message stringlengths 2 15.4k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 45 107k | cluster float64 21 21 | __index_level_0__ int64 90 214k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As Will is stuck in the Upside Down, he can still communicate with his mom, Joyce, through the Christmas lights (he can turn them on and off with his mind). He can't directly tell his mom where he is, because the monster that took him to the Upside Down will know and relocate him.
<image>
Thus, he came up with a puzzle to tell his mom his coordinates. His coordinates are the answer to the following problem.
A string consisting only of parentheses ('(' and ')') is called a bracket sequence. Some bracket sequence are called correct bracket sequences. More formally:
* Empty string is a correct bracket sequence.
* if s is a correct bracket sequence, then (s) is also a correct bracket sequence.
* