no such string.
Examples
Input
baa
5
1 2 ba
2 3 a
1 2 b
2 3 aa
1 3 b
Output
-1
aa
ba
-1
ba
Input
bacb
4
1 2 ba
2 3 ac
1 3 ac
3 4 c
Output
-1
c
b
cb
Input
bba
1
1 1 b
Output
-1
Note
Consider the first example.
The string s is "baa". The queries are as follows.
1. We consider the substring s_1 s_2 that is "ba". It has substrings "b", "a" and "ba", since none of them is greater than the query string "ba", the answer is -1.
2. We consider substring "aa". Among its substrings only "aa" is greater than the query string "a". So the answer is "aa".
3. We consider substring "ba". Out of "b", "a" and "ba" only "ba" is greater than the query string "b", so the answer is "ba".
4. We consider substring "aa". No substring of "aa" is greater than the query string "aa" so the answer is -1.
5. We consider substring "baa" and it has (among others) substrings "ba", "baa" which are greater than the query string "b". Since "ba" is lexicographically smaller than "baa", the answer is "ba".
Submitted Solution:
```
def retcount(li, x):
L = []
for i in range(len(li)):
for j in range(i, len(li)):
if i == j:
L.append(li[i])
else:
L.append(li[i]+li[j])
L.append(''.join(li))
L = [w for w in L if len(w)>x]
return L
s = input()
n = int(input())
for i in range(n):
l, r, q = map(str,input().split())
li = list(s[int(l)-1:int(r)])
n = len(q)
a = retcount(li, n)
if len(a) < 1:
print('-1')
else:
print(a[0])
```
No
| 37,750 | [
0.182861328125,
-0.09967041015625,
0.28759765625,
0.306884765625,
-0.75537109375,
-0.2464599609375,
-0.0214996337890625,
0.12152099609375,
-0.0662841796875,
0.90673828125,
0.6923828125,
-0.0071258544921875,
-0.14599609375,
-0.83642578125,
-0.8779296875,
0.27685546875,
-0.65576171875,... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vova has won n trophies in different competitions. Each trophy is either golden or silver. The trophies are arranged in a row.
The beauty of the arrangement is the length of the longest subsegment consisting of golden trophies. Vova wants to swap two trophies (not necessarily adjacent ones) to make the arrangement as beautiful as possible — that means, to maximize the length of the longest such subsegment.
Help Vova! Tell him the maximum possible beauty of the arrangement if he is allowed to do at most one swap.
Input
The first line contains one integer n (2 ≤ n ≤ 10^5) — the number of trophies.
The second line contains n characters, each of them is either G or S. If the i-th character is G, then the i-th trophy is a golden one, otherwise it's a silver trophy.
Output
Print the maximum possible length of a subsegment of golden trophies, if Vova is allowed to do at most one swap.
Examples
Input
10
GGGSGGGSGG
Output
7
Input
4
GGGG
Output
4
Input
3
SSS
Output
0
Note
In the first example Vova has to swap trophies with indices 4 and 10. Thus he will obtain the sequence "GGGGGGGSGS", the length of the longest subsegment of golden trophies is 7.
In the second example Vova can make no swaps at all. The length of the longest subsegment of golden trophies in the sequence is 4.
In the third example Vova cannot do anything to make the length of the longest subsegment of golden trophies in the sequence greater than 0.
Submitted Solution:
```
x=int(input())
string=input()
last=0
new=0
maximum=0
if string.count("G")==0:
print(0)
else:
for i in string:
if i=="G":
new+=1
else:
maximum=new+last
last=new
new=0
if maximum==0:
print(x)
else:
if string.count("S")==1:
print(max(maximum,new+last))
else:
print(max(maximum+1,new+last+1))
```
No
| 37,780 | [
0.364990234375,
-0.060638427734375,
0.09735107421875,
0.8916015625,
-0.364501953125,
-0.4677734375,
-0.01450347900390625,
0.465576171875,
0.137939453125,
0.67431640625,
0.5810546875,
0.08447265625,
-0.0697021484375,
-0.84912109375,
-0.486083984375,
0.297607421875,
-0.37939453125,
-... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vova has won n trophies in different competitions. Each trophy is either golden or silver. The trophies are arranged in a row.
The beauty of the arrangement is the length of the longest subsegment consisting of golden trophies. Vova wants to swap two trophies (not necessarily adjacent ones) to make the arrangement as beautiful as possible — that means, to maximize the length of the longest such subsegment.
Help Vova! Tell him the maximum possible beauty of the arrangement if he is allowed to do at most one swap.
Input
The first line contains one integer n (2 ≤ n ≤ 10^5) — the number of trophies.
The second line contains n characters, each of them is either G or S. If the i-th character is G, then the i-th trophy is a golden one, otherwise it's a silver trophy.
Output
Print the maximum possible length of a subsegment of golden trophies, if Vova is allowed to do at most one swap.
Examples
Input
10
GGGSGGGSGG
Output
7
Input
4
GGGG
Output
4
Input
3
SSS
Output
0
Note
In the first example Vova has to swap trophies with indices 4 and 10. Thus he will obtain the sequence "GGGGGGGSGS", the length of the longest subsegment of golden trophies is 7.
In the second example Vova can make no swaps at all. The length of the longest subsegment of golden trophies in the sequence is 4.
In the third example Vova cannot do anything to make the length of the longest subsegment of golden trophies in the sequence greater than 0.
Submitted Solution:
```
x=int(input())
string=input()
last=0
new=0
maximum=0
if string.count("G")==0:
print(0)
else:
for i in string:
if i=="G":
new+=1
else:
maximum=new+last
last=new
new=0
if maximum==0:
print(x-string.count("S"))
else:
if string.count("S")==1:
print(max(maximum,new+last))
else:
maximum=max(maximum+1,new+last+1)
if maximum>=string.count("G"):
print(string.count("G"))
```
No
| 37,783 | [
0.356689453125,
-0.0611572265625,
0.12469482421875,
0.88134765625,
-0.387939453125,
-0.47802734375,
-0.0205230712890625,
0.453857421875,
0.14404296875,
0.6787109375,
0.583984375,
0.0765380859375,
-0.07177734375,
-0.8408203125,
-0.484375,
0.2958984375,
-0.3759765625,
-0.84228515625,... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a permutation p consisting of exactly 26 integers from 1 to 26 (since it is a permutation, each integer from 1 to 26 occurs in p exactly once) and two strings s and t consisting of lowercase Latin letters.
A substring t' of string t is an occurence of string s if the following conditions are met:
1. |t'| = |s|;
2. for each i ∈ [1, |s|], either s_i = t'_i, or p_{idx(s_i)} = idx(t'_i), where idx(c) is the index of character c in Latin alphabet (idx(a) = 1, idx(b) = 2, idx(z) = 26).
For example, if p_1 = 2, p_2 = 3, p_3 = 1, s = abc, t = abcaaba, then three substrings of t are occurences of s (they are t' = abc, t' = bca and t' = aba).
For each substring of t having length equal to |s|, check if it is an occurence of s.
Input
The first line contains 26 integers p_1, p_2, ..., p_{26} (1 ≤ p_i ≤ 26, all these integers are pairwise distinct).
The second line contains one string s, and the third line contains one string t (2 ≤ |s| ≤ |t| ≤ 2 ⋅ 10^5) both consisting of lowercase Latin letters.
Output
Print a string of |t| - |s| + 1 characters, each character should be either 0 or 1. The i-th character should be 1 if and only if the substring of t starting with the i-th character and ending with the (i + |s| - 1)-th character (inclusive) is an occurence of s.
Example
Input
2 3 1 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
abc
abcaaba
Output
11001
Submitted Solution:
```
print("bing bing bong bong")
```
No
| 37,856 | [
0.185791015625,
0.04779052734375,
0.0201873779296875,
0.1807861328125,
-0.61328125,
-0.36767578125,
-0.38525390625,
0.08563232421875,
0.2186279296875,
1.107421875,
0.6064453125,
-0.117919921875,
-0.02972412109375,
-0.75732421875,
-0.7763671875,
0.0787353515625,
-0.7685546875,
-0.73... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a permutation p consisting of exactly 26 integers from 1 to 26 (since it is a permutation, each integer from 1 to 26 occurs in p exactly once) and two strings s and t consisting of lowercase Latin letters.
A substring t' of string t is an occurence of string s if the following conditions are met:
1. |t'| = |s|;
2. for each i ∈ [1, |s|], either s_i = t'_i, or p_{idx(s_i)} = idx(t'_i), where idx(c) is the index of character c in Latin alphabet (idx(a) = 1, idx(b) = 2, idx(z) = 26).
For example, if p_1 = 2, p_2 = 3, p_3 = 1, s = abc, t = abcaaba, then three substrings of t are occurences of s (they are t' = abc, t' = bca and t' = aba).
For each substring of t having length equal to |s|, check if it is an occurence of s.
Input
The first line contains 26 integers p_1, p_2, ..., p_{26} (1 ≤ p_i ≤ 26, all these integers are pairwise distinct).
The second line contains one string s, and the third line contains one string t (2 ≤ |s| ≤ |t| ≤ 2 ⋅ 10^5) both consisting of lowercase Latin letters.
Output
Print a string of |t| - |s| + 1 characters, each character should be either 0 or 1. The i-th character should be 1 if and only if the substring of t starting with the i-th character and ending with the (i + |s| - 1)-th character (inclusive) is an occurence of s.
Example
Input
2 3 1 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
abc
abcaaba
Output
11001
Submitted Solution:
```
r=[0]
idx=dict()
j=0
for c in "abcdefghijklmnopqrstuvwxyz":
j=j+1
idx[c]=j
p=r+list(map(int,input().split()))
s=input("entrer s ")
t=input("entrer t" )
def testv ( ch,s , p ):
i=0
test= (ch[i]==s[i]) or(p[idx[s[i]]]==idx[ch[i]])
while ((test== True) and (i<len(s))):
test= (ch[i]==s[i]) or(p[idx[s[i]]]==idx[ch[i]])
i=i+1
return test
def cc (j):
if j<0 :
return ''
else:
return cc(j-1)+str(int((testv(t[j:j+len(s)],s,p))))
print(cc(len(t)-len(s)))
```
No
| 37,857 | [
0.1478271484375,
0.061309814453125,
0.0189056396484375,
0.08380126953125,
-0.611328125,
-0.42578125,
-0.323974609375,
0.02764892578125,
0.1959228515625,
1.1337890625,
0.58251953125,
-0.1378173828125,
-0.0582275390625,
-0.76806640625,
-0.82177734375,
0.03826904296875,
-0.7763671875,
... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a permutation p consisting of exactly 26 integers from 1 to 26 (since it is a permutation, each integer from 1 to 26 occurs in p exactly once) and two strings s and t consisting of lowercase Latin letters.
A substring t' of string t is an occurence of string s if the following conditions are met:
1. |t'| = |s|;
2. for each i ∈ [1, |s|], either s_i = t'_i, or p_{idx(s_i)} = idx(t'_i), where idx(c) is the index of character c in Latin alphabet (idx(a) = 1, idx(b) = 2, idx(z) = 26).
For example, if p_1 = 2, p_2 = 3, p_3 = 1, s = abc, t = abcaaba, then three substrings of t are occurences of s (they are t' = abc, t' = bca and t' = aba).
For each substring of t having length equal to |s|, check if it is an occurence of s.
Input
The first line contains 26 integers p_1, p_2, ..., p_{26} (1 ≤ p_i ≤ 26, all these integers are pairwise distinct).
The second line contains one string s, and the third line contains one string t (2 ≤ |s| ≤ |t| ≤ 2 ⋅ 10^5) both consisting of lowercase Latin letters.
Output
Print a string of |t| - |s| + 1 characters, each character should be either 0 or 1. The i-th character should be 1 if and only if the substring of t starting with the i-th character and ending with the (i + |s| - 1)-th character (inclusive) is an occurence of s.
Example
Input
2 3 1 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
abc
abcaaba
Output
11001
Submitted Solution:
```
a=list(map(int,input().split()))
l=list(input())
k=list(input())
x=len(l)
s=k[0:x]
y=len(s)
final=''
ind=y
for i in range(len(k)-x+1):
c=0
for j in range(y):
if(s[j]==l[j])or(ord(s[j])-97==a[j]-1):
c+=1
if(c==x):
final=final+'1'
else:
final=final+'0'
s.pop(0)
if(i!=len(k)-x):
s.append(k[ind])
ind+=1
print(final)
```
No
| 37,858 | [
0.19091796875,
0.1173095703125,
0.02099609375,
0.08306884765625,
-0.6396484375,
-0.46630859375,
-0.321533203125,
0.0606689453125,
0.17431640625,
1.09375,
0.6201171875,
-0.11346435546875,
-0.006866455078125,
-0.79541015625,
-0.79931640625,
0.02520751953125,
-0.73779296875,
-0.741210... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a permutation p consisting of exactly 26 integers from 1 to 26 (since it is a permutation, each integer from 1 to 26 occurs in p exactly once) and two strings s and t consisting of lowercase Latin letters.
A substring t' of string t is an occurence of string s if the following conditions are met:
1. |t'| = |s|;
2. for each i ∈ [1, |s|], either s_i = t'_i, or p_{idx(s_i)} = idx(t'_i), where idx(c) is the index of character c in Latin alphabet (idx(a) = 1, idx(b) = 2, idx(z) = 26).
For example, if p_1 = 2, p_2 = 3, p_3 = 1, s = abc, t = abcaaba, then three substrings of t are occurences of s (they are t' = abc, t' = bca and t' = aba).
For each substring of t having length equal to |s|, check if it is an occurence of s.
Input
The first line contains 26 integers p_1, p_2, ..., p_{26} (1 ≤ p_i ≤ 26, all these integers are pairwise distinct).
The second line contains one string s, and the third line contains one string t (2 ≤ |s| ≤ |t| ≤ 2 ⋅ 10^5) both consisting of lowercase Latin letters.
Output
Print a string of |t| - |s| + 1 characters, each character should be either 0 or 1. The i-th character should be 1 if and only if the substring of t starting with the i-th character and ending with the (i + |s| - 1)-th character (inclusive) is an occurence of s.
Example
Input
2 3 1 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
abc
abcaaba
Output
11001
Submitted Solution:
```
p=[int(x) for x in input().split()]
s=input()
t=input()
S=len(s);T=len(t)
for i in range(T-S+1):
k=t[i:i+S]
f=0
for j in range(len(k)):
ca=p[ord(s[j])-96-1]
cb=ord(k[j])-96
#print(ca,cb)
if(s[j]==k[j] or ca==cb):
f+=1
if(f==3):
print(1,end='')
else:
print(0,end='')
```
No
| 37,859 | [
0.1904296875,
0.11346435546875,
0.038604736328125,
0.09063720703125,
-0.61962890625,
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0.0523681640625,
0.1768798828125,
1.0849609375,
0.62890625,
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-0.04718017578125,
-0.8076171875,
-0.80322265625,
0.03302001953125,
-0.724609375,
-0... | 0 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n strings s1, s2, ..., sn consisting of characters 0 and 1. m operations are performed, on each of them you concatenate two existing strings into a new one. On the i-th operation the concatenation saisbi is saved into a new string sn + i (the operations are numbered starting from 1). After each operation you need to find the maximum positive integer k such that all possible strings consisting of 0 and 1 of length k (there are 2k such strings) are substrings of the new string. If there is no such k, print 0.
Input
The first line contains single integer n (1 ≤ n ≤ 100) — the number of strings. The next n lines contain strings s1, s2, ..., sn (1 ≤ |si| ≤ 100), one per line. The total length of strings is not greater than 100.
The next line contains single integer m (1 ≤ m ≤ 100) — the number of operations. m lines follow, each of them contains two integers ai abd bi (1 ≤ ai, bi ≤ n + i - 1) — the number of strings that are concatenated to form sn + i.
Output
Print m lines, each should contain one integer — the answer to the question after the corresponding operation.
Example
Input
5
01
10
101
11111
0
3
1 2
6 5
4 4
Output
1
2
0
Note
On the first operation, a new string "0110" is created. For k = 1 the two possible binary strings of length k are "0" and "1", they are substrings of the new string. For k = 2 and greater there exist strings of length k that do not appear in this string (for k = 2 such string is "00"). So the answer is 1.
On the second operation the string "01100" is created. Now all strings of length k = 2 are present.
On the third operation the string "1111111111" is created. There is no zero, so the answer is 0.
Tags: bitmasks, brute force, dp, implementation, strings
Correct Solution:
```
from sys import stdin, stdout
K = 20
def findAllStrings(s):
n = len(s)
sDict = {}
for i in range(1,K+1):
sDict[i]=set()
for x in range(n-i+1):
sDict[i].add(s[x:x+i])
return sDict
n = int(stdin.readline().rstrip())
stringDicts = []
stringEnd = []
stringBegin = []
for i in range(n):
s = stdin.readline().rstrip()
stringDicts.append(findAllStrings(s))
if len(s)<K:
stringEnd.append(s)
stringBegin.append(s)
else:
stringEnd.append(s[-20:])
stringBegin.append(s[:20])
m = int(stdin.readline().rstrip())
for _ in range(m):
a,b = map(int,stdin.readline().rstrip().split())
a-=1
b-=1
sDict1 = findAllStrings(stringEnd[a]+stringBegin[b])
sDict2 = stringDicts[a]
sDict3 = stringDicts[b]
sDict={}
for i in range(1,K+1):
sDict[i] = sDict1[i]|sDict2[i]|sDict3[i]
stringDicts.append(sDict)
for i in range(1,K+1):
if len(sDict[i])!=2**i:
print(i-1)
break
if len(stringBegin[a])<K and len(stringBegin[a])+len(stringBegin[b])<K:
stringBegin.append(stringBegin[a]+stringBegin[b])
elif len(stringBegin[a])<K:
s = stringBegin[a]+stringBegin[b]
stringBegin.append(s[:K])
else:
stringBegin.append(stringBegin[a])
if len(stringEnd[b])<K and len(stringEnd[a])+len(stringEnd[b])<K:
stringEnd.append(stringEnd[a]+stringEnd[b])
elif len(stringEnd[b])<K:
s = stringEnd[a]+stringEnd[b]
stringEnd.append(s[-K:])
else:
stringEnd.append(stringEnd[b])
```
| 38,264 | [
0.06707763671875,
-0.127685546875,
-0.0675048828125,
0.15185546875,
-0.365234375,
-0.9267578125,
-0.3447265625,
0.132568359375,
0.13232421875,
0.8515625,
0.6259765625,
-0.20556640625,
0.07666015625,
-0.8564453125,
-0.474365234375,
0.09869384765625,
-0.32177734375,
-0.58642578125,
... | 0 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n strings s1, s2, ..., sn consisting of characters 0 and 1. m operations are performed, on each of them you concatenate two existing strings into a new one. On the i-th operation the concatenation saisbi is saved into a new string sn + i (the operations are numbered starting from 1). After each operation you need to find the maximum positive integer k such that all possible strings consisting of 0 and 1 of length k (there are 2k such strings) are substrings of the new string. If there is no such k, print 0.
Input
The first line contains single integer n (1 ≤ n ≤ 100) — the number of strings. The next n lines contain strings s1, s2, ..., sn (1 ≤ |si| ≤ 100), one per line. The total length of strings is not greater than 100.
The next line contains single integer m (1 ≤ m ≤ 100) — the number of operations. m lines follow, each of them contains two integers ai abd bi (1 ≤ ai, bi ≤ n + i - 1) — the number of strings that are concatenated to form sn + i.
Output
Print m lines, each should contain one integer — the answer to the question after the corresponding operation.
Example
Input
5
01
10
101
11111
0
3
1 2
6 5
4 4
Output
1
2
0
Note
On the first operation, a new string "0110" is created. For k = 1 the two possible binary strings of length k are "0" and "1", they are substrings of the new string. For k = 2 and greater there exist strings of length k that do not appear in this string (for k = 2 such string is "00"). So the answer is 1.
On the second operation the string "01100" is created. Now all strings of length k = 2 are present.
On the third operation the string "1111111111" is created. There is no zero, so the answer is 0.
Tags: bitmasks, brute force, dp, implementation, strings
Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**15
mod = 10**9+7
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def main():
n = I()
t = [[S(), [set() for _ in range(151)]] for _ in range(n)]
m = I()
q = [LI_() for _ in range(m)]
for c in t:
s = c[0]
l = len(s)
for i in range(1,min(151,l+1)):
for j in range(l-i+1):
c[1][i].add(s[j:j+i])
rr = []
for li,ri in q:
l = t[li]
r = t[ri]
c = ['',[l[1][i]|r[1][i] for i in range(151)]]
for i in range(1,min(150,len(l[0])+1)):
ls = l[0][-i:]
for j in range(1,min(151-i,len(r[0])+1)):
c[1][i+j].add(ls+r[0][:j])
c[0] = l[0] + r[0]
if len(c[0]) > 300:
c[0] = c[0][:150] + c[0][-150:]
t.append(c)
r = 0
for i in range(1,151):
tt = len(c[1][i])
if tt != 2**i:
break
r = i
rr.append(r)
return '\n'.join(map(str,rr))
print(main())
```
| 38,265 | [
0.1201171875,
-0.11456298828125,
-0.17724609375,
0.229736328125,
-0.410888671875,
-0.86865234375,
-0.302978515625,
0.047821044921875,
0.1689453125,
0.82177734375,
0.69921875,
-0.1748046875,
0.08270263671875,
-0.900390625,
-0.50927734375,
0.12060546875,
-0.38916015625,
-0.5971679687... | 0 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n strings s1, s2, ..., sn consisting of characters 0 and 1. m operations are performed, on each of them you concatenate two existing strings into a new one. On the i-th operation the concatenation saisbi is saved into a new string sn + i (the operations are numbered starting from 1). After each operation you need to find the maximum positive integer k such that all possible strings consisting of 0 and 1 of length k (there are 2k such strings) are substrings of the new string. If there is no such k, print 0.
Input
The first line contains single integer n (1 ≤ n ≤ 100) — the number of strings. The next n lines contain strings s1, s2, ..., sn (1 ≤ |si| ≤ 100), one per line. The total length of strings is not greater than 100.
The next line contains single integer m (1 ≤ m ≤ 100) — the number of operations. m lines follow, each of them contains two integers ai abd bi (1 ≤ ai, bi ≤ n + i - 1) — the number of strings that are concatenated to form sn + i.
Output
Print m lines, each should contain one integer — the answer to the question after the corresponding operation.
Example
Input
5
01
10
101
11111
0
3
1 2
6 5
4 4
Output
1
2
0
Note
On the first operation, a new string "0110" is created. For k = 1 the two possible binary strings of length k are "0" and "1", they are substrings of the new string. For k = 2 and greater there exist strings of length k that do not appear in this string (for k = 2 such string is "00"). So the answer is 1.
On the second operation the string "01100" is created. Now all strings of length k = 2 are present.
On the third operation the string "1111111111" is created. There is no zero, so the answer is 0.
Tags: bitmasks, brute force, dp, implementation, strings
Correct Solution:
```
from sys import stdin, stdout
k = 20
def findAllStrings(s):
n = len(s)
sdict = {}
for i in range(1,k+1):
sdict[i]=set()
for x in range(n-i+1):
sdict[i].add(s[x:x+i])
return sdict
n = int(stdin.readline().rstrip())
strdict = []
stringend = []
stringbegin = []
for i in range(n):
s = stdin.readline().rstrip()
strdict.append(findAllStrings(s))
if len(s)<k:
stringend.append(s)
stringbegin.append(s)
else:
stringend.append(s[-20:])
stringbegin.append(s[:20])
m = int(stdin.readline().rstrip())
for _ in range(m):
a,b = map(int,stdin.readline().rstrip().split())
a-=1
b-=1
sdict1 = findAllStrings(stringend[a]+stringbegin[b])
sdict2 = strdict[a]
sdict3 = strdict[b]
sdict={}
for i in range(1,k+1):
sdict[i] = sdict1[i]|sdict2[i]|sdict3[i]
strdict.append(sdict)
for i in range(1,k+1):
if len(sdict[i])!=2**i:
print(i-1)
break
if len(stringbegin[a])<k and len(stringbegin[a])+len(stringbegin[b])<k:
stringbegin.append(stringbegin[a]+stringbegin[b])
elif len(stringbegin[a])<k:
s = stringbegin[a]+stringbegin[b]
stringbegin.append(s[:k])
else:
stringbegin.append(stringbegin[a])
if len(stringend[b])<k and len(stringend[a])+len(stringend[b])<k:
stringend.append(stringend[a]+stringend[b])
elif len(stringend[b])<k:
s = stringend[a]+stringend[b]
stringend.append(s[-k:])
else:
stringend.append(stringend[b])
```
| 38,266 | [
0.06707763671875,
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0.15185546875,
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-0.8564453125,
-0.474365234375,
0.09869384765625,
-0.32177734375,
-0.58642578125,
... | 0 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n strings s1, s2, ..., sn consisting of characters 0 and 1. m operations are performed, on each of them you concatenate two existing strings into a new one. On the i-th operation the concatenation saisbi is saved into a new string sn + i (the operations are numbered starting from 1). After each operation you need to find the maximum positive integer k such that all possible strings consisting of 0 and 1 of length k (there are 2k such strings) are substrings of the new string. If there is no such k, print 0.
Input
The first line contains single integer n (1 ≤ n ≤ 100) — the number of strings. The next n lines contain strings s1, s2, ..., sn (1 ≤ |si| ≤ 100), one per line. The total length of strings is not greater than 100.
The next line contains single integer m (1 ≤ m ≤ 100) — the number of operations. m lines follow, each of them contains two integers ai abd bi (1 ≤ ai, bi ≤ n + i - 1) — the number of strings that are concatenated to form sn + i.
Output
Print m lines, each should contain one integer — the answer to the question after the corresponding operation.
Example
Input
5
01
10
101
11111
0
3
1 2
6 5
4 4
Output
1
2
0
Note
On the first operation, a new string "0110" is created. For k = 1 the two possible binary strings of length k are "0" and "1", they are substrings of the new string. For k = 2 and greater there exist strings of length k that do not appear in this string (for k = 2 such string is "00"). So the answer is 1.
On the second operation the string "01100" is created. Now all strings of length k = 2 are present.
On the third operation the string "1111111111" is created. There is no zero, so the answer is 0.
Tags: bitmasks, brute force, dp, implementation, strings
Correct Solution:
```
from math import log
n = int(input())
p = [bin(p)[2:] for p in range(0,512)]
def mset(s):
ss = set()
for k in range(0,10):
for pi in range(0,2 ** k):
cs = p[pi]
cs = (k - len(cs)) * "0" + cs
if cs in s:
ss.add(cs)
return ss
def q(s):
for k in range(0,10):
for pi in range(0,2 ** k):
cs = p[pi]
cs = (k - len(cs)) * "0" + cs
if not cs in s:
return k - 1
s = [[v[:9], v[-9:], mset(v)] for v in [input() for i in range(n)]]
for qa, qb in [[int(v) - 1 for v in input().split()] for i in range(int(input()))]:
v = [s[qa][0], s[qb][1], mset(s[qa][1] + s[qb][0]) | s[qa][2] | s[qb][2]]
if len(v[0]) < 9:
v[0] = (v[0] + s[qb][0])[:9]
if len(v[1]) < 9:
v[1] = (s[qa][1] + s[qb][1])[-9:]
s += [v]
print(max(q(v[2]),0))
```
| 38,267 | [
0.1353759765625,
-0.171142578125,
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0.1455078125,
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0.11395263671875,
0.09893798828125,
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0.11181640625,
-0.96533203125,
-0.4619140625,
0.12445068359375,
-0.326171875,
-0.576... | 0 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n strings s1, s2, ..., sn consisting of characters 0 and 1. m operations are performed, on each of them you concatenate two existing strings into a new one. On the i-th operation the concatenation saisbi is saved into a new string sn + i (the operations are numbered starting from 1). After each operation you need to find the maximum positive integer k such that all possible strings consisting of 0 and 1 of length k (there are 2k such strings) are substrings of the new string. If there is no such k, print 0.
Input
The first line contains single integer n (1 ≤ n ≤ 100) — the number of strings. The next n lines contain strings s1, s2, ..., sn (1 ≤ |si| ≤ 100), one per line. The total length of strings is not greater than 100.
The next line contains single integer m (1 ≤ m ≤ 100) — the number of operations. m lines follow, each of them contains two integers ai abd bi (1 ≤ ai, bi ≤ n + i - 1) — the number of strings that are concatenated to form sn + i.
Output
Print m lines, each should contain one integer — the answer to the question after the corresponding operation.
Example
Input
5
01
10
101
11111
0
3
1 2
6 5
4 4
Output
1
2
0
Note
On the first operation, a new string "0110" is created. For k = 1 the two possible binary strings of length k are "0" and "1", they are substrings of the new string. For k = 2 and greater there exist strings of length k that do not appear in this string (for k = 2 such string is "00"). So the answer is 1.
On the second operation the string "01100" is created. Now all strings of length k = 2 are present.
On the third operation the string "1111111111" is created. There is no zero, so the answer is 0.
Tags: bitmasks, brute force, dp, implementation, strings
Correct Solution:
```
# -*- coding: utf-8 -*-
import math
import collections
import bisect
import heapq
import time
import random
"""
created by shhuan at 2017/10/5 15:00
"""
N = int(input())
S = ['']
for i in range(N):
S.append(input())
M = int(input())
# t0 = time.time()
# N = 3
# S = ["", "00010110000", "110101110101101010101101101110100010000001101101011000010001011000010101", "11100101100111010"]
# M = 1000
A = [[set() for _ in range(10)] for _ in range(M+N+1)]
D = collections.defaultdict(int)
for i in range(1, N+1):
for j in range(1, 10):
s = S[i]
if j > len(s):
break
for k in range(len(s)-j+1):
A[i][j].add(int(s[k:k+j], 2))
if all(v in A[i][j] for v in range(2**j)):
D[i] = j
for i in range(M):
# a, b = random.randint(1, i+N), random.randint(1, i+N)
a, b = map(int, input().split())
s, sa, sb = S[a] + S[b], S[a], S[b]
if len(s) > 30:
S.append(s[:10] + s[-10:])
else:
S.append(s)
ai = i+N+1
d = max(D[a], D[b]) + 1
for dv in range(d, 10):
if len(sa) + len(sb) < dv:
break
A[ai][dv] = A[a][dv] | A[b][dv] | {int(v, 2) for v in {sa[-i:] + sb[:dv-i] for i in range(1, dv+1)} if len(v) == dv}
ans = d-1
for dv in range(d, 10):
if any(v not in A[ai][dv] for v in range(2**dv)):
break
ans = dv
print(ans)
D[ai] = ans
# print(time.time() - t0)
```
| 38,268 | [
0.08038330078125,
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-0.55859375,
0.06884765625,
-0.396484375,
-0.609375,
-0.11181640... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n strings s1, s2, ..., sn consisting of characters 0 and 1. m operations are performed, on each of them you concatenate two existing strings into a new one. On the i-th operation the concatenation saisbi is saved into a new string sn + i (the operations are numbered starting from 1). After each operation you need to find the maximum positive integer k such that all possible strings consisting of 0 and 1 of length k (there are 2k such strings) are substrings of the new string. If there is no such k, print 0.
Input
The first line contains single integer n (1 ≤ n ≤ 100) — the number of strings. The next n lines contain strings s1, s2, ..., sn (1 ≤ |si| ≤ 100), one per line. The total length of strings is not greater than 100.
The next line contains single integer m (1 ≤ m ≤ 100) — the number of operations. m lines follow, each of them contains two integers ai abd bi (1 ≤ ai, bi ≤ n + i - 1) — the number of strings that are concatenated to form sn + i.
Output
Print m lines, each should contain one integer — the answer to the question after the corresponding operation.
Example
Input
5
01
10
101
11111
0
3
1 2
6 5
4 4
Output
1
2
0
Note
On the first operation, a new string "0110" is created. For k = 1 the two possible binary strings of length k are "0" and "1", they are substrings of the new string. For k = 2 and greater there exist strings of length k that do not appear in this string (for k = 2 such string is "00"). So the answer is 1.
On the second operation the string "01100" is created. Now all strings of length k = 2 are present.
On the third operation the string "1111111111" is created. There is no zero, so the answer is 0.
Submitted Solution:
```
import sys
read = sys.stdin.readline
def solve(n, numbers, m, concatenation):
for a, b in concatenation:
numbers.append(numbers[a-1] + numbers[b-1])
for i in range(n, n+m):
cur = numbers[i]
str_zero = "0"
str_one = "1"
find = cur.find(str_zero)
while find != -1:
str_zero += "0"
find = cur.find(str_zero)
find = cur.find(str_one)
while find != -1:
str_one += "1"
find = cur.find(str_one)
print(str(min(len(str_zero)-1, len(str_one)-1)))
def run():
n = int(read().replace("\n", "").replace("\r\n", ""))
numbers = []
for _ in range(n):
numbers.append(read().replace("\n", "").replace("\r\n", ""))
m = int(read().replace("\n", "").replace("\r\n", ""))
concatenation = []
for _ in range(m):
concatenation.append(list(map(int, read().replace("\n", "").replace("\r\n", "").split())))
solve(n, numbers, m, concatenation)
run()
```
No
| 38,269 | [
0.18115234375,
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0.1922607421875,
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0.2529296875,
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-0.489013671875,
-0.10638427734375,
-0.360107421875,
-0.624... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n strings s1, s2, ..., sn consisting of characters 0 and 1. m operations are performed, on each of them you concatenate two existing strings into a new one. On the i-th operation the concatenation saisbi is saved into a new string sn + i (the operations are numbered starting from 1). After each operation you need to find the maximum positive integer k such that all possible strings consisting of 0 and 1 of length k (there are 2k such strings) are substrings of the new string. If there is no such k, print 0.
Input
The first line contains single integer n (1 ≤ n ≤ 100) — the number of strings. The next n lines contain strings s1, s2, ..., sn (1 ≤ |si| ≤ 100), one per line. The total length of strings is not greater than 100.
The next line contains single integer m (1 ≤ m ≤ 100) — the number of operations. m lines follow, each of them contains two integers ai abd bi (1 ≤ ai, bi ≤ n + i - 1) — the number of strings that are concatenated to form sn + i.
Output
Print m lines, each should contain one integer — the answer to the question after the corresponding operation.
Example
Input
5
01
10
101
11111
0
3
1 2
6 5
4 4
Output
1
2
0
Note
On the first operation, a new string "0110" is created. For k = 1 the two possible binary strings of length k are "0" and "1", they are substrings of the new string. For k = 2 and greater there exist strings of length k that do not appear in this string (for k = 2 such string is "00"). So the answer is 1.
On the second operation the string "01100" is created. Now all strings of length k = 2 are present.
On the third operation the string "1111111111" is created. There is no zero, so the answer is 0.
Submitted Solution:
```
def kont(a,b):
ed=max(sp[a][1],sp[b][1])
ze=max(sp[a][0],sp[b][0])
if sp[a][4]==sp[b][2]:
if sp[a][4]=='1':
ed=max(ed,sp[a][5]+sp[b][3])
else:
ze=max(ed,sp[a][5]+sp[b][3])
if sp[b][6]==max(sp[b][3],sp[b][5]) and sp[b][4]==sp[a][2]:
kend=sp[b][6]+sp[a][5]
end=sp[b][2]
else:
kend=sp[b][5]
end=sp[b][4]
if sp[a][6]==max(sp[a][3],sp[a][5]) and sp[b][2]==sp[a][4]:
kbe=sp[a][6]+sp[b][3]
be=sp[b][2]
else:
kbe=sp[a][3]
be=sp[a][2]
sp.append((ze,ed,be,kbe,end,kend,sp[a][6]+sp[b][6]))
print(min(ed,ze))
n=int(input())
sp=[]
for o in range(n):
s=input()
ze=0
ed=0
el=s[0]
k=1
p=len(s)
for i in range(1,p):
if s[i]==el:
k+=1
else:
if el=='0':
ze=max(ze,k)
else:
ed=max(ed,k)
k=1
el=s[i]
if el=='0':
ze=max(ze,k)
else:
ed=max(ed,k)
end=el
kend=k
el=s[0]
k=1
for i in range(1,p):
if s[i]==el:
k+=1
else:
break
sp.append((ze,ed,el,k,end,kend,p))
n=int(input())
for i in range(n):
a,b=map(int,input().split())
a-=1
b-=1
kont(a,b)
```
No
| 38,270 | [
0.209716796875,
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0.23974609375,
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-0.390380859375,
... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n strings s1, s2, ..., sn consisting of characters 0 and 1. m operations are performed, on each of them you concatenate two existing strings into a new one. On the i-th operation the concatenation saisbi is saved into a new string sn + i (the operations are numbered starting from 1). After each operation you need to find the maximum positive integer k such that all possible strings consisting of 0 and 1 of length k (there are 2k such strings) are substrings of the new string. If there is no such k, print 0.
Input
The first line contains single integer n (1 ≤ n ≤ 100) — the number of strings. The next n lines contain strings s1, s2, ..., sn (1 ≤ |si| ≤ 100), one per line. The total length of strings is not greater than 100.
The next line contains single integer m (1 ≤ m ≤ 100) — the number of operations. m lines follow, each of them contains two integers ai abd bi (1 ≤ ai, bi ≤ n + i - 1) — the number of strings that are concatenated to form sn + i.
Output
Print m lines, each should contain one integer — the answer to the question after the corresponding operation.
Example
Input
5
01
10
101
11111
0
3
1 2
6 5
4 4
Output
1
2
0
Note
On the first operation, a new string "0110" is created. For k = 1 the two possible binary strings of length k are "0" and "1", they are substrings of the new string. For k = 2 and greater there exist strings of length k that do not appear in this string (for k = 2 such string is "00"). So the answer is 1.
On the second operation the string "01100" is created. Now all strings of length k = 2 are present.
On the third operation the string "1111111111" is created. There is no zero, so the answer is 0.
Submitted Solution:
```
from math import log
n = int(input())
p = [bin(p)[2:] for p in range(0,256)]
def q(s):
for k in range(0,10):
for pi in range(0,2**k):
cs = p[pi]
cs = (k-len(cs))*"0"+cs
if not cs in s:
return k - 1
s = [(v, q(v)) for v in [input() for i in range(n)]]
for qa, qb in [[int(v) - 1 for v in input().split()] for i in range(int(input()))]:
vs = s[qa][0] + s[qb][0]
qs = q(vs)
if len(vs) > 100:
vs = vs[:50] + vs[-50:]
v = (vs, max(qs, s[qa][1], s[qb][1], 0))
s += [v]
print(v[1])
```
No
| 38,271 | [
0.2032470703125,
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0.1456298828125,
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0.036346435546875,
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-0.44140625,
-0.07720947265625,
-0.378662109375,
-0.6166... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n strings s1, s2, ..., sn consisting of characters 0 and 1. m operations are performed, on each of them you concatenate two existing strings into a new one. On the i-th operation the concatenation saisbi is saved into a new string sn + i (the operations are numbered starting from 1). After each operation you need to find the maximum positive integer k such that all possible strings consisting of 0 and 1 of length k (there are 2k such strings) are substrings of the new string. If there is no such k, print 0.
Input
The first line contains single integer n (1 ≤ n ≤ 100) — the number of strings. The next n lines contain strings s1, s2, ..., sn (1 ≤ |si| ≤ 100), one per line. The total length of strings is not greater than 100.
The next line contains single integer m (1 ≤ m ≤ 100) — the number of operations. m lines follow, each of them contains two integers ai abd bi (1 ≤ ai, bi ≤ n + i - 1) — the number of strings that are concatenated to form sn + i.
Output
Print m lines, each should contain one integer — the answer to the question after the corresponding operation.
Example
Input
5
01
10
101
11111
0
3
1 2
6 5
4 4
Output
1
2
0
Note
On the first operation, a new string "0110" is created. For k = 1 the two possible binary strings of length k are "0" and "1", they are substrings of the new string. For k = 2 and greater there exist strings of length k that do not appear in this string (for k = 2 such string is "00"). So the answer is 1.
On the second operation the string "01100" is created. Now all strings of length k = 2 are present.
On the third operation the string "1111111111" is created. There is no zero, so the answer is 0.
Submitted Solution:
```
from math import log
n = int(input())
p = [bin(p)[2:] for p in range(0,256)]
def q(s):
for k in range(0,10):
for pi in range(0,2**k):
cs = p[pi]
cs = (k-len(cs))*"0"+cs
if not cs in s:
return k - 1
node = lambda s: (s[:9],s[-9:],q(s))
s = [node(v) for v in [input() for i in range(n)]]
for qa, qb in [[int(v) - 1 for v in input().split()] for i in range(int(input()))]:
vs = s[qa][0] + "#" + s[qa][1] + s[qb][0] + "#" + s[qb][1]
v = (vs.replace("#","")[:9], vs.replace("#","")[-9:], max(0, s[qa][2], s[qb][2], q(vs)))
s += [v]
print(v)
```
No
| 38,272 | [
0.2032470703125,
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0.1456298828125,
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0.036346435546875,
-0.93115234375,
-0.44140625,
-0.07720947265625,
-0.378662109375,
-0.6166... | 0 |
Provide tags and a correct Python 3 solution for this coding contest problem.
"QAQ" is a word to denote an expression of crying. Imagine "Q" as eyes with tears and "A" as a mouth.
Now Diamond has given Bort a string consisting of only uppercase English letters of length n. There is a great number of "QAQ" in the string (Diamond is so cute!).
<image> illustration by 猫屋 https://twitter.com/nekoyaliu
Bort wants to know how many subsequences "QAQ" are in the string Diamond has given. Note that the letters "QAQ" don't have to be consecutive, but the order of letters should be exact.
Input
The only line contains a string of length n (1 ≤ n ≤ 100). It's guaranteed that the string only contains uppercase English letters.
Output
Print a single integer — the number of subsequences "QAQ" in the string.
Examples
Input
QAQAQYSYIOIWIN
Output
4
Input
QAQQQZZYNOIWIN
Output
3
Note
In the first example there are 4 subsequences "QAQ": "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN".
Tags: brute force, dp
Correct Solution:
```
"""
tb
"""
s=input()
c=0
l=len(s)
arr=[0]
for i in s:
if i=='Q':
arr.append(arr[-1]+1)
else:
arr.append(arr[-1])
for i in range(l):
if s[i]=='A':
c=c+arr[i+1]*(arr[-1]-arr[i+1])
print(c)
```
| 38,273 | [
0.2098388671875,
-0.1234130859375,
0.13916015625,
0.318115234375,
-0.369384765625,
-0.7177734375,
0.0848388671875,
0.09613037109375,
0.4345703125,
0.82470703125,
0.6552734375,
-0.090576171875,
-0.1639404296875,
-0.76318359375,
-0.429931640625,
-0.09185791015625,
-0.3115234375,
-0.8... | 0 |
Provide tags and a correct Python 3 solution for this coding contest problem.
"QAQ" is a word to denote an expression of crying. Imagine "Q" as eyes with tears and "A" as a mouth.
Now Diamond has given Bort a string consisting of only uppercase English letters of length n. There is a great number of "QAQ" in the string (Diamond is so cute!).
<image> illustration by 猫屋 https://twitter.com/nekoyaliu
Bort wants to know how many subsequences "QAQ" are in the string Diamond has given. Note that the letters "QAQ" don't have to be consecutive, but the order of letters should be exact.
Input
The only line contains a string of length n (1 ≤ n ≤ 100). It's guaranteed that the string only contains uppercase English letters.
Output
Print a single integer — the number of subsequences "QAQ" in the string.
Examples
Input
QAQAQYSYIOIWIN
Output
4
Input
QAQQQZZYNOIWIN
Output
3
Note
In the first example there are 4 subsequences "QAQ": "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN".
Tags: brute force, dp
Correct Solution:
```
s = input()
res = 0
for i in range(0, len(s)):
if s[i] == 'Q':
for i in range(i + 1, len(s)):
if s[i] == 'A':
for i in range(i + 1, len(s)):
if s[i] == 'Q':
res += 1
print(res)
```
| 38,274 | [
0.21728515625,
-0.1729736328125,
0.1414794921875,
0.3193359375,
-0.40234375,
-0.69677734375,
0.1187744140625,
0.10296630859375,
0.4482421875,
0.80712890625,
0.6337890625,
-0.07452392578125,
-0.13720703125,
-0.72802734375,
-0.370361328125,
-0.0977783203125,
-0.26904296875,
-0.836914... | 0 |
Provide tags and a correct Python 3 solution for this coding contest problem.
"QAQ" is a word to denote an expression of crying. Imagine "Q" as eyes with tears and "A" as a mouth.
Now Diamond has given Bort a string consisting of only uppercase English letters of length n. There is a great number of "QAQ" in the string (Diamond is so cute!).
<image> illustration by 猫屋 https://twitter.com/nekoyaliu
Bort wants to know how many subsequences "QAQ" are in the string Diamond has given. Note that the letters "QAQ" don't have to be consecutive, but the order of letters should be exact.
Input
The only line contains a string of length n (1 ≤ n ≤ 100). It's guaranteed that the string only contains uppercase English letters.
Output
Print a single integer — the number of subsequences "QAQ" in the string.
Examples
Input
QAQAQYSYIOIWIN
Output
4
Input
QAQQQZZYNOIWIN
Output
3
Note
In the first example there are 4 subsequences "QAQ": "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN".
Tags: brute force, dp
Correct Solution:
```
a=[]
for i in input():
if i in "AQ":
a.append(i)
n=len(a)
count=0
for i in range(n):
for j in range(n):
for k in range(n):
if a[i]==a[k]=="Q" and a[j]=="A" and i<j<k:
count+=1
print(count)
```
| 38,275 | [
0.2347412109375,
-0.15673828125,
0.133056640625,
0.341064453125,
-0.402099609375,
-0.73193359375,
0.1092529296875,
0.1365966796875,
0.47216796875,
0.8359375,
0.60595703125,
-0.09136962890625,
-0.146240234375,
-0.73388671875,
-0.377197265625,
-0.07177734375,
-0.2802734375,
-0.850097... | 0 |
Provide tags and a correct Python 3 solution for this coding contest problem.
"QAQ" is a word to denote an expression of crying. Imagine "Q" as eyes with tears and "A" as a mouth.
Now Diamond has given Bort a string consisting of only uppercase English letters of length n. There is a great number of "QAQ" in the string (Diamond is so cute!).
<image> illustration by 猫屋 https://twitter.com/nekoyaliu
Bort wants to know how many subsequences "QAQ" are in the string Diamond has given. Note that the letters "QAQ" don't have to be consecutive, but the order of letters should be exact.
Input
The only line contains a string of length n (1 ≤ n ≤ 100). It's guaranteed that the string only contains uppercase English letters.
Output
Print a single integer — the number of subsequences "QAQ" in the string.
Examples
Input
QAQAQYSYIOIWIN
Output
4
Input
QAQQQZZYNOIWIN
Output
3
Note
In the first example there are 4 subsequences "QAQ": "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN".
Tags: brute force, dp
Correct Solution:
```
s = input()
res = 0
while True:
a = s.find("A")
if a == -1:
break
res += s[:a].count("Q") * s[a + 1:].count("Q")
s = s[:a] + s[a + 1:]
print(res)
```
| 38,276 | [
0.251220703125,
-0.1351318359375,
0.0953369140625,
0.317138671875,
-0.37353515625,
-0.7177734375,
0.061798095703125,
0.1357421875,
0.443359375,
0.7939453125,
0.5830078125,
-0.07647705078125,
-0.1397705078125,
-0.671875,
-0.334716796875,
-0.10302734375,
-0.2119140625,
-0.8125,
-0.... | 0 |
Provide tags and a correct Python 3 solution for this coding contest problem.
"QAQ" is a word to denote an expression of crying. Imagine "Q" as eyes with tears and "A" as a mouth.
Now Diamond has given Bort a string consisting of only uppercase English letters of length n. There is a great number of "QAQ" in the string (Diamond is so cute!).
<image> illustration by 猫屋 https://twitter.com/nekoyaliu
Bort wants to know how many subsequences "QAQ" are in the string Diamond has given. Note that the letters "QAQ" don't have to be consecutive, but the order of letters should be exact.
Input
The only line contains a string of length n (1 ≤ n ≤ 100). It's guaranteed that the string only contains uppercase English letters.
Output
Print a single integer — the number of subsequences "QAQ" in the string.
Examples
Input
QAQAQYSYIOIWIN
Output
4
Input
QAQQQZZYNOIWIN
Output
3
Note
In the first example there are 4 subsequences "QAQ": "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN".
Tags: brute force, dp
Correct Solution:
```
s = input()
cnt = 0
for i in range(len(s)):
for j in range(i + 1, len(s)):
for k in range(j + 1, len(s)):
if s[i] == s[k] == 'Q' and s[j] == 'A':
cnt += 1
print(cnt)
```
| 38,277 | [
0.22998046875,
-0.1505126953125,
0.1580810546875,
0.30029296875,
-0.387939453125,
-0.69921875,
0.11395263671875,
0.09429931640625,
0.439453125,
0.83642578125,
0.62744140625,
-0.061798095703125,
-0.125244140625,
-0.75341796875,
-0.38623046875,
-0.08197021484375,
-0.267822265625,
-0.... | 0 |
Provide tags and a correct Python 3 solution for this coding contest problem.
"QAQ" is a word to denote an expression of crying. Imagine "Q" as eyes with tears and "A" as a mouth.
Now Diamond has given Bort a string consisting of only uppercase English letters of length n. There is a great number of "QAQ" in the string (Diamond is so cute!).
<image> illustration by 猫屋 https://twitter.com/nekoyaliu
Bort wants to know how many subsequences "QAQ" are in the string Diamond has given. Note that the letters "QAQ" don't have to be consecutive, but the order of letters should be exact.
Input
The only line contains a string of length n (1 ≤ n ≤ 100). It's guaranteed that the string only contains uppercase English letters.
Output
Print a single integer — the number of subsequences "QAQ" in the string.
Examples
Input
QAQAQYSYIOIWIN
Output
4
Input
QAQQQZZYNOIWIN
Output
3
Note
In the first example there are 4 subsequences "QAQ": "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN".
Tags: brute force, dp
Correct Solution:
```
s = str(input())
c=0
for i in range(len(s)):
for j in range(i,len(s)):
for k in range(j,len(s)):
if s[i]=="Q" and s[j]=="A" and s[k]=="Q": c+=1
print(c)
```
| 38,278 | [
0.222412109375,
-0.18212890625,
0.1402587890625,
0.300537109375,
-0.422119140625,
-0.6923828125,
0.08355712890625,
0.11700439453125,
0.423828125,
0.79345703125,
0.62548828125,
-0.0882568359375,
-0.135986328125,
-0.7177734375,
-0.37353515625,
-0.0848388671875,
-0.29541015625,
-0.842... | 0 |
Provide tags and a correct Python 3 solution for this coding contest problem.
"QAQ" is a word to denote an expression of crying. Imagine "Q" as eyes with tears and "A" as a mouth.
Now Diamond has given Bort a string consisting of only uppercase English letters of length n. There is a great number of "QAQ" in the string (Diamond is so cute!).
<image> illustration by 猫屋 https://twitter.com/nekoyaliu
Bort wants to know how many subsequences "QAQ" are in the string Diamond has given. Note that the letters "QAQ" don't have to be consecutive, but the order of letters should be exact.
Input
The only line contains a string of length n (1 ≤ n ≤ 100). It's guaranteed that the string only contains uppercase English letters.
Output
Print a single integer — the number of subsequences "QAQ" in the string.
Examples
Input
QAQAQYSYIOIWIN
Output
4
Input
QAQQQZZYNOIWIN
Output
3
Note
In the first example there are 4 subsequences "QAQ": "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN".
Tags: brute force, dp
Correct Solution:
```
a = 0
d = 0
ans = 0
for i in input():
if i == "Q":
ans += d
a += 1
elif i== "A":
d += a
print(ans)
```
| 38,279 | [
0.264892578125,
-0.1590576171875,
0.143310546875,
0.333251953125,
-0.37548828125,
-0.7138671875,
0.10528564453125,
0.11346435546875,
0.46142578125,
0.78369140625,
0.59814453125,
-0.08563232421875,
-0.145751953125,
-0.66650390625,
-0.366943359375,
-0.08526611328125,
-0.226318359375,
... | 0 |
Provide tags and a correct Python 3 solution for this coding contest problem.
"QAQ" is a word to denote an expression of crying. Imagine "Q" as eyes with tears and "A" as a mouth.
Now Diamond has given Bort a string consisting of only uppercase English letters of length n. There is a great number of "QAQ" in the string (Diamond is so cute!).
<image> illustration by 猫屋 https://twitter.com/nekoyaliu
Bort wants to know how many subsequences "QAQ" are in the string Diamond has given. Note that the letters "QAQ" don't have to be consecutive, but the order of letters should be exact.
Input
The only line contains a string of length n (1 ≤ n ≤ 100). It's guaranteed that the string only contains uppercase English letters.
Output
Print a single integer — the number of subsequences "QAQ" in the string.
Examples
Input
QAQAQYSYIOIWIN
Output
4
Input
QAQQQZZYNOIWIN
Output
3
Note
In the first example there are 4 subsequences "QAQ": "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN".
Tags: brute force, dp
Correct Solution:
```
s=list(input())
cnt=0
for i in range(len(s)):
if s[i]=='Q':
for j in range(i,len(s)):
if s[j]=='A':
for k in range(j,len(s)):
if s[k]=='Q':
cnt+=1
print(cnt)
```
| 38,280 | [
0.2154541015625,
-0.168212890625,
0.150390625,
0.314697265625,
-0.3740234375,
-0.7060546875,
0.11688232421875,
0.11041259765625,
0.4677734375,
0.79736328125,
0.630859375,
-0.07330322265625,
-0.1317138671875,
-0.72509765625,
-0.399658203125,
-0.076416015625,
-0.287109375,
-0.8422851... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
"QAQ" is a word to denote an expression of crying. Imagine "Q" as eyes with tears and "A" as a mouth.
Now Diamond has given Bort a string consisting of only uppercase English letters of length n. There is a great number of "QAQ" in the string (Diamond is so cute!).
<image> illustration by 猫屋 https://twitter.com/nekoyaliu
Bort wants to know how many subsequences "QAQ" are in the string Diamond has given. Note that the letters "QAQ" don't have to be consecutive, but the order of letters should be exact.
Input
The only line contains a string of length n (1 ≤ n ≤ 100). It's guaranteed that the string only contains uppercase English letters.
Output
Print a single integer — the number of subsequences "QAQ" in the string.
Examples
Input
QAQAQYSYIOIWIN
Output
4
Input
QAQQQZZYNOIWIN
Output
3
Note
In the first example there are 4 subsequences "QAQ": "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN".
Submitted Solution:
```
s = str(input())
counter = 0
for i in range(len(s)):
if s[i] == 'Q':
for j in range(i, len(s)):
if s[j] == 'A':
for k in range(j, len(s)):
if s[k] == 'Q':
counter += 1
print(counter)
```
Yes
| 38,281 | [
0.2420654296875,
-0.1434326171875,
0.082763671875,
0.2275390625,
-0.51611328125,
-0.62841796875,
0.0236663818359375,
0.154052734375,
0.363525390625,
0.798828125,
0.57177734375,
-0.107421875,
-0.157958984375,
-0.744140625,
-0.469970703125,
-0.1263427734375,
-0.324951171875,
-0.88574... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
"QAQ" is a word to denote an expression of crying. Imagine "Q" as eyes with tears and "A" as a mouth.
Now Diamond has given Bort a string consisting of only uppercase English letters of length n. There is a great number of "QAQ" in the string (Diamond is so cute!).
<image> illustration by 猫屋 https://twitter.com/nekoyaliu
Bort wants to know how many subsequences "QAQ" are in the string Diamond has given. Note that the letters "QAQ" don't have to be consecutive, but the order of letters should be exact.
Input
The only line contains a string of length n (1 ≤ n ≤ 100). It's guaranteed that the string only contains uppercase English letters.
Output
Print a single integer — the number of subsequences "QAQ" in the string.
Examples
Input
QAQAQYSYIOIWIN
Output
4
Input
QAQQQZZYNOIWIN
Output
3
Note
In the first example there are 4 subsequences "QAQ": "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN".
Submitted Solution:
```
from itertools import combinations as c
print(sum(map(lambda x: x == ('Q', 'A', 'Q'), c((i for i in input() if i in 'QA'), 3))))
```
Yes
| 38,282 | [
0.2880859375,
-0.16259765625,
0.09185791015625,
0.2171630859375,
-0.493896484375,
-0.6513671875,
-0.006458282470703125,
0.1583251953125,
0.322021484375,
0.7705078125,
0.5751953125,
-0.124755859375,
-0.1654052734375,
-0.67138671875,
-0.440185546875,
-0.11737060546875,
-0.266357421875,... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
"QAQ" is a word to denote an expression of crying. Imagine "Q" as eyes with tears and "A" as a mouth.
Now Diamond has given Bort a string consisting of only uppercase English letters of length n. There is a great number of "QAQ" in the string (Diamond is so cute!).
<image> illustration by 猫屋 https://twitter.com/nekoyaliu
Bort wants to know how many subsequences "QAQ" are in the string Diamond has given. Note that the letters "QAQ" don't have to be consecutive, but the order of letters should be exact.
Input
The only line contains a string of length n (1 ≤ n ≤ 100). It's guaranteed that the string only contains uppercase English letters.
Output
Print a single integer — the number of subsequences "QAQ" in the string.
Examples
Input
QAQAQYSYIOIWIN
Output
4
Input
QAQQQZZYNOIWIN
Output
3
Note
In the first example there are 4 subsequences "QAQ": "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN".
Submitted Solution:
```
from functools import lru_cache
def subseqsearch(string,substr):
substrset=set(substr)
#fixs has only element in substr
fixs = [i for i in string if i in substrset]
@lru_cache(maxsize=None) #memoisation decorator applyed to recs()
def recs(fi=0,si=0):
if si >= len(substr):
return 1
r=0
for i in range(fi,len(fixs)):
if substr[si] == fixs[i]:
r+=recs(i+1,si+1)
return r
return recs()
#test
from functools import reduce
def flat(i) : return reduce(lambda x,y:x+y,i,[])
string = input()
substr = "QAQ"
#print("".join(str(i) for i in string),"substr:","".join(str(i) for i in substr),sep="\n")
print(subseqsearch(string,substr))
```
Yes
| 38,283 | [
0.2454833984375,
-0.1966552734375,
0.03448486328125,
0.1968994140625,
-0.54443359375,
-0.489501953125,
-0.005084991455078125,
0.130615234375,
0.41845703125,
0.7724609375,
0.595703125,
-0.2100830078125,
-0.062347412109375,
-0.65478515625,
-0.498046875,
-0.05712890625,
-0.30419921875,
... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
"QAQ" is a word to denote an expression of crying. Imagine "Q" as eyes with tears and "A" as a mouth.
Now Diamond has given Bort a string consisting of only uppercase English letters of length n. There is a great number of "QAQ" in the string (Diamond is so cute!).
<image> illustration by 猫屋 https://twitter.com/nekoyaliu
Bort wants to know how many subsequences "QAQ" are in the string Diamond has given. Note that the letters "QAQ" don't have to be consecutive, but the order of letters should be exact.
Input
The only line contains a string of length n (1 ≤ n ≤ 100). It's guaranteed that the string only contains uppercase English letters.
Output
Print a single integer — the number of subsequences "QAQ" in the string.
Examples
Input
QAQAQYSYIOIWIN
Output
4
Input
QAQQQZZYNOIWIN
Output
3
Note
In the first example there are 4 subsequences "QAQ": "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN".
Submitted Solution:
```
x=input()
c=0
for i in range(0,len(x)):
if x[i]=='Q':
for j in range(i,len(x)):
if x[j]=='A':
for k in range(j,len(x)):
if x[k]=='Q':
c+=1
print(c)
```
Yes
| 38,284 | [
0.2998046875,
-0.150146484375,
0.09735107421875,
0.25244140625,
-0.495849609375,
-0.6572265625,
0.045440673828125,
0.156494140625,
0.350830078125,
0.8369140625,
0.5869140625,
-0.08428955078125,
-0.174072265625,
-0.7646484375,
-0.470703125,
-0.10797119140625,
-0.302978515625,
-0.882... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
"QAQ" is a word to denote an expression of crying. Imagine "Q" as eyes with tears and "A" as a mouth.
Now Diamond has given Bort a string consisting of only uppercase English letters of length n. There is a great number of "QAQ" in the string (Diamond is so cute!).
<image> illustration by 猫屋 https://twitter.com/nekoyaliu
Bort wants to know how many subsequences "QAQ" are in the string Diamond has given. Note that the letters "QAQ" don't have to be consecutive, but the order of letters should be exact.
Input
The only line contains a string of length n (1 ≤ n ≤ 100). It's guaranteed that the string only contains uppercase English letters.
Output
Print a single integer — the number of subsequences "QAQ" in the string.
Examples
Input
QAQAQYSYIOIWIN
Output
4
Input
QAQQQZZYNOIWIN
Output
3
Note
In the first example there are 4 subsequences "QAQ": "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN".
Submitted Solution:
```
import math as mt
import sys,string,bisect
input=sys.stdin.readline
import random
from collections import deque,defaultdict
L=lambda : list(map(int,input().split()))
Ls=lambda : list(input().split())
M=lambda : map(int,input().split())
I=lambda :int(input())
s=input().strip()
i=0
ca=0
cq=0
f=0
ans=0
while(i<len(s)):
if(f==0):
if(s[i]=="Q"):
f=1
cq+=1
else:
if(s[i]=="Q"):
ans=ca
cq+=1
elif(s[i]=="A"):
ca+=1
i+=1
print((cq-1)*ans)
```
No
| 38,285 | [
0.2181396484375,
-0.03564453125,
0.050750732421875,
0.190673828125,
-0.51416015625,
-0.5205078125,
-0.0175933837890625,
0.17919921875,
0.370361328125,
0.83447265625,
0.59033203125,
-0.1307373046875,
-0.158203125,
-0.74951171875,
-0.484619140625,
-0.07391357421875,
-0.275634765625,
... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
"QAQ" is a word to denote an expression of crying. Imagine "Q" as eyes with tears and "A" as a mouth.
Now Diamond has given Bort a string consisting of only uppercase English letters of length n. There is a great number of "QAQ" in the string (Diamond is so cute!).
<image> illustration by 猫屋 https://twitter.com/nekoyaliu
Bort wants to know how many subsequences "QAQ" are in the string Diamond has given. Note that the letters "QAQ" don't have to be consecutive, but the order of letters should be exact.
Input
The only line contains a string of length n (1 ≤ n ≤ 100). It's guaranteed that the string only contains uppercase English letters.
Output
Print a single integer — the number of subsequences "QAQ" in the string.
Examples
Input
QAQAQYSYIOIWIN
Output
4
Input
QAQQQZZYNOIWIN
Output
3
Note
In the first example there are 4 subsequences "QAQ": "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN".
Submitted Solution:
```
s = input()
cnt = 0
lst = []
n = len(s)
for i in s:
if i == "Q":
cnt += 1
lst.append(cnt)
ans = 0
for i in range(n):
if s[i] == "A":
ans += (lst[i]*(lst[n - i - 1] - lst[i]))
print(ans)
```
No
| 38,286 | [
0.25390625,
-0.131591796875,
0.1591796875,
0.193359375,
-0.4892578125,
-0.6416015625,
0.043853759765625,
0.179443359375,
0.380859375,
0.80029296875,
0.556640625,
-0.1097412109375,
-0.1632080078125,
-0.73095703125,
-0.484130859375,
-0.1192626953125,
-0.30712890625,
-0.853515625,
-... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
"QAQ" is a word to denote an expression of crying. Imagine "Q" as eyes with tears and "A" as a mouth.
Now Diamond has given Bort a string consisting of only uppercase English letters of length n. There is a great number of "QAQ" in the string (Diamond is so cute!).
<image> illustration by 猫屋 https://twitter.com/nekoyaliu
Bort wants to know how many subsequences "QAQ" are in the string Diamond has given. Note that the letters "QAQ" don't have to be consecutive, but the order of letters should be exact.
Input
The only line contains a string of length n (1 ≤ n ≤ 100). It's guaranteed that the string only contains uppercase English letters.
Output
Print a single integer — the number of subsequences "QAQ" in the string.
Examples
Input
QAQAQYSYIOIWIN
Output
4
Input
QAQQQZZYNOIWIN
Output
3
Note
In the first example there are 4 subsequences "QAQ": "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN".
Submitted Solution:
```
n=input()
i=len(n)
l=[0]*(i)
c=0
for i in range(0,len(n)):
if n[i]=='Q':
c=c+1
l[i]=c
ans=0
total=l[len(n)-1]
for i in range(0,len(n)):
if n[i]=='A':
f1=l[i]
f2=total-f1
ans=ans+max(f1,f2)
print(ans)
```
No
| 38,288 | [
0.2440185546875,
-0.1290283203125,
0.116455078125,
0.237548828125,
-0.461181640625,
-0.62158203125,
0.07806396484375,
0.150146484375,
0.374755859375,
0.81298828125,
0.59326171875,
-0.08966064453125,
-0.145263671875,
-0.734375,
-0.4755859375,
-0.04638671875,
-0.290771484375,
-0.8935... | 0 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a string S of length 2N, containing N occurrences of `a` and N occurrences of `b`.
You will choose some of the characters in S. Here, for each i = 1,2,...,N, it is not allowed to choose exactly one of the following two: the i-th occurrence of `a` and the i-th occurrence of `b`. (That is, you can only choose both or neither.) Then, you will concatenate the chosen characters (without changing the order).
Find the lexicographically largest string that can be obtained in this way.
Constraints
* 1 \leq N \leq 3000
* S is a string of length 2N containing N occurrences of `a` and N occurrences of `b`.
Input
Input is given from Standard Input in the following format:
N
S
Output
Print the lexicographically largest string that satisfies the condition.
Examples
Input
3
aababb
Output
abab
Input
3
bbabaa
Output
bbabaa
Input
6
bbbaabbabaaa
Output
bbbabaaa
Input
9
abbbaababaababbaba
Output
bbaababababa
"Correct Solution:
```
from itertools import accumulate
def search_ab(sss, cursor):
# print('a', cursor)
ai = aaa.index(cursor)
tmp_cur = bbb[ai]
max_cur = sss.index(0, cursor)
repeat = 1
while tmp_cur < max_cur:
cur = s.find('a', tmp_cur, max_cur)
if cur == -1:
break
ai = aaa.index(cur, ai)
tmp_cur = bbb[ai]
repeat += 1
return repeat, max_cur + 1
def search_ba(sss, cursor):
# print('b', cursor)
first_bi = bbb.index(cursor)
max_cursor = sss.index(0, cursor)
last_bi = aaa.index(max_cursor)
tmp_buf = [''] * (last_bi - first_bi + 1) * 2
tmp_max = ''
for i in range(last_bi, first_bi - 1, -1):
tmp_buf[aaa[i] - cursor] = 'a'
tmp_buf[bbb[i] - cursor] = 'b'
tmp = ''.join(tmp_buf)
if tmp > tmp_max:
tmp_max = tmp
return tmp_max, max_cursor + 1
def integrate(parts_b):
tmp_max = ''
for pb in reversed(parts_b):
tmp = pb + tmp_max
if tmp > tmp_max:
tmp_max = tmp
return tmp_max
n = int(input())
s = input()
n2 = n * 2
sss = []
aaa, bbb = [], []
for i, c in enumerate(s):
if c == 'a':
aaa.append(i)
sss.append(-1)
else:
bbb.append(i)
sss.append(1)
sss = list(accumulate(sss))
repeats_a = []
parts_b = []
last_b_cur = 0
cursor = 0
while cursor < n2:
c = sss[cursor]
if c < 0:
repeat, cursor = search_ab(sss, cursor)
repeats_a.append((cursor, repeat))
else:
tmp, cursor = search_ba(sss, cursor)
parts_b.append(tmp)
last_b_cur = cursor
print(integrate(parts_b) + 'ab' * sum(r for c, r in repeats_a if c > last_b_cur))
```
| 38,408 | [
0.258544921875,
-0.10662841796875,
0.2010498046875,
0.446533203125,
-0.50048828125,
-0.1768798828125,
-0.3955078125,
0.036651611328125,
0.25537109375,
0.57421875,
0.6513671875,
-0.54052734375,
-0.1126708984375,
-0.87451171875,
-0.63818359375,
-0.1505126953125,
-0.80517578125,
-0.55... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string S of length 2N, containing N occurrences of `a` and N occurrences of `b`.
You will choose some of the characters in S. Here, for each i = 1,2,...,N, it is not allowed to choose exactly one of the following two: the i-th occurrence of `a` and the i-th occurrence of `b`. (That is, you can only choose both or neither.) Then, you will concatenate the chosen characters (without changing the order).
Find the lexicographically largest string that can be obtained in this way.
Constraints
* 1 \leq N \leq 3000
* S is a string of length 2N containing N occurrences of `a` and N occurrences of `b`.
Input
Input is given from Standard Input in the following format:
N
S
Output
Print the lexicographically largest string that satisfies the condition.
Examples
Input
3
aababb
Output
abab
Input
3
bbabaa
Output
bbabaa
Input
6
bbbaabbabaaa
Output
bbbabaaa
Input
9
abbbaababaababbaba
Output
bbaababababa
Submitted Solution:
```
n = int(input())
s = input()
ab_index = {'a': 0, 'b': 0}
ab_dic = {}
i = 0
while i < (2 * n):
ab_dic[i] = (s[i], ab_index[s[i]])
ab_index[s[i]] += 1
i += 1
import copy
class Info:
def __init__(self, string, includeds, excludeds):
self.string = string
self.includeds = includeds
self.excludeds = excludeds
def include(self, s, i):
dic = copy.copy(self.includeds)
dic[(s, i)] = True
return Info(self.string + s, dic, self.excludeds)
def exclude(self, s, i):
dic = copy.copy(self.excludeds)
dic[(s, i)] = True
return Info(self.string, self.includeds, dic)
def __repr__(self):
return '({0}, {1}, {2})'.format(self.string, self.includeds, self.excludeds)
dp: [[Info]] = [[None for _ in range(len(s) + 1)] for _ in range(len(s) + 1)]
dp[0][0] = Info('', {}, {})
for i in range(len(s)):
ab, abi = ab_dic[i]
for j in range(i + 1):
inf = dp[i][j]
if inf is None:
continue
if ab == 'a':
if inf.includeds.get(('b', abi)):
dp[i + 1][j + 1] = inf.include('a', abi)
elif inf.excludeds.get(('b', abi)):
candidate = inf.exclude('a', abi)
if dp[i + 1][j] is None:
dp[i + 1][j] = candidate
else:
if dp[i + 1][j].string <= candidate.string:
dp[i + 1][j] = candidate
else:
pass
else:
dp[i + 1][j + 1] = inf.include('a', abi)
candidate = inf.exclude('a', abi)
if dp[i + 1][j] is None:
dp[i + 1][j] = candidate
else:
if dp[i + 1][j].string <= candidate.string:
dp[i + 1][j] = candidate
else:
pass
if ab == 'b':
if inf.includeds.get(('a', abi)):
dp[i + 1][j + 1] = inf.include('b', abi)
elif inf.excludeds.get(('a', abi)):
candidate = inf.exclude('b', abi)
if dp[i + 1][j] is None:
dp[i + 1][j] = candidate
else:
if dp[i + 1][j].string < candidate.string:
dp[i + 1][j] = candidate
else:
pass
else:
dp[i + 1][j + 1] = inf.include('b', abi)
candidate = inf.exclude('b', abi)
if dp[i + 1][j] is None:
dp[i + 1][j] = candidate
else:
if dp[i + 1][j].string < candidate.string:
dp[i + 1][j] = candidate
else:
pass
print(sorted(map(lambda x: x.string, filter(lambda x: x is not None, dp[-1])))[-1])
```
No
| 38,409 | [
0.346923828125,
-0.2269287109375,
0.29638671875,
0.280029296875,
-0.2724609375,
-0.2418212890625,
-0.2763671875,
0.06427001953125,
0.25927734375,
0.8056640625,
0.52392578125,
-0.452392578125,
0.08074951171875,
-0.89990234375,
-0.669921875,
-0.1824951171875,
-0.87158203125,
-0.46777... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string S of length 2N, containing N occurrences of `a` and N occurrences of `b`.
You will choose some of the characters in S. Here, for each i = 1,2,...,N, it is not allowed to choose exactly one of the following two: the i-th occurrence of `a` and the i-th occurrence of `b`. (That is, you can only choose both or neither.) Then, you will concatenate the chosen characters (without changing the order).
Find the lexicographically largest string that can be obtained in this way.
Constraints
* 1 \leq N \leq 3000
* S is a string of length 2N containing N occurrences of `a` and N occurrences of `b`.
Input
Input is given from Standard Input in the following format:
N
S
Output
Print the lexicographically largest string that satisfies the condition.
Examples
Input
3
aababb
Output
abab
Input
3
bbabaa
Output
bbabaa
Input
6
bbbaabbabaaa
Output
bbbabaaa
Input
9
abbbaababaababbaba
Output
bbaababababa
Submitted Solution:
```
import bisect
from itertools import accumulate, combinations
def search_ab(sss, cursor):
ai = aaa.index(cursor)
tmp_cur = bbb[ai]
repeat = 1
while sss[tmp_cur] < 0:
repeat += 1
ai = bisect.bisect(aaa, tmp_cur, lo=ai)
tmp_cur = bbb[ai]
return 'ab' * repeat, tmp_cur + 1
def search_ba(sss, cursor):
first_bi = bbb.index(cursor)
last_cursor = sss.index(0, cursor)
last_bi = aaa.index(last_cursor)
tmp_buf = [''] * (last_bi - first_bi + 1) * 2
tmp_max = ''
for i in range(last_bi, first_bi - 1, -1):
tmp_buf[aaa[i] - cursor] = 'a'
tmp_buf[bbb[i] - cursor] = 'b'
tmp = ''.join(tmp_buf)
if tmp > tmp_max:
tmp_max = tmp
return tmp_max, last_cursor + 1
def integrate(parts_a, parts_b):
if not parts_b:
return ''.join(parts_a)
tmp_max = ''
for k in range(1, len(parts_b) + 1):
for cmb in combinations(parts_b, k):
tmp = ''.join(cmb)
if tmp > tmp_max:
tmp_max = tmp
return tmp_max
n = int(input())
n2 = n * 2
s = input()
sss = []
aaa, bbb = [], []
for i, c in enumerate(s):
if c == 'a':
aaa.append(i)
sss.append(-1)
else:
bbb.append(i)
sss.append(1)
sss = list(accumulate(sss))
parts_a = []
parts_b = []
cursor = 0
while cursor < n2:
c = sss[cursor]
if c < 0:
tmp, cursor = search_ab(sss, cursor)
parts_a.append(tmp)
else:
tmp, cursor = search_ba(sss, cursor)
parts_b.append(tmp)
print(integrate(parts_a, parts_b))
```
No
| 38,410 | [
0.34033203125,
-0.12200927734375,
0.2060546875,
0.44580078125,
-0.50634765625,
-0.180908203125,
-0.35546875,
0.07891845703125,
0.19580078125,
0.65087890625,
0.56591796875,
-0.432373046875,
-0.073486328125,
-0.869140625,
-0.6806640625,
-0.174560546875,
-0.802734375,
-0.587890625,
... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string S of length 2N, containing N occurrences of `a` and N occurrences of `b`.
You will choose some of the characters in S. Here, for each i = 1,2,...,N, it is not allowed to choose exactly one of the following two: the i-th occurrence of `a` and the i-th occurrence of `b`. (That is, you can only choose both or neither.) Then, you will concatenate the chosen characters (without changing the order).
Find the lexicographically largest string that can be obtained in this way.
Constraints
* 1 \leq N \leq 3000
* S is a string of length 2N containing N occurrences of `a` and N occurrences of `b`.
Input
Input is given from Standard Input in the following format:
N
S
Output
Print the lexicographically largest string that satisfies the condition.
Examples
Input
3
aababb
Output
abab
Input
3
bbabaa
Output
bbabaa
Input
6
bbbaabbabaaa
Output
bbbabaaa
Input
9
abbbaababaababbaba
Output
bbaababababa
Submitted Solution:
```
from itertools import accumulate
def search_ab(sss, cursor):
# print('a', cursor)
ai = aaa.index(cursor)
tmp_cur = bbb[ai]
max_cur = sss.index(0, cursor)
repeat = 1
while tmp_cur < max_cur:
cur = s.find('a', tmp_cur, max_cur)
if cur == -1:
break
ai = aaa.index(cur, ai)
tmp_cur = bbb[ai]
repeat += 1
return repeat, max_cur + 1
def search_ba(sss, cursor):
# print('b', cursor)
first_bi = bbb.index(cursor)
max_cursor = sss.index(0, cursor)
last_bi = aaa.index(max_cursor)
tmp_buf = [''] * (last_bi - first_bi + 1) * 2
tmp_max = ''
for i in range(last_bi, first_bi - 1, -1):
tmp_buf[aaa[i] - cursor] = 'a'
tmp_buf[bbb[i] - cursor] = 'b'
tmp = ''.join(tmp_buf)
if tmp > tmp_max:
tmp_max = tmp
return tmp_max, max_cursor + 1
def integrate(parts_b):
tmp_max = ''
for pb in reversed(parts_b):
tmp = pb + tmp_max
if tmp > tmp_max:
tmp_max = tmp
return tmp_max
n = int(input())
s = input()
n2 = n * 2
sss = []
aaa, bbb = [], []
for i, c in enumerate(s):
if c == 'a':
aaa.append(i)
sss.append(-1)
else:
bbb.append(i)
sss.append(1)
sss = list(accumulate(sss))
repeat_a = 0
parts_b = []
cursor = 0
while cursor < n2:
c = sss[cursor]
if c < 0:
repeat, cursor = search_ab(sss, cursor)
repeat_a += repeat
else:
tmp, cursor = search_ba(sss, cursor)
parts_b.append(tmp)
if parts_b:
print(integrate(parts_b))
else:
print('ab' * repeat_a)
```
No
| 38,411 | [
0.358642578125,
-0.06939697265625,
0.185791015625,
0.5009765625,
-0.45849609375,
-0.091796875,
-0.338623046875,
0.1148681640625,
0.2335205078125,
0.61181640625,
0.5556640625,
-0.429443359375,
-0.1834716796875,
-0.8505859375,
-0.609375,
-0.190185546875,
-0.75732421875,
-0.5341796875... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string S of length 2N, containing N occurrences of `a` and N occurrences of `b`.
You will choose some of the characters in S. Here, for each i = 1,2,...,N, it is not allowed to choose exactly one of the following two: the i-th occurrence of `a` and the i-th occurrence of `b`. (That is, you can only choose both or neither.) Then, you will concatenate the chosen characters (without changing the order).
Find the lexicographically largest string that can be obtained in this way.
Constraints
* 1 \leq N \leq 3000
* S is a string of length 2N containing N occurrences of `a` and N occurrences of `b`.
Input
Input is given from Standard Input in the following format:
N
S
Output
Print the lexicographically largest string that satisfies the condition.
Examples
Input
3
aababb
Output
abab
Input
3
bbabaa
Output
bbabaa
Input
6
bbbaabbabaaa
Output
bbbabaaa
Input
9
abbbaababaababbaba
Output
bbaababababa
Submitted Solution:
```
n = int(input())
s = input()
ab_index = {'a': 0, 'b': 0}
ab_dic = {}
i = 0
while i < (2 * n):
ab_dic[i] = (s[i], ab_index[s[i]])
ab_index[s[i]] += 1
i += 1
import copy
class Info:
def __init__(self, string, includeds, excludeds):
self.string = string
self.includeds = includeds
self.excludeds = excludeds
def include(self, s, i):
dic = copy.copy(self.includeds)
dic[(s, i)] = True
return Info(self.string + s, dic, self.excludeds)
def exclude(self, s, i):
dic = copy.copy(self.excludeds)
dic[(s, i)] = True
return Info(self.string, self.includeds, dic)
def __repr__(self):
return '({0}, {1}, {2})'.format(self.string, self.includeds, self.excludeds)
dp = [[None for _ in range(len(s) + 1)] for _ in range(len(s) + 1)]
dp[0][0] = Info('', {}, {})
for i in range(len(s)):
ab, abi = ab_dic[i]
for j in range(i + 1):
inf = dp[i][j]
if inf is None:
continue
if inf.string == 'aa':
dp[i][j] = None
continue
if ab == 'a':
if inf.includeds.get(('b', abi)):
dp[i + 1][j + 1] = inf.include('a', abi)
elif inf.excludeds.get(('b', abi)):
candidate = inf.exclude('a', abi)
if dp[i + 1][j] is None:
dp[i + 1][j] = candidate
else:
if dp[i + 1][j].string <= candidate.string:
dp[i + 1][j] = candidate
else:
pass
else:
dp[i + 1][j + 1] = inf.include('a', abi)
candidate = inf.exclude('a', abi)
if dp[i + 1][j] is None:
dp[i + 1][j] = candidate
else:
if dp[i + 1][j].string <= candidate.string:
dp[i + 1][j] = candidate
else:
pass
if ab == 'b':
if inf.includeds.get(('a', abi)):
dp[i + 1][j + 1] = inf.include('b', abi)
elif inf.excludeds.get(('a', abi)):
candidate = inf.exclude('b', abi)
if dp[i + 1][j] is None:
dp[i + 1][j] = candidate
else:
if dp[i + 1][j].string < candidate.string:
dp[i + 1][j] = candidate
else:
pass
else:
dp[i + 1][j + 1] = inf.include('b', abi)
candidate = inf.exclude('b', abi)
if dp[i + 1][j] is None:
dp[i + 1][j] = candidate
else:
if dp[i + 1][j].string < candidate.string:
dp[i + 1][j] = candidate
else:
pass
print(sorted(map(lambda x: x.string, filter(lambda x: x is not None, dp[-1])))[-1])
```
No
| 38,412 | [
0.346923828125,
-0.2269287109375,
0.29638671875,
0.280029296875,
-0.2724609375,
-0.2418212890625,
-0.2763671875,
0.06427001953125,
0.25927734375,
0.8056640625,
0.52392578125,
-0.452392578125,
0.08074951171875,
-0.89990234375,
-0.669921875,
-0.1824951171875,
-0.87158203125,
-0.46777... | 0 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of n lowercase Latin letters.
You have to remove at most one (i.e. zero or one) character of this string in such a way that the string you obtain will be lexicographically smallest among all strings that can be obtained using this operation.
String s = s_1 s_2 ... s_n is lexicographically smaller than string t = t_1 t_2 ... t_m if n < m and s_1 = t_1, s_2 = t_2, ..., s_n = t_n or there exists a number p such that p ≤ min(n, m) and s_1 = t_1, s_2 = t_2, ..., s_{p-1} = t_{p-1} and s_p < t_p.
For example, "aaa" is smaller than "aaaa", "abb" is smaller than "abc", "pqr" is smaller than "z".
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of s.
The second line of the input contains exactly n lowercase Latin letters — the string s.
Output
Print one string — the smallest possible lexicographically string that can be obtained by removing at most one character from the string s.
Examples
Input
3
aaa
Output
aa
Input
5
abcda
Output
abca
Note
In the first example you can remove any character of s to obtain the string "aa".
In the second example "abca" < "abcd" < "abcda" < "abda" < "acda" < "bcda".
Tags: greedy, strings
Correct Solution:
```
n=int(input())
s=input()
f=1
for i in range(n-1):
if ord(s[i])>ord(s[i+1]) and f:
f=0
# print("DF",s[i])
pass
else:
print(s[i],end="")
if f==0:
print(s[i+1])
```
| 38,650 | [
0.064697265625,
-0.004138946533203125,
0.327880859375,
0.2352294921875,
-0.619140625,
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0.1353759765625,
0.15185546875,
0.08221435546875,
0.54541015625,
0.97021484375,
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-0.54345703125,
-0.86572265625,
-0.84423828125,
0.10333251953125,
-0.403564453125,
-... | 0 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of n lowercase Latin letters.
You have to remove at most one (i.e. zero or one) character of this string in such a way that the string you obtain will be lexicographically smallest among all strings that can be obtained using this operation.
String s = s_1 s_2 ... s_n is lexicographically smaller than string t = t_1 t_2 ... t_m if n < m and s_1 = t_1, s_2 = t_2, ..., s_n = t_n or there exists a number p such that p ≤ min(n, m) and s_1 = t_1, s_2 = t_2, ..., s_{p-1} = t_{p-1} and s_p < t_p.
For example, "aaa" is smaller than "aaaa", "abb" is smaller than "abc", "pqr" is smaller than "z".
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of s.
The second line of the input contains exactly n lowercase Latin letters — the string s.
Output
Print one string — the smallest possible lexicographically string that can be obtained by removing at most one character from the string s.
Examples
Input
3
aaa
Output
aa
Input
5
abcda
Output
abca
Note
In the first example you can remove any character of s to obtain the string "aa".
In the second example "abca" < "abcd" < "abcda" < "abda" < "acda" < "bcda".
Tags: greedy, strings
Correct Solution:
```
def ii():
return int(input())
def mi():
return map(int, input().split())
def li():
return list(mi())
n = ii()
s = input().strip()
i = 0
while i < n - 1:
if ord(s[i]) > ord(s[i + 1]):
break
i += 1
ans = s[:i] + s[i + 1:]
print(ans)
```
| 38,651 | [
0.102294921875,
0.0218353271484375,
0.344482421875,
0.1898193359375,
-0.62841796875,
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0.0767822265625,
0.10723876953125,
0.112548828125,
0.5869140625,
0.939453125,
-0.330322265625,
-0.56640625,
-0.8251953125,
-0.79345703125,
0.0823974609375,
-0.37451171875,
-0.5878906... | 0 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of n lowercase Latin letters.
You have to remove at most one (i.e. zero or one) character of this string in such a way that the string you obtain will be lexicographically smallest among all strings that can be obtained using this operation.
String s = s_1 s_2 ... s_n is lexicographically smaller than string t = t_1 t_2 ... t_m if n < m and s_1 = t_1, s_2 = t_2, ..., s_n = t_n or there exists a number p such that p ≤ min(n, m) and s_1 = t_1, s_2 = t_2, ..., s_{p-1} = t_{p-1} and s_p < t_p.
For example, "aaa" is smaller than "aaaa", "abb" is smaller than "abc", "pqr" is smaller than "z".
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of s.
The second line of the input contains exactly n lowercase Latin letters — the string s.
Output
Print one string — the smallest possible lexicographically string that can be obtained by removing at most one character from the string s.
Examples
Input
3
aaa
Output
aa
Input
5
abcda
Output
abca
Note
In the first example you can remove any character of s to obtain the string "aa".
In the second example "abca" < "abcd" < "abcda" < "abda" < "acda" < "bcda".
Tags: greedy, strings
Correct Solution:
```
n = int(input())
s = input()
for i in range(1, n):
if s[i] < s[i - 1]:
print(s[:i - 1] + s[i:])
break
elif i == n - 1:
print(s[:-1])
```
| 38,652 | [
0.13330078125,
0.03717041015625,
0.34375,
0.205322265625,
-0.5888671875,
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0.045501708984375,
0.08306884765625,
0.053955078125,
0.56201171875,
0.9560546875,
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-0.86767578125,
-0.830078125,
0.029449462890625,
-0.374755859375,
-0.59912109375,
... | 0 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of n lowercase Latin letters.
You have to remove at most one (i.e. zero or one) character of this string in such a way that the string you obtain will be lexicographically smallest among all strings that can be obtained using this operation.
String s = s_1 s_2 ... s_n is lexicographically smaller than string t = t_1 t_2 ... t_m if n < m and s_1 = t_1, s_2 = t_2, ..., s_n = t_n or there exists a number p such that p ≤ min(n, m) and s_1 = t_1, s_2 = t_2, ..., s_{p-1} = t_{p-1} and s_p < t_p.
For example, "aaa" is smaller than "aaaa", "abb" is smaller than "abc", "pqr" is smaller than "z".
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of s.
The second line of the input contains exactly n lowercase Latin letters — the string s.
Output
Print one string — the smallest possible lexicographically string that can be obtained by removing at most one character from the string s.
Examples
Input
3
aaa
Output
aa
Input
5
abcda
Output
abca
Note
In the first example you can remove any character of s to obtain the string "aa".
In the second example "abca" < "abcd" < "abcda" < "abda" < "acda" < "bcda".
Tags: greedy, strings
Correct Solution:
```
n = int(input())
text = input()
remove_idx = -1
for i in range(len(text) - 1):
if ord(text[i]) - ord(text[i + 1]) > 0:
remove_idx = i
break
if remove_idx >= 0:
print(text[:remove_idx] + text[remove_idx + 1:])
else:
print(text[:-1])
```
| 38,653 | [
-0.0108642578125,
-0.0662841796875,
0.390380859375,
0.34375,
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0.07720947265625,
0.1424560546875,
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0.8447265625,
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-0.8369140625,
-0.82568359375,
0.109619140625,
-0.42041015625,
-0.65087890625,
-0.23... | 0 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of n lowercase Latin letters.
You have to remove at most one (i.e. zero or one) character of this string in such a way that the string you obtain will be lexicographically smallest among all strings that can be obtained using this operation.
String s = s_1 s_2 ... s_n is lexicographically smaller than string t = t_1 t_2 ... t_m if n < m and s_1 = t_1, s_2 = t_2, ..., s_n = t_n or there exists a number p such that p ≤ min(n, m) and s_1 = t_1, s_2 = t_2, ..., s_{p-1} = t_{p-1} and s_p < t_p.
For example, "aaa" is smaller than "aaaa", "abb" is smaller than "abc", "pqr" is smaller than "z".
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of s.
The second line of the input contains exactly n lowercase Latin letters — the string s.
Output
Print one string — the smallest possible lexicographically string that can be obtained by removing at most one character from the string s.
Examples
Input
3
aaa
Output
aa
Input
5
abcda
Output
abca
Note
In the first example you can remove any character of s to obtain the string "aa".
In the second example "abca" < "abcd" < "abcda" < "abda" < "acda" < "bcda".
Tags: greedy, strings
Correct Solution:
```
n = int(input())
#a = [int(i) for i in input().split()]
s = input()
max = 0
num = 0
for i in range(n-1):
if s[i] > s[i+1]:
print(s[:i]+s[i+1:])
exit()
print(s[:-1])
```
| 38,654 | [
0.10919189453125,
0.0732421875,
0.351318359375,
0.2139892578125,
-0.60205078125,
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0.06634521484375,
0.11968994140625,
0.060394287109375,
0.5517578125,
0.93701171875,
-0.294921875,
-0.5673828125,
-0.88232421875,
-0.8173828125,
0.0298004150390625,
-0.363525390625,
-0.... | 0 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of n lowercase Latin letters.
You have to remove at most one (i.e. zero or one) character of this string in such a way that the string you obtain will be lexicographically smallest among all strings that can be obtained using this operation.
String s = s_1 s_2 ... s_n is lexicographically smaller than string t = t_1 t_2 ... t_m if n < m and s_1 = t_1, s_2 = t_2, ..., s_n = t_n or there exists a number p such that p ≤ min(n, m) and s_1 = t_1, s_2 = t_2, ..., s_{p-1} = t_{p-1} and s_p < t_p.
For example, "aaa" is smaller than "aaaa", "abb" is smaller than "abc", "pqr" is smaller than "z".
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of s.
The second line of the input contains exactly n lowercase Latin letters — the string s.
Output
Print one string — the smallest possible lexicographically string that can be obtained by removing at most one character from the string s.
Examples
Input
3
aaa
Output
aa
Input
5
abcda
Output
abca
Note
In the first example you can remove any character of s to obtain the string "aa".
In the second example "abca" < "abcd" < "abcda" < "abda" < "acda" < "bcda".
Tags: greedy, strings
Correct Solution:
```
n=int(input())
s=input()
index=-1
for i in range(1,n):
if s[i]<s[i-1]:
index=i-1
break
if index==-1:
print(s[:-1])
else:
print(s[:index]+s[index+1:])
```
| 38,655 | [
0.0640869140625,
0.07025146484375,
0.347412109375,
0.201416015625,
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0.0557861328125,
0.109619140625,
0.06488037109375,
0.59228515625,
0.96728515625,
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-0.5380859375,
-0.90673828125,
-0.82958984375,
0.06695556640625,
-0.38671875,
-0.611... | 0 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of n lowercase Latin letters.
You have to remove at most one (i.e. zero or one) character of this string in such a way that the string you obtain will be lexicographically smallest among all strings that can be obtained using this operation.
String s = s_1 s_2 ... s_n is lexicographically smaller than string t = t_1 t_2 ... t_m if n < m and s_1 = t_1, s_2 = t_2, ..., s_n = t_n or there exists a number p such that p ≤ min(n, m) and s_1 = t_1, s_2 = t_2, ..., s_{p-1} = t_{p-1} and s_p < t_p.
For example, "aaa" is smaller than "aaaa", "abb" is smaller than "abc", "pqr" is smaller than "z".
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of s.
The second line of the input contains exactly n lowercase Latin letters — the string s.
Output
Print one string — the smallest possible lexicographically string that can be obtained by removing at most one character from the string s.
Examples
Input
3
aaa
Output
aa
Input
5
abcda
Output
abca
Note
In the first example you can remove any character of s to obtain the string "aa".
In the second example "abca" < "abcd" < "abcda" < "abda" < "acda" < "bcda".
Tags: greedy, strings
Correct Solution:
```
import sys
# import heapq, collections
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**15
mod = 10**9+7
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def solve(n, s):
_set = set(list(s))
ret = s
for ch in _set:
i = s.index(ch)
ret = min(ret, s[:i] + s[i+1:])
return ret
def main():
n = I()
s = S()
print(solve(n, s))
main()
```
| 38,656 | [
0.101318359375,
0.0119781494140625,
0.415771484375,
0.29296875,
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0.0408935546875,
0.01580810546875,
0.0556640625,
0.611328125,
0.90185546875,
-0.36083984375,
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-0.88037109375,
-0.82080078125,
0.07421875,
-0.384521484375,
-0.64990234375,
-0.3381... | 0 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of n lowercase Latin letters.
You have to remove at most one (i.e. zero or one) character of this string in such a way that the string you obtain will be lexicographically smallest among all strings that can be obtained using this operation.
String s = s_1 s_2 ... s_n is lexicographically smaller than string t = t_1 t_2 ... t_m if n < m and s_1 = t_1, s_2 = t_2, ..., s_n = t_n or there exists a number p such that p ≤ min(n, m) and s_1 = t_1, s_2 = t_2, ..., s_{p-1} = t_{p-1} and s_p < t_p.
For example, "aaa" is smaller than "aaaa", "abb" is smaller than "abc", "pqr" is smaller than "z".
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of s.
The second line of the input contains exactly n lowercase Latin letters — the string s.
Output
Print one string — the smallest possible lexicographically string that can be obtained by removing at most one character from the string s.
Examples
Input
3
aaa
Output
aa
Input
5
abcda
Output
abca
Note
In the first example you can remove any character of s to obtain the string "aa".
In the second example "abca" < "abcd" < "abcda" < "abda" < "acda" < "bcda".
Tags: greedy, strings
Correct Solution:
```
n = int(input())
s = input()
k = n-1
for i in range(1,n):
if s[i]<s[i-1]:
k = i-1
break
for i in range(n):
if i==k:
continue
print(s[i],end="")
```
| 38,657 | [
0.11322021484375,
0.040557861328125,
0.37060546875,
0.2244873046875,
-0.58984375,
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0.04327392578125,
0.099365234375,
0.03973388671875,
0.55419921875,
0.96142578125,
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-0.5517578125,
-0.85888671875,
-0.82470703125,
0.061431884765625,
-0.393798828125,
-0.6362... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of n lowercase Latin letters.
You have to remove at most one (i.e. zero or one) character of this string in such a way that the string you obtain will be lexicographically smallest among all strings that can be obtained using this operation.
String s = s_1 s_2 ... s_n is lexicographically smaller than string t = t_1 t_2 ... t_m if n < m and s_1 = t_1, s_2 = t_2, ..., s_n = t_n or there exists a number p such that p ≤ min(n, m) and s_1 = t_1, s_2 = t_2, ..., s_{p-1} = t_{p-1} and s_p < t_p.
For example, "aaa" is smaller than "aaaa", "abb" is smaller than "abc", "pqr" is smaller than "z".
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of s.
The second line of the input contains exactly n lowercase Latin letters — the string s.
Output
Print one string — the smallest possible lexicographically string that can be obtained by removing at most one character from the string s.
Examples
Input
3
aaa
Output
aa
Input
5
abcda
Output
abca
Note
In the first example you can remove any character of s to obtain the string "aa".
In the second example "abca" < "abcd" < "abcda" < "abda" < "acda" < "bcda".
Submitted Solution:
```
def get_smallest(string):
for idx in range(len(string) - 1):
if string[idx] > string [idx+1]:
return string[:idx] + string[idx+1:]
return string[:-1]
n = int(input())
string = input()
result = get_smallest(string)
print(result)
```
Yes
| 38,658 | [
0.12060546875,
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0.28173828125,
0.1435546875,
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0.142578125,
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-0.85888671875,
-0.83056640625,
0.05340576171875,
-0.390869140625,
-0.53271484375,
... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of n lowercase Latin letters.
You have to remove at most one (i.e. zero or one) character of this string in such a way that the string you obtain will be lexicographically smallest among all strings that can be obtained using this operation.
String s = s_1 s_2 ... s_n is lexicographically smaller than string t = t_1 t_2 ... t_m if n < m and s_1 = t_1, s_2 = t_2, ..., s_n = t_n or there exists a number p such that p ≤ min(n, m) and s_1 = t_1, s_2 = t_2, ..., s_{p-1} = t_{p-1} and s_p < t_p.
For example, "aaa" is smaller than "aaaa", "abb" is smaller than "abc", "pqr" is smaller than "z".
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of s.
The second line of the input contains exactly n lowercase Latin letters — the string s.
Output
Print one string — the smallest possible lexicographically string that can be obtained by removing at most one character from the string s.
Examples
Input
3
aaa
Output
aa
Input
5
abcda
Output
abca
Note
In the first example you can remove any character of s to obtain the string "aa".
In the second example "abca" < "abcd" < "abcda" < "abda" < "acda" < "bcda".
Submitted Solution:
```
n = int(input())
string = input()
done = 0
for ind in range (len(string)-1):
if string[ind]>string[ind+1]:
string = string[:ind] + string[ind+1:]
done = 1
break
if done==0:
string = string[:-1]
print(string)
```
Yes
| 38,659 | [
0.06884765625,
-0.0190582275390625,
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0.133056640625,
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0.1473388671875,
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-0.55224609375,
-0.91162109375,
-0.79052734375,
0.07208251953125,
-0.432861328125,
... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of n lowercase Latin letters.
You have to remove at most one (i.e. zero or one) character of this string in such a way that the string you obtain will be lexicographically smallest among all strings that can be obtained using this operation.
String s = s_1 s_2 ... s_n is lexicographically smaller than string t = t_1 t_2 ... t_m if n < m and s_1 = t_1, s_2 = t_2, ..., s_n = t_n or there exists a number p such that p ≤ min(n, m) and s_1 = t_1, s_2 = t_2, ..., s_{p-1} = t_{p-1} and s_p < t_p.
For example, "aaa" is smaller than "aaaa", "abb" is smaller than "abc", "pqr" is smaller than "z".
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of s.
The second line of the input contains exactly n lowercase Latin letters — the string s.
Output
Print one string — the smallest possible lexicographically string that can be obtained by removing at most one character from the string s.
Examples
Input
3
aaa
Output
aa
Input
5
abcda
Output
abca
Note
In the first example you can remove any character of s to obtain the string "aa".
In the second example "abca" < "abcd" < "abcda" < "abda" < "acda" < "bcda".
Submitted Solution:
```
def a():
n = int(input())
s = input()
for i in range(n-1):
if s[i] > s[i+1]:
return s[:i]+s[i+1:]
return s[:-1]
print(a())
```
Yes
| 38,660 | [
0.12469482421875,
-0.02862548828125,
0.307861328125,
0.153564453125,
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0.1968994140625,
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0.8876953125,
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-0.59375,
-0.9013671875,
-0.82958984375,
0.049774169921875,
-0.42138671875,
-0.... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of n lowercase Latin letters.
You have to remove at most one (i.e. zero or one) character of this string in such a way that the string you obtain will be lexicographically smallest among all strings that can be obtained using this operation.
String s = s_1 s_2 ... s_n is lexicographically smaller than string t = t_1 t_2 ... t_m if n < m and s_1 = t_1, s_2 = t_2, ..., s_n = t_n or there exists a number p such that p ≤ min(n, m) and s_1 = t_1, s_2 = t_2, ..., s_{p-1} = t_{p-1} and s_p < t_p.
For example, "aaa" is smaller than "aaaa", "abb" is smaller than "abc", "pqr" is smaller than "z".
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of s.
The second line of the input contains exactly n lowercase Latin letters — the string s.
Output
Print one string — the smallest possible lexicographically string that can be obtained by removing at most one character from the string s.
Examples
Input
3
aaa
Output
aa
Input
5
abcda
Output
abca
Note
In the first example you can remove any character of s to obtain the string "aa".
In the second example "abca" < "abcd" < "abcda" < "abda" < "acda" < "bcda".
Submitted Solution:
```
n = int(input())
s = str(input())
for i in range(n-1):
if s[i] > s[i+1]:
print(s[0:i] + s[i+1:])
raise SystemExit
print(s[0:n-1])
```
Yes
| 38,661 | [
0.0950927734375,
-0.03912353515625,
0.2978515625,
0.158203125,
-0.6162109375,
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0.18408203125,
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0.599609375,
0.876953125,
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-0.58349609375,
-0.90966796875,
-0.80810546875,
0.06988525390625,
-0.37548828125,
-0.5507... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of n lowercase Latin letters.
You have to remove at most one (i.e. zero or one) character of this string in such a way that the string you obtain will be lexicographically smallest among all strings that can be obtained using this operation.
String s = s_1 s_2 ... s_n is lexicographically smaller than string t = t_1 t_2 ... t_m if n < m and s_1 = t_1, s_2 = t_2, ..., s_n = t_n or there exists a number p such that p ≤ min(n, m) and s_1 = t_1, s_2 = t_2, ..., s_{p-1} = t_{p-1} and s_p < t_p.
For example, "aaa" is smaller than "aaaa", "abb" is smaller than "abc", "pqr" is smaller than "z".
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of s.
The second line of the input contains exactly n lowercase Latin letters — the string s.
Output
Print one string — the smallest possible lexicographically string that can be obtained by removing at most one character from the string s.
Examples
Input
3
aaa
Output
aa
Input
5
abcda
Output
abca
Note
In the first example you can remove any character of s to obtain the string "aa".
In the second example "abca" < "abcd" < "abcda" < "abda" < "acda" < "bcda".
Submitted Solution:
```
n,s=int(input()),input()
for i in range(1,n):
if s[i]<s[i-1]:break
print(s[:i-1]+s[i:])
```
No
| 38,662 | [
0.11041259765625,
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0.3076171875,
0.1397705078125,
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0.2032470703125,
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0.57470703125,
0.89892578125,
-0.21533203125,
-0.56298828125,
-0.9091796875,
-0.8193359375,
0.06329345703125,
-0.415771484375,
... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of n lowercase Latin letters.
You have to remove at most one (i.e. zero or one) character of this string in such a way that the string you obtain will be lexicographically smallest among all strings that can be obtained using this operation.
String s = s_1 s_2 ... s_n is lexicographically smaller than string t = t_1 t_2 ... t_m if n < m and s_1 = t_1, s_2 = t_2, ..., s_n = t_n or there exists a number p such that p ≤ min(n, m) and s_1 = t_1, s_2 = t_2, ..., s_{p-1} = t_{p-1} and s_p < t_p.
For example, "aaa" is smaller than "aaaa", "abb" is smaller than "abc", "pqr" is smaller than "z".
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of s.
The second line of the input contains exactly n lowercase Latin letters — the string s.
Output
Print one string — the smallest possible lexicographically string that can be obtained by removing at most one character from the string s.
Examples
Input
3
aaa
Output
aa
Input
5
abcda
Output
abca
Note
In the first example you can remove any character of s to obtain the string "aa".
In the second example "abca" < "abcd" < "abcda" < "abda" < "acda" < "bcda".
Submitted Solution:
```
n=int(input())
s=input()
t=-1
for i in range(n-1):
if s[i]>s[i+1]:
t=i
break
if t==-1:t=n-1
print(s[:i]+s[i+1:])
```
No
| 38,663 | [
0.11541748046875,
-0.0184173583984375,
0.302001953125,
0.1546630859375,
-0.6103515625,
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0.191650390625,
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0.61376953125,
0.87744140625,
-0.2020263671875,
-0.5654296875,
-0.92529296875,
-0.82373046875,
0.053985595703125,
-0.395751953125... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of n lowercase Latin letters.
You have to remove at most one (i.e. zero or one) character of this string in such a way that the string you obtain will be lexicographically smallest among all strings that can be obtained using this operation.
String s = s_1 s_2 ... s_n is lexicographically smaller than string t = t_1 t_2 ... t_m if n < m and s_1 = t_1, s_2 = t_2, ..., s_n = t_n or there exists a number p such that p ≤ min(n, m) and s_1 = t_1, s_2 = t_2, ..., s_{p-1} = t_{p-1} and s_p < t_p.
For example, "aaa" is smaller than "aaaa", "abb" is smaller than "abc", "pqr" is smaller than "z".
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of s.
The second line of the input contains exactly n lowercase Latin letters — the string s.
Output
Print one string — the smallest possible lexicographically string that can be obtained by removing at most one character from the string s.
Examples
Input
3
aaa
Output
aa
Input
5
abcda
Output
abca
Note
In the first example you can remove any character of s to obtain the string "aa".
In the second example "abca" < "abcd" < "abcda" < "abda" < "acda" < "bcda".
Submitted Solution:
```
n=int(input())
s=input().strip()
x=n-1
while x>0:
if s[x]<s[x-1]:
s=s[:x-1]+s[x:]
break
x=x-1
if x==0:
s=s[:-1]
print(s)
```
No
| 38,664 | [
0.14111328125,
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0.335205078125,
0.1495361328125,
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-0.203125,
0.189453125,
-0.037811279296875,
0.61376953125,
0.91455078125,
-0.1728515625,
-0.56494140625,
-0.92236328125,
-0.794921875,
0.08941650390625,
-0.423095703125,
-0.611816... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of n lowercase Latin letters.
You have to remove at most one (i.e. zero or one) character of this string in such a way that the string you obtain will be lexicographically smallest among all strings that can be obtained using this operation.
String s = s_1 s_2 ... s_n is lexicographically smaller than string t = t_1 t_2 ... t_m if n < m and s_1 = t_1, s_2 = t_2, ..., s_n = t_n or there exists a number p such that p ≤ min(n, m) and s_1 = t_1, s_2 = t_2, ..., s_{p-1} = t_{p-1} and s_p < t_p.
For example, "aaa" is smaller than "aaaa", "abb" is smaller than "abc", "pqr" is smaller than "z".
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of s.
The second line of the input contains exactly n lowercase Latin letters — the string s.
Output
Print one string — the smallest possible lexicographically string that can be obtained by removing at most one character from the string s.
Examples
Input
3
aaa
Output
aa
Input
5
abcda
Output
abca
Note
In the first example you can remove any character of s to obtain the string "aa".
In the second example "abca" < "abcd" < "abcda" < "abda" < "acda" < "bcda".
Submitted Solution:
```
n=int(input())
a=input()
x=0
for i in range(n-1):
if ord(a[i])<=ord(a[i+1]):
pass
else:
x=1
break
if x==1:
print(a.replace(a[i],''))
else:
print(a[:-1])
```
No
| 38,665 | [
0.1160888671875,
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0.29248046875,
0.15625,
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0.1962890625,
0.0081634521484375,
0.56396484375,
0.96923828125,
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-0.57275390625,
-0.97998046875,
-0.80908203125,
0.11090087890625,
-0.40625,
-0.6044921875,... | 0 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings s and t, both consisting of exactly k lowercase Latin letters, s is lexicographically less than t.
Let's consider list of all strings consisting of exactly k lowercase Latin letters, lexicographically not less than s and not greater than t (including s and t) in lexicographical order. For example, for k=2, s="az" and t="bf" the list will be ["az", "ba", "bb", "bc", "bd", "be", "bf"].
Your task is to print the median (the middle element) of this list. For the example above this will be "bc".
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Input
The first line of the input contains one integer k (1 ≤ k ≤ 2 ⋅ 10^5) — the length of strings.
The second line of the input contains one string s consisting of exactly k lowercase Latin letters.
The third line of the input contains one string t consisting of exactly k lowercase Latin letters.
It is guaranteed that s is lexicographically less than t.
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Output
Print one string consisting exactly of k lowercase Latin letters — the median (the middle element) of list of strings of length k lexicographically not less than s and not greater than t.
Examples
Input
2
az
bf
Output
bc
Input
5
afogk
asdji
Output
alvuw
Input
6
nijfvj
tvqhwp
Output
qoztvz
Tags: bitmasks, math, number theory, strings
Correct Solution:
```
def plus(a, b):
c = ""
i = len(a)-1
arg = 0
while i >= 0:
k = (arg + ord(a[i]) + ord(b[i]) - 194)
arg = 1
if k < 26:
arg = 0
i -= 1
c = chr(k % 26) + c
c = chr(1) * arg + c
return c
def minus(a):
c = ""
i = 0
arg = 0
while i < len(a):
c += chr((arg * 26 + ord(a[i]))//2)
arg = ord(a[i]) % 2
i += 1
if arg:
c = c[:len(c)-1] + chr(ord(c[len(c)-1])+1)
return c
n = int(input())
a1 = ""
i = 0
while i < n:
a1 += "s"
i += 1
c = plus(input() ,input())
c = minus(c)
c1 = ""
for q in c:
c1 += chr(ord(q) + 97)
print(c1[len(c1)-n:])
```
| 38,698 | [
0.20166015625,
-0.0200042724609375,
0.376953125,
0.402587890625,
-0.478515625,
-0.414794921875,
-0.037841796875,
-0.03167724609375,
-0.1768798828125,
1.1171875,
0.564453125,
-0.1907958984375,
-0.1378173828125,
-0.9423828125,
-0.74853515625,
0.19384765625,
-0.3828125,
-0.6865234375,... | 0 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings s and t, both consisting of exactly k lowercase Latin letters, s is lexicographically less than t.
Let's consider list of all strings consisting of exactly k lowercase Latin letters, lexicographically not less than s and not greater than t (including s and t) in lexicographical order. For example, for k=2, s="az" and t="bf" the list will be ["az", "ba", "bb", "bc", "bd", "be", "bf"].
Your task is to print the median (the middle element) of this list. For the example above this will be "bc".
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Input
The first line of the input contains one integer k (1 ≤ k ≤ 2 ⋅ 10^5) — the length of strings.
The second line of the input contains one string s consisting of exactly k lowercase Latin letters.
The third line of the input contains one string t consisting of exactly k lowercase Latin letters.
It is guaranteed that s is lexicographically less than t.
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Output
Print one string consisting exactly of k lowercase Latin letters — the median (the middle element) of list of strings of length k lexicographically not less than s and not greater than t.
Examples
Input
2
az
bf
Output
bc
Input
5
afogk
asdji
Output
alvuw
Input
6
nijfvj
tvqhwp
Output
qoztvz
Tags: bitmasks, math, number theory, strings
Correct Solution:
```
def plus(a, b):
c = ""
i = len(a)-1
arg = 0
while i >= 0:
k = (arg + ord(a[i]) + ord(b[i]) - 194)
arg = 1
if k < 26:
arg = 0
i -= 1
c = chr(k % 26) + c
c = chr(1) * arg + c
return c
def minus(a):
c = ""
arg = 0
for q in a:
c += chr((arg * 26 + ord(q))//2)
arg = ord(q) % 2
if arg:
c = c[:len(c)-1] + chr(ord(c[len(c)-1])+1)
return c
n = int(input())
a1 = ""
i = 0
while i < n:
a1 += "s"
i += 1
c = minus(plus(input() ,input()))
c1 = ""
for q in c:
c1 += chr(ord(q) + 97)
print(c1[len(c1)-n:])
```
| 38,699 | [
0.2061767578125,
-0.02191162109375,
0.375732421875,
0.403564453125,
-0.485107421875,
-0.416748046875,
-0.037567138671875,
-0.02728271484375,
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1.111328125,
0.568359375,
-0.1959228515625,
-0.1392822265625,
-0.95166015625,
-0.7509765625,
0.198486328125,
-0.385498046875,... | 0 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings s and t, both consisting of exactly k lowercase Latin letters, s is lexicographically less than t.
Let's consider list of all strings consisting of exactly k lowercase Latin letters, lexicographically not less than s and not greater than t (including s and t) in lexicographical order. For example, for k=2, s="az" and t="bf" the list will be ["az", "ba", "bb", "bc", "bd", "be", "bf"].
Your task is to print the median (the middle element) of this list. For the example above this will be "bc".
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Input
The first line of the input contains one integer k (1 ≤ k ≤ 2 ⋅ 10^5) — the length of strings.
The second line of the input contains one string s consisting of exactly k lowercase Latin letters.
The third line of the input contains one string t consisting of exactly k lowercase Latin letters.
It is guaranteed that s is lexicographically less than t.
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Output
Print one string consisting exactly of k lowercase Latin letters — the median (the middle element) of list of strings of length k lexicographically not less than s and not greater than t.
Examples
Input
2
az
bf
Output
bc
Input
5
afogk
asdji
Output
alvuw
Input
6
nijfvj
tvqhwp
Output
qoztvz
Tags: bitmasks, math, number theory, strings
Correct Solution:
```
def arr(ls):
l = [0]*len(ls)
for i,v in enumerate(ls):
l[i] = ord(v)-97
return l
def div2(ls, n):
carry = 0
for i in range(n):
rem = ls[i]%2 == 1
ls[i] = ls[i]//2 + carry
if rem: carry = 13
else: carry = 0
return ls
def subtr(ls1, ls2,n):
carry = 0
for i in range(n-1, -1, -1):
ls1[i]-=carry
carry = int(ls1[i] < ls2[i])
ls1[i] = (ls1[i]-ls2[i])%26
return ls1
def add(ls1, ls2, n):
for i in range(n-1, -1, -1):
ls1[i] += ls2[i]
if ls1[i] > 25:
ls1[i-1]+=1
ls1[i]-=26
return ls1
import io, os
# input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
# input = io.StringIO(os.read(0, os.fstat(0).st_size).decode()).readline
kk=lambda:map(int,input().split())
k2=lambda:map(lambda x:int(x)-1, input().split())
ll=lambda:list(kk())
n,w1, w2 =int(input()), list(input().strip()), list(input().strip())
ls1, ls2 = arr(w1), arr(w2)
subtr(ls2, ls1, n)
div2(ls2, n)
add(ls1, ls2, n)
print("".join(map(lambda x: chr(97+x), [x for x in ls1])))
```
| 38,700 | [
0.208984375,
-0.01473236083984375,
0.40771484375,
0.39013671875,
-0.3984375,
-0.383056640625,
-0.146240234375,
-0.03253173828125,
-0.112060546875,
1.1337890625,
0.5400390625,
-0.1734619140625,
-0.09637451171875,
-0.9560546875,
-0.79296875,
0.18701171875,
-0.432373046875,
-0.6528320... | 0 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings s and t, both consisting of exactly k lowercase Latin letters, s is lexicographically less than t.
Let's consider list of all strings consisting of exactly k lowercase Latin letters, lexicographically not less than s and not greater than t (including s and t) in lexicographical order. For example, for k=2, s="az" and t="bf" the list will be ["az", "ba", "bb", "bc", "bd", "be", "bf"].
Your task is to print the median (the middle element) of this list. For the example above this will be "bc".
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Input
The first line of the input contains one integer k (1 ≤ k ≤ 2 ⋅ 10^5) — the length of strings.
The second line of the input contains one string s consisting of exactly k lowercase Latin letters.
The third line of the input contains one string t consisting of exactly k lowercase Latin letters.
It is guaranteed that s is lexicographically less than t.
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Output
Print one string consisting exactly of k lowercase Latin letters — the median (the middle element) of list of strings of length k lexicographically not less than s and not greater than t.
Examples
Input
2
az
bf
Output
bc
Input
5
afogk
asdji
Output
alvuw
Input
6
nijfvj
tvqhwp
Output
qoztvz
Tags: bitmasks, math, number theory, strings
Correct Solution:
```
def divide_2(a, m):
r = 0
q = []
for x in a:
cur = r * m + x
q.append(cur // 2)
r = cur % 2
return q
def add(s, t, m):
r = 0
a = []
for x, y in zip(s[::-1], t[::-1]):
cur = r+x+y
a.append(cur % m )
r = cur // m
if r != 0:
a.append(r)
return a[::-1]
def to_num(s):
a = []
for x in s:
a.append(ord(x)-ord('a'))
return a
def to_char(s, k):
a = []
for x in s[-k:]:
a.append(chr(x+ord('a')))
return ''.join(a)
k = int(input())
x = to_num(input())
y = to_num(input())
print(to_char(divide_2(add(x, y, 26), 26), k))
```
| 38,701 | [
0.1434326171875,
-0.0261993408203125,
0.375,
0.4287109375,
-0.46484375,
-0.393310546875,
-0.041168212890625,
-0.03753662109375,
-0.1605224609375,
1.1171875,
0.5556640625,
-0.234375,
-0.130615234375,
-0.95361328125,
-0.71044921875,
0.171142578125,
-0.419921875,
-0.7587890625,
-0.2... | 0 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings s and t, both consisting of exactly k lowercase Latin letters, s is lexicographically less than t.
Let's consider list of all strings consisting of exactly k lowercase Latin letters, lexicographically not less than s and not greater than t (including s and t) in lexicographical order. For example, for k=2, s="az" and t="bf" the list will be ["az", "ba", "bb", "bc", "bd", "be", "bf"].
Your task is to print the median (the middle element) of this list. For the example above this will be "bc".
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Input
The first line of the input contains one integer k (1 ≤ k ≤ 2 ⋅ 10^5) — the length of strings.
The second line of the input contains one string s consisting of exactly k lowercase Latin letters.
The third line of the input contains one string t consisting of exactly k lowercase Latin letters.
It is guaranteed that s is lexicographically less than t.
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Output
Print one string consisting exactly of k lowercase Latin letters — the median (the middle element) of list of strings of length k lexicographically not less than s and not greater than t.
Examples
Input
2
az
bf
Output
bc
Input
5
afogk
asdji
Output
alvuw
Input
6
nijfvj
tvqhwp
Output
qoztvz
Tags: bitmasks, math, number theory, strings
Correct Solution:
```
def arr(ls):
l = [0]*len(ls)
for i,v in enumerate(ls):
l[i] = v-97
return l
def div2(ls, n):
carry = 0
for i in range(n):
rem = ls[i]%2 == 1
ls[i] = ls[i]//2 + carry
if rem: carry = 13
else: carry = 0
return ls
def subtr(ls1, ls2,n):
carry = 0
for i in range(n-1, -1, -1):
ls1[i]-=carry
carry = int(ls1[i] < ls2[i])
ls1[i] = (ls1[i]-ls2[i])%26
return ls1
def add(ls1, ls2, n):
for i in range(n-1, -1, -1):
ls1[i] += ls2[i]
if ls1[i] > 25:
ls1[i-1]+=1
ls1[i]-=26
return ls1
import io, os
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
# input = io.StringIO(os.read(0, os.fstat(0).st_size).decode()).readline
kk=lambda:map(int,input().split())
k2=lambda:map(lambda x:int(x)-1, input().split())
ll=lambda:list(kk())
n,w1, w2 =int(input()), list(input().strip()), list(input().strip())
ls1, ls2 = arr(w1), arr(w2)
subtr(ls2, ls1, n)
div2(ls2, n)
add(ls1, ls2, n)
print("".join(map(lambda x: chr(97+x), [x for x in ls1])))
```
| 38,702 | [
0.2159423828125,
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0.40234375,
0.375732421875,
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1.1396484375,
0.54443359375,
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-0.08514404296875,
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0.179443359375,
-0.42724609375,
... | 0 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings s and t, both consisting of exactly k lowercase Latin letters, s is lexicographically less than t.
Let's consider list of all strings consisting of exactly k lowercase Latin letters, lexicographically not less than s and not greater than t (including s and t) in lexicographical order. For example, for k=2, s="az" and t="bf" the list will be ["az", "ba", "bb", "bc", "bd", "be", "bf"].
Your task is to print the median (the middle element) of this list. For the example above this will be "bc".
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Input
The first line of the input contains one integer k (1 ≤ k ≤ 2 ⋅ 10^5) — the length of strings.
The second line of the input contains one string s consisting of exactly k lowercase Latin letters.
The third line of the input contains one string t consisting of exactly k lowercase Latin letters.
It is guaranteed that s is lexicographically less than t.
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Output
Print one string consisting exactly of k lowercase Latin letters — the median (the middle element) of list of strings of length k lexicographically not less than s and not greater than t.
Examples
Input
2
az
bf
Output
bc
Input
5
afogk
asdji
Output
alvuw
Input
6
nijfvj
tvqhwp
Output
qoztvz
Tags: bitmasks, math, number theory, strings
Correct Solution:
```
def plus(a, b):
c = ""
i = len(a)-1
arg = 0
while i >= 0:
k = (arg + ord(a[i]) + ord(b[i]) - 194)
arg = k >= 26
i -= 1
c = chr(k % 26) + c
c = chr(1) * arg + c
return c
def minus(a):
c = ""
arg = 0
for q in a:
c += chr(ord(q)//2 + arg*13)
arg = ord(q) % 2
if arg:
c = c[:len(c)-1] + chr(ord(c[len(c)-1])+1)
return c
n = int(input())
a1 = ""
a1 = "s"*n
c = minus(plus(input() ,input()))
c1 = ""
for q in c:
c1 += chr(ord(q) + 97)
print(c1[len(c1)-n:])
```
| 38,703 | [
0.20947265625,
-0.0228118896484375,
0.369140625,
0.405029296875,
-0.477294921875,
-0.416748046875,
-0.03662109375,
-0.034820556640625,
-0.1856689453125,
1.109375,
0.56103515625,
-0.200927734375,
-0.1513671875,
-0.9501953125,
-0.74609375,
0.1826171875,
-0.37158203125,
-0.68115234375... | 0 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings s and t, both consisting of exactly k lowercase Latin letters, s is lexicographically less than t.
Let's consider list of all strings consisting of exactly k lowercase Latin letters, lexicographically not less than s and not greater than t (including s and t) in lexicographical order. For example, for k=2, s="az" and t="bf" the list will be ["az", "ba", "bb", "bc", "bd", "be", "bf"].
Your task is to print the median (the middle element) of this list. For the example above this will be "bc".
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Input
The first line of the input contains one integer k (1 ≤ k ≤ 2 ⋅ 10^5) — the length of strings.
The second line of the input contains one string s consisting of exactly k lowercase Latin letters.
The third line of the input contains one string t consisting of exactly k lowercase Latin letters.
It is guaranteed that s is lexicographically less than t.
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Output
Print one string consisting exactly of k lowercase Latin letters — the median (the middle element) of list of strings of length k lexicographically not less than s and not greater than t.
Examples
Input
2
az
bf
Output
bc
Input
5
afogk
asdji
Output
alvuw
Input
6
nijfvj
tvqhwp
Output
qoztvz
Tags: bitmasks, math, number theory, strings
Correct Solution:
```
def plus(a, b):
c = ""
i = len(a)-1
arg = 0
while i >= 0:
k = (arg + ord(a[i]) + ord(b[i]) - 194)
arg = 1
if k < 26:
arg = 0
i -= 1
c = chr(k % 26) + c
c = chr(1) * arg + c
return c
def minus(a):
c = ""
arg = 0
for q in a:
c += chr(ord(q)//2 + arg*13)
arg = ord(q) % 2
if arg:
c = c[:len(c)-1] + chr(ord(c[len(c)-1])+1)
return c
n = int(input())
a1 = ""
i = 0
while i < n:
a1 += "s"
i += 1
c = minus(plus(input() ,input()))
c1 = ""
for q in c:
c1 += chr(ord(q) + 97)
print(c1[len(c1)-n:])
```
| 38,704 | [
0.2061767578125,
-0.02191162109375,
0.375732421875,
0.403564453125,
-0.485107421875,
-0.416748046875,
-0.037567138671875,
-0.02728271484375,
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1.111328125,
0.568359375,
-0.1959228515625,
-0.1392822265625,
-0.95166015625,
-0.7509765625,
0.198486328125,
-0.385498046875,... | 0 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings s and t, both consisting of exactly k lowercase Latin letters, s is lexicographically less than t.
Let's consider list of all strings consisting of exactly k lowercase Latin letters, lexicographically not less than s and not greater than t (including s and t) in lexicographical order. For example, for k=2, s="az" and t="bf" the list will be ["az", "ba", "bb", "bc", "bd", "be", "bf"].
Your task is to print the median (the middle element) of this list. For the example above this will be "bc".
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Input
The first line of the input contains one integer k (1 ≤ k ≤ 2 ⋅ 10^5) — the length of strings.
The second line of the input contains one string s consisting of exactly k lowercase Latin letters.
The third line of the input contains one string t consisting of exactly k lowercase Latin letters.
It is guaranteed that s is lexicographically less than t.
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Output
Print one string consisting exactly of k lowercase Latin letters — the median (the middle element) of list of strings of length k lexicographically not less than s and not greater than t.
Examples
Input
2
az
bf
Output
bc
Input
5
afogk
asdji
Output
alvuw
Input
6
nijfvj
tvqhwp
Output
qoztvz
Tags: bitmasks, math, number theory, strings
Correct Solution:
```
def plus(a, b):
c = ""
i = len(a)-1
arg = 0
while i >= 0:
k = (arg + ord(a[i]) + ord(b[i]) - 194)
arg = 1
if k < 26:
arg = 0
i -= 1
c = chr(k % 26) + c
c = chr(1) * arg + c
return c
def minus(a):
c = ""
i = 0
arg = 0
while i < len(a):
c += chr((arg * 26 + ord(a[i]))//2)
arg = ord(a[i]) % 2
i += 1
if arg:
c = c[:len(c)-1] + chr(ord(c[len(c)-1])+1)
return c
n = int(input())
aa = input()
bb = input()
a1 = ""
b1 = ""
i = 0
while i < n:
a1 += "s"
i += 1
c = plus(aa,bb)
c = minus(c)
c1 = ""
for q in c:
c1 += chr(ord(q) + 97)
print(c1[len(c1)-n:])
```
| 38,705 | [
0.20166015625,
-0.0200042724609375,
0.376953125,
0.402587890625,
-0.478515625,
-0.414794921875,
-0.037841796875,
-0.03167724609375,
-0.1768798828125,
1.1171875,
0.564453125,
-0.1907958984375,
-0.1378173828125,
-0.9423828125,
-0.74853515625,
0.19384765625,
-0.3828125,
-0.6865234375,... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t, both consisting of exactly k lowercase Latin letters, s is lexicographically less than t.
Let's consider list of all strings consisting of exactly k lowercase Latin letters, lexicographically not less than s and not greater than t (including s and t) in lexicographical order. For example, for k=2, s="az" and t="bf" the list will be ["az", "ba", "bb", "bc", "bd", "be", "bf"].
Your task is to print the median (the middle element) of this list. For the example above this will be "bc".
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Input
The first line of the input contains one integer k (1 ≤ k ≤ 2 ⋅ 10^5) — the length of strings.
The second line of the input contains one string s consisting of exactly k lowercase Latin letters.
The third line of the input contains one string t consisting of exactly k lowercase Latin letters.
It is guaranteed that s is lexicographically less than t.
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Output
Print one string consisting exactly of k lowercase Latin letters — the median (the middle element) of list of strings of length k lexicographically not less than s and not greater than t.
Examples
Input
2
az
bf
Output
bc
Input
5
afogk
asdji
Output
alvuw
Input
6
nijfvj
tvqhwp
Output
qoztvz
Submitted Solution:
```
from math import ceil
k = int(input())
s1 = input()
s2 = input()
so1 = []
so2 = []
for i in range(k):
so1.append(ord(s1[i]) - ord("a"))
so2.append(ord(s2[i]) - ord("a"))
for i in range(k-1, -1, -1):
so1[i] += so2[i]
if i != 0:
so1[i - 1] += so1[i] // 26
so1[i] %= 26
med = ""
for i in range(k):
v = so1[i] // 2
if i < k - 1:
so1[i + 1] += (so1[i] % 2) * 26
med += chr(ord("a") + v)
print(med)
```
Yes
| 38,706 | [
0.2215576171875,
-0.053497314453125,
0.305908203125,
0.361572265625,
-0.5908203125,
-0.1893310546875,
-0.1480712890625,
0.10601806640625,
-0.148681640625,
1.0810546875,
0.44482421875,
-0.2880859375,
-0.209228515625,
-0.8681640625,
-0.70703125,
0.0096282958984375,
-0.6279296875,
-0.... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t, both consisting of exactly k lowercase Latin letters, s is lexicographically less than t.
Let's consider list of all strings consisting of exactly k lowercase Latin letters, lexicographically not less than s and not greater than t (including s and t) in lexicographical order. For example, for k=2, s="az" and t="bf" the list will be ["az", "ba", "bb", "bc", "bd", "be", "bf"].
Your task is to print the median (the middle element) of this list. For the example above this will be "bc".
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Input
The first line of the input contains one integer k (1 ≤ k ≤ 2 ⋅ 10^5) — the length of strings.
The second line of the input contains one string s consisting of exactly k lowercase Latin letters.
The third line of the input contains one string t consisting of exactly k lowercase Latin letters.
It is guaranteed that s is lexicographically less than t.
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Output
Print one string consisting exactly of k lowercase Latin letters — the median (the middle element) of list of strings of length k lexicographically not less than s and not greater than t.
Examples
Input
2
az
bf
Output
bc
Input
5
afogk
asdji
Output
alvuw
Input
6
nijfvj
tvqhwp
Output
qoztvz
Submitted Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import random
import heapq, bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
import sys
import threading
from collections import defaultdict
#threading.stack_size(10**8)
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
#sys.setrecursionlimit(300000)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# -------------------game starts now----------------------------------------------------import math
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
"""
Side effect!!! Changes node. Node should have exactly one child
"""
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError("Out of ranges")
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
# -----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default=2**30, func=lambda a, b: min(a , b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b:a + b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] < key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def binary(x, length):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
def countGreater(arr, n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] > k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary------------------------------------
class TrieNode:
def __init__(self):
self.children = [None] * 26
self.isEndOfWord = False
class Trie:
def __init__(self):
self.root = self.getNode()
def getNode(self):
return TrieNode()
def _charToIndex(self, ch):
return ord(ch) - ord('a')
def insert(self, key):
pCrawl = self.root
length = len(key)
for level in range(length):
index = self._charToIndex(key[level])
if not pCrawl.children[index]:
pCrawl.children[index] = self.getNode()
pCrawl = pCrawl.children[index]
pCrawl.isEndOfWord = True
def search(self, key):
pCrawl = self.root
length = len(key)
for level in range(length):
index = self._charToIndex(key[level])
if not pCrawl.children[index]:
return False
pCrawl = pCrawl.children[index]
return pCrawl != None and pCrawl.isEndOfWord
#-----------------------------------------trie---------------------------------
class Node:
def __init__(self, data):
self.data = data
self.count=0
self.left = None # left node for 0
self.right = None # right node for 1
class BinaryTrie:
def __init__(self):
self.root = Node(0)
def insert(self, pre_xor):
self.temp = self.root
for i in range(31, -1, -1):
val = pre_xor & (1 << i)
if val:
if not self.temp.right:
self.temp.right = Node(0)
self.temp = self.temp.right
self.temp.count+=1
if not val:
if not self.temp.left:
self.temp.left = Node(0)
self.temp = self.temp.left
self.temp.count += 1
self.temp.data = pre_xor
def query(self, xor):
self.temp = self.root
for i in range(31, -1, -1):
val = xor & (1 << i)
if not val:
if self.temp.left and self.temp.left.count>0:
self.temp = self.temp.left
elif self.temp.right:
self.temp = self.temp.right
else:
if self.temp.right and self.temp.right.count>0:
self.temp = self.temp.right
elif self.temp.left:
self.temp = self.temp.left
self.temp.count-=1
return xor ^ self.temp.data
#-------------------------bin trie-------------------------------------------
n=int(input())
s=input()
t=input()
d1=defaultdict(chr)
d=defaultdict(int)
for i in range(97,123):
d[chr(i)]=i-97
d1[i-97]=chr(i)
carry=0
ans=deque([])
for i in range(n-1,-1,-1):
su=carry+d[s[i]]+d[t[i]]
ans.appendleft(d1[su%26])
carry=su//26
if carry!=0:
ans.appendleft(d1[carry])
rem=0
fi=[]
#print(ans)
for i in range(len(ans)):
qu=(rem*26+d[ans[i]])//2
rem=d[ans[i]]%2
if i==0 and qu==0 and d[s[i]]!=0:
continue
fi.append(d1[qu])
print(*fi,sep="")
```
Yes
| 38,707 | [
0.15087890625,
-0.00946807861328125,
0.2305908203125,
0.39453125,
-0.505859375,
-0.1292724609375,
-0.207763671875,
0.0634765625,
-0.10626220703125,
1.1787109375,
0.52490234375,
-0.306884765625,
-0.1658935546875,
-0.89892578125,
-0.71337890625,
0.06402587890625,
-0.65625,
-0.8813476... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t, both consisting of exactly k lowercase Latin letters, s is lexicographically less than t.
Let's consider list of all strings consisting of exactly k lowercase Latin letters, lexicographically not less than s and not greater than t (including s and t) in lexicographical order. For example, for k=2, s="az" and t="bf" the list will be ["az", "ba", "bb", "bc", "bd", "be", "bf"].
Your task is to print the median (the middle element) of this list. For the example above this will be "bc".
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Input
The first line of the input contains one integer k (1 ≤ k ≤ 2 ⋅ 10^5) — the length of strings.
The second line of the input contains one string s consisting of exactly k lowercase Latin letters.
The third line of the input contains one string t consisting of exactly k lowercase Latin letters.
It is guaranteed that s is lexicographically less than t.
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Output
Print one string consisting exactly of k lowercase Latin letters — the median (the middle element) of list of strings of length k lexicographically not less than s and not greater than t.
Examples
Input
2
az
bf
Output
bc
Input
5
afogk
asdji
Output
alvuw
Input
6
nijfvj
tvqhwp
Output
qoztvz
Submitted Solution:
```
def plus(a, b):
c = ""
i = len(a)-1
arg = 0
while i >= 0:
k = (arg + ord(a[i]) + ord(b[i]) - 194)
arg = 1
if k < 26:
arg = 0
i -= 1
c = chr(k % 26) + c
c = chr(1) * arg + c
return c
def minus(a):
c = ""
i = 0
arg = 0
while i < len(a):
c += chr((arg * 26 + ord(a[i]))//2)
arg = ord(a[i]) % 2
i += 1
if arg:
c = c[:len(c)-1] + chr(ord(c[len(c)-1])+1)
return c
n = int(input())
aa = input()
bb = input()
a1 = ""
b1 = ""
i = 0
while i < n:
a1 += aa[i]
i += 1
c = plus(aa,bb)
c = minus(c)
c1 = ""
for q in c:
c1 += chr(ord(q) + ord('a'))
print(c1[len(c1)-n:])
```
Yes
| 38,708 | [
0.239501953125,
0.0264739990234375,
0.31982421875,
0.323486328125,
-0.5791015625,
-0.1934814453125,
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0.059783935546875,
-0.2066650390625,
1.072265625,
0.49072265625,
-0.2227783203125,
-0.185791015625,
-0.900390625,
-0.71875,
0.00301361083984375,
-0.6259765625,
-0.777... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t, both consisting of exactly k lowercase Latin letters, s is lexicographically less than t.
Let's consider list of all strings consisting of exactly k lowercase Latin letters, lexicographically not less than s and not greater than t (including s and t) in lexicographical order. For example, for k=2, s="az" and t="bf" the list will be ["az", "ba", "bb", "bc", "bd", "be", "bf"].
Your task is to print the median (the middle element) of this list. For the example above this will be "bc".
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Input
The first line of the input contains one integer k (1 ≤ k ≤ 2 ⋅ 10^5) — the length of strings.
The second line of the input contains one string s consisting of exactly k lowercase Latin letters.
The third line of the input contains one string t consisting of exactly k lowercase Latin letters.
It is guaranteed that s is lexicographically less than t.
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Output
Print one string consisting exactly of k lowercase Latin letters — the median (the middle element) of list of strings of length k lexicographically not less than s and not greater than t.
Examples
Input
2
az
bf
Output
bc
Input
5
afogk
asdji
Output
alvuw
Input
6
nijfvj
tvqhwp
Output
qoztvz
Submitted Solution:
```
k = int(input())
s = input()
t = input()
val = [0 for i in range(0, k)]
nval = [0 for i in range(0, k)]
for i in range(0, k):
val[i] = ord(s[i]) - ord('a') + ord(t[i]) - ord('a')
ans = ['a' for i in range(0, k)]
carry = 0
for i in range(0, k):
ncarry = 0
t = val[i] // 2 + carry
if val[i] % 2 == 1:
ncarry += 13
nval[i] = t
carry = ncarry
for i in range(k - 1, -1, -1):
if nval[i] >= 26:
nval[i - 1] += nval[i] // 26
nval[i] %= 26
ans[i] = chr(ord('a') + nval[i])
print("".join(x for x in ans))
```
Yes
| 38,709 | [
0.19921875,
-0.03289794921875,
0.2880859375,
0.3837890625,
-0.57373046875,
-0.212158203125,
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0.05718994140625,
-0.158935546875,
1.06640625,
0.45458984375,
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-0.91015625,
-0.66357421875,
0.0257110595703125,
-0.61865234375,
-0.7841796875,
... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t, both consisting of exactly k lowercase Latin letters, s is lexicographically less than t.
Let's consider list of all strings consisting of exactly k lowercase Latin letters, lexicographically not less than s and not greater than t (including s and t) in lexicographical order. For example, for k=2, s="az" and t="bf" the list will be ["az", "ba", "bb", "bc", "bd", "be", "bf"].
Your task is to print the median (the middle element) of this list. For the example above this will be "bc".
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Input
The first line of the input contains one integer k (1 ≤ k ≤ 2 ⋅ 10^5) — the length of strings.
The second line of the input contains one string s consisting of exactly k lowercase Latin letters.
The third line of the input contains one string t consisting of exactly k lowercase Latin letters.
It is guaranteed that s is lexicographically less than t.
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Output
Print one string consisting exactly of k lowercase Latin letters — the median (the middle element) of list of strings of length k lexicographically not less than s and not greater than t.
Examples
Input
2
az
bf
Output
bc
Input
5
afogk
asdji
Output
alvuw
Input
6
nijfvj
tvqhwp
Output
qoztvz
Submitted Solution:
```
k=int(input())
s1=str(input())
s2=str(input())
ans=''
rem=0
s='abcdefghijklmnopqrstuvwxyz'
for i in range(len(s1)):
val1=ord(s1[i])-96
val2=ord(s2[i])-96
val=(val1+val2)+(26*rem)
#print(val)
if(val>53):
indexx=len(ans)-1
flag=0
while(flag==0):
index1=s.index(ans[indexx])
if(index1==25):
ans=ans[:indexx]+'a'+ans[indexx+1:]
indexx-=1
else:
ans=ans[:indexx]+s[index1+1]+ans[indexx+1:]
flag=1
break
rem+=(val//26)
val=val%26
rem+=val%2
val=val//2
ans+=s[val-1]
print(ans)
```
No
| 38,710 | [
0.1878662109375,
0.00128173828125,
0.30810546875,
0.31005859375,
-0.56689453125,
-0.1851806640625,
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0.038818359375,
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1.048828125,
0.49267578125,
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-0.1795654296875,
-0.8994140625,
-0.71337890625,
0.0198822021484375,
-0.62939453125,
-0.8... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t, both consisting of exactly k lowercase Latin letters, s is lexicographically less than t.
Let's consider list of all strings consisting of exactly k lowercase Latin letters, lexicographically not less than s and not greater than t (including s and t) in lexicographical order. For example, for k=2, s="az" and t="bf" the list will be ["az", "ba", "bb", "bc", "bd", "be", "bf"].
Your task is to print the median (the middle element) of this list. For the example above this will be "bc".
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Input
The first line of the input contains one integer k (1 ≤ k ≤ 2 ⋅ 10^5) — the length of strings.
The second line of the input contains one string s consisting of exactly k lowercase Latin letters.
The third line of the input contains one string t consisting of exactly k lowercase Latin letters.
It is guaranteed that s is lexicographically less than t.
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Output
Print one string consisting exactly of k lowercase Latin letters — the median (the middle element) of list of strings of length k lexicographically not less than s and not greater than t.
Examples
Input
2
az
bf
Output
bc
Input
5
afogk
asdji
Output
alvuw
Input
6
nijfvj
tvqhwp
Output
qoztvz
Submitted Solution:
```
k=int(input())
a=input("")
b=input("")
xx=1
aa=0
bb=0
for i in range(k-1,-1,-1):
aa+=xx*(ord(a[i])-ord('a'))
xx*=26
xx=1
for i in range(k-1,-1,-1):
bb+=xx*(ord(b[i])-ord('a'))
xx*=26
cc=(int)(aa+bb)
cc=(int)(cc/2)
xx=1
print(aa)
print(bb)
for i in range(k-1):
xx*=26
for i in range(k):
yy=(int)(cc/xx)
print((chr)(ord('a')+yy),end='')
cc%=xx
xx=(int)(xx/26)
```
No
| 38,711 | [
0.2027587890625,
-0.049407958984375,
0.298583984375,
0.37939453125,
-0.56640625,
-0.19091796875,
-0.09808349609375,
0.055450439453125,
-0.160400390625,
1.0693359375,
0.486083984375,
-0.28076171875,
-0.23193359375,
-0.90283203125,
-0.69482421875,
-0.00030112266540527344,
-0.6201171875... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t, both consisting of exactly k lowercase Latin letters, s is lexicographically less than t.
Let's consider list of all strings consisting of exactly k lowercase Latin letters, lexicographically not less than s and not greater than t (including s and t) in lexicographical order. For example, for k=2, s="az" and t="bf" the list will be ["az", "ba", "bb", "bc", "bd", "be", "bf"].
Your task is to print the median (the middle element) of this list. For the example above this will be "bc".
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Input
The first line of the input contains one integer k (1 ≤ k ≤ 2 ⋅ 10^5) — the length of strings.
The second line of the input contains one string s consisting of exactly k lowercase Latin letters.
The third line of the input contains one string t consisting of exactly k lowercase Latin letters.
It is guaranteed that s is lexicographically less than t.
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Output
Print one string consisting exactly of k lowercase Latin letters — the median (the middle element) of list of strings of length k lexicographically not less than s and not greater than t.
Examples
Input
2
az
bf
Output
bc
Input
5
afogk
asdji
Output
alvuw
Input
6
nijfvj
tvqhwp
Output
qoztvz
Submitted Solution:
```
k = int(input())
a = input()
b = input()
o = ord('a')
f = 0
s = 0
if (k == 159000):
exit(0)
for i in a:
f = f * 26 + (ord(i) - o)
for i in b:
s = s * 26 + (ord(i) - o)
t = (f + s) // 2
str = []
while k > 0:
str.append(chr(t % 26 + o))
t //= 26
k -= 1
print(*str[::-1], sep="")
```
No
| 38,712 | [
0.208984375,
-0.05316162109375,
0.30126953125,
0.377685546875,
-0.57275390625,
-0.2230224609375,
-0.1226806640625,
0.04986572265625,
-0.136474609375,
1.068359375,
0.5029296875,
-0.284423828125,
-0.1981201171875,
-0.89794921875,
-0.72265625,
0.0123291015625,
-0.62939453125,
-0.80810... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t, both consisting of exactly k lowercase Latin letters, s is lexicographically less than t.
Let's consider list of all strings consisting of exactly k lowercase Latin letters, lexicographically not less than s and not greater than t (including s and t) in lexicographical order. For example, for k=2, s="az" and t="bf" the list will be ["az", "ba", "bb", "bc", "bd", "be", "bf"].
Your task is to print the median (the middle element) of this list. For the example above this will be "bc".
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Input
The first line of the input contains one integer k (1 ≤ k ≤ 2 ⋅ 10^5) — the length of strings.
The second line of the input contains one string s consisting of exactly k lowercase Latin letters.
The third line of the input contains one string t consisting of exactly k lowercase Latin letters.
It is guaranteed that s is lexicographically less than t.
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Output
Print one string consisting exactly of k lowercase Latin letters — the median (the middle element) of list of strings of length k lexicographically not less than s and not greater than t.
Examples
Input
2
az
bf
Output
bc
Input
5
afogk
asdji
Output
alvuw
Input
6
nijfvj
tvqhwp
Output
qoztvz
Submitted Solution:
```
n = int(input())
a = [ord(c) - 96 for c in input()]
b = [ord(c) - 96 for c in input()]
ans = []
for i in range(n):
sum = a[i] + b[i]
if i+1<n:
a[i+1]+=(sum%2)*26
sum//=2
ans.append(sum)
for i in range(n-1,-1,-1):
ans[i]%=26
if ans[i]==0:
ans[i]=26
if i-1 >= 0:
ans[i-1] += (ans[i]//26)
ans = [chr(x+96) for x in ans]
print(''.join(ans))
```
No
| 38,713 | [
0.1920166015625,
-0.0262908935546875,
0.314453125,
0.35595703125,
-0.51416015625,
-0.207275390625,
-0.09954833984375,
0.09625244140625,
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0.9970703125,
0.50390625,
-0.256591796875,
-0.1810302734375,
-0.87939453125,
-0.75341796875,
-0.01068115234375,
-0.65478515625,
... | 0 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a uppercase Latin letters 'A' and b letters 'B'.
The period of the string is the smallest such positive integer k that s_i = s_{i~mod~k} (0-indexed) for each i. Note that this implies that k won't always divide a+b = |s|.
For example, the period of string "ABAABAA" is 3, the period of "AAAA" is 1, and the period of "AABBB" is 5.
Find the number of different periods over all possible strings with a letters 'A' and b letters 'B'.
Input
The first line contains two integers a and b (1 ≤ a, b ≤ 10^9) — the number of letters 'A' and 'B', respectively.
Output
Print the number of different periods over all possible strings with a letters 'A' and b letters 'B'.
Examples
Input
2 4
Output
4
Input
5 3
Output
5
Note
All the possible periods for the first example:
* 3 "BBABBA"
* 4 "BBAABB"
* 5 "BBBAAB"
* 6 "AABBBB"
All the possible periods for the second example:
* 3 "BAABAABA"
* 5 "BAABABAA"
* 6 "BABAAABA"
* 7 "BAABAAAB"
* 8 "AAAAABBB"
Note that these are not the only possible strings for the given periods.
Tags: binary search, implementation, math
Correct Solution:
```
a,b= map(int,input().split())
ans,l,n=0,1,a+b
while l<=n:
g= n//l
if a<g or b<g:
l= (n//g) +1
continue
r= n//g
al = (a+g)//(g+1)
ah = a//g
bl=(b+g)//(g+1)
bh = b//g
if (al <= ah and bl <= bh):
ans += max(0,min(r,ah+bh)- max(l,al +bl)+1)
l=r+1
print(ans)
```
| 38,730 | [
0.420654296875,
-0.054229736328125,
0.1583251953125,
-0.05499267578125,
-0.49853515625,
-0.24267578125,
-0.2259521484375,
0.01361846923828125,
-0.0232391357421875,
0.97998046875,
0.64599609375,
-0.3427734375,
-0.1026611328125,
-0.84228515625,
-0.560546875,
0.44873046875,
-0.963867187... | 0 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a uppercase Latin letters 'A' and b letters 'B'.
The period of the string is the smallest such positive integer k that s_i = s_{i~mod~k} (0-indexed) for each i. Note that this implies that k won't always divide a+b = |s|.
For example, the period of string "ABAABAA" is 3, the period of "AAAA" is 1, and the period of "AABBB" is 5.
Find the number of different periods over all possible strings with a letters 'A' and b letters 'B'.
Input
The first line contains two integers a and b (1 ≤ a, b ≤ 10^9) — the number of letters 'A' and 'B', respectively.
Output
Print the number of different periods over all possible strings with a letters 'A' and b letters 'B'.
Examples
Input
2 4
Output
4
Input
5 3
Output
5
Note
All the possible periods for the first example:
* 3 "BBABBA"
* 4 "BBAABB"
* 5 "BBBAAB"
* 6 "AABBBB"
All the possible periods for the second example:
* 3 "BAABAABA"
* 5 "BAABABAA"
* 6 "BABAAABA"
* 7 "BAABAAAB"
* 8 "AAAAABBB"
Note that these are not the only possible strings for the given periods.
Tags: binary search, implementation, math
Correct Solution:
```
import math
a,b= map(int,input().split())
n=a+b
ans,l=0,1
while l<=n:
g= n//l
if a<g or b<g:
l= (n//g) +1
continue
r= n//g
a_low = (a+g)//(g+1)
a_high = a//g
b_low=(b+g)//(g+1)
b_high = b//g
if (a_low <= a_high and b_low <= b_high):
ans += max(0,min(r,a_high+b_high)- max(l,a_low +b_low)+1)
l=r+1
print(ans)
```
| 38,731 | [
0.443359375,
-0.044952392578125,
0.166748046875,
-0.0777587890625,
-0.51025390625,
-0.252197265625,
-0.239501953125,
0.00507354736328125,
-0.01010894775390625,
0.99658203125,
0.62451171875,
-0.31298828125,
-0.08380126953125,
-0.828125,
-0.53369140625,
0.43896484375,
-0.96142578125,
... | 0 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a uppercase Latin letters 'A' and b letters 'B'.
The period of the string is the smallest such positive integer k that s_i = s_{i~mod~k} (0-indexed) for each i. Note that this implies that k won't always divide a+b = |s|.
For example, the period of string "ABAABAA" is 3, the period of "AAAA" is 1, and the period of "AABBB" is 5.
Find the number of different periods over all possible strings with a letters 'A' and b letters 'B'.
Input
The first line contains two integers a and b (1 ≤ a, b ≤ 10^9) — the number of letters 'A' and 'B', respectively.
Output
Print the number of different periods over all possible strings with a letters 'A' and b letters 'B'.
Examples
Input
2 4
Output
4
Input
5 3
Output
5
Note
All the possible periods for the first example:
* 3 "BBABBA"
* 4 "BBAABB"
* 5 "BBBAAB"
* 6 "AABBBB"
All the possible periods for the second example:
* 3 "BAABAABA"
* 5 "BAABABAA"
* 6 "BABAAABA"
* 7 "BAABAAAB"
* 8 "AAAAABBB"
Note that these are not the only possible strings for the given periods.
Tags: binary search, implementation, math
Correct Solution:
```
import math
a, b = [int(x) for x in input().split()]
cnt = 0
ab = a + b
k = 2
while k <= ab:
t = ab // k
aPeriod = a // t
bPeriod = b // t
aLeft = (a + t) // (t + 1)
bLeft = (b + t) // (t + 1)
arem = a - t * aPeriod
brem = b - t * bPeriod
if aPeriod >= arem and bPeriod >= brem:
validK = aPeriod + bPeriod - aLeft - bLeft + 1
if (t + 1) * (aLeft + bLeft) == ab:
validK -= 1
if validK > 0:
cnt += validK
k = ab // t + 1
print(cnt)
```
| 38,732 | [
0.427490234375,
-0.07318115234375,
0.1640625,
-0.08245849609375,
-0.52734375,
-0.2098388671875,
-0.17529296875,
-0.01432037353515625,
0.01071929931640625,
0.947265625,
0.6611328125,
-0.396240234375,
-0.0068359375,
-0.7998046875,
-0.59130859375,
0.53466796875,
-0.9697265625,
-0.6044... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a uppercase Latin letters 'A' and b letters 'B'.
The period of the string is the smallest such positive integer k that s_i = s_{i~mod~k} (0-indexed) for each i. Note that this implies that k won't always divide a+b = |s|.
For example, the period of string "ABAABAA" is 3, the period of "AAAA" is 1, and the period of "AABBB" is 5.
Find the number of different periods over all possible strings with a letters 'A' and b letters 'B'.
Input
The first line contains two integers a and b (1 ≤ a, b ≤ 10^9) — the number of letters 'A' and 'B', respectively.
Output
Print the number of different periods over all possible strings with a letters 'A' and b letters 'B'.
Examples
Input
2 4
Output
4
Input
5 3
Output
5
Note
All the possible periods for the first example:
* 3 "BBABBA"
* 4 "BBAABB"
* 5 "BBBAAB"
* 6 "AABBBB"
All the possible periods for the second example:
* 3 "BAABAABA"
* 5 "BAABABAA"
* 6 "BABAAABA"
* 7 "BAABAAAB"
* 8 "AAAAABBB"
Note that these are not the only possible strings for the given periods.
Submitted Solution:
```
a = [int(i) for i in input().split()]
print(max(a))
```
No
| 38,733 | [
0.4375,
-0.0203704833984375,
0.2073974609375,
-0.052398681640625,
-0.58837890625,
-0.1368408203125,
-0.294189453125,
-0.0012607574462890625,
-0.00994873046875,
0.97607421875,
0.5595703125,
-0.316650390625,
-0.1962890625,
-0.8017578125,
-0.56201171875,
0.34228515625,
-0.9658203125,
... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a uppercase Latin letters 'A' and b letters 'B'.
The period of the string is the smallest such positive integer k that s_i = s_{i~mod~k} (0-indexed) for each i. Note that this implies that k won't always divide a+b = |s|.
For example, the period of string "ABAABAA" is 3, the period of "AAAA" is 1, and the period of "AABBB" is 5.
Find the number of different periods over all possible strings with a letters 'A' and b letters 'B'.
Input
The first line contains two integers a and b (1 ≤ a, b ≤ 10^9) — the number of letters 'A' and 'B', respectively.
Output
Print the number of different periods over all possible strings with a letters 'A' and b letters 'B'.
Examples
Input
2 4
Output
4
Input
5 3
Output
5
Note
All the possible periods for the first example:
* 3 "BBABBA"
* 4 "BBAABB"
* 5 "BBBAAB"
* 6 "AABBBB"
All the possible periods for the second example:
* 3 "BAABAABA"
* 5 "BAABABAA"
* 6 "BABAAABA"
* 7 "BAABAAAB"
* 8 "AAAAABBB"
Note that these are not the only possible strings for the given periods.
Submitted Solution:
```
import fileinput
def calc(a, b):
ans = 0
l = a + b
i = 1
while i < l:
P = l // i
r = l // P
if P <= a and P <= b:
an = (a + P) // (P + 1)
ax = a // P
bn = (b + P) // (P + 1)
bx = b // P
if an <= ax and bn <= bx:
ans += min(ax + bx, r) - max(an + bn, i) + 1
i = r + 1
return ans
if __name__ == '__main__':
it = fileinput.input()
a, b = [int(x) for x in next(it).split()]
print(calc(a, b))
```
No
| 38,734 | [
0.364990234375,
0.06768798828125,
0.296142578125,
-0.05828857421875,
-0.59375,
-0.1295166015625,
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0.017974853515625,
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0.95849609375,
0.57080078125,
-0.324462890625,
-0.18798828125,
-0.8115234375,
-0.58154296875,
0.355712890625,
-0.96142578125,
-0.76855... | 0 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a uppercase Latin letters 'A' and b letters 'B'.
The period of the string is the smallest such positive integer k that s_i = s_{i~mod~k} (0-indexed) for each i. Note that this implies that k won't always divide a+b = |s|.
For example, the period of string "ABAABAA" is 3, the period of "AAAA" is 1, and the period of "AABBB" is 5.
Find the number of different periods over all possible strings with a letters 'A' and b letters 'B'.
Input
The first line contains two integers a and b (1 ≤ a, b ≤ 10^9) — the number of letters 'A' and 'B', respectively.
Output
Print the number of different periods over all possible strings with a letters 'A' and b letters 'B'.
Examples
Input
2 4
Output
4
Input
5 3
Output
5
Note
All the possible periods for the first example:
* 3 "BBABBA"
* 4 "BBAABB"
* 5 "BBBAAB"
* 6 "AABBBB"
All the possible periods for the second example:
* 3 "BAABAABA"
* 5 "BAABABAA"
* 6 "BABAAABA"
* 7 "BAABAAAB"
* 8 "AAAAABBB"
Note that these are not the only possible strings for the given periods.
Submitted Solution:
```
a,b = map(int, input().split())
total = a+b
result = 0
i = 1
while i <= total:
groups = total//i
rem = total%i
if a%groups <= rem:
if rem > groups:
result += rem//groups
i += rem//groups
else:
result += 1
i += 1
else:
i += 1
print(result)
```
No
| 38,735 | [
0.458740234375,
0.0472412109375,
0.19091796875,
-0.138671875,
-0.6220703125,
-0.10540771484375,
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0.004016876220703125,
0.057952880859375,
1.04296875,
0.6630859375,
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-0.77294921875,
-0.59033203125,
0.346923828125,
-0.9619140625,
-0.8125,
-0.3... | 0 |
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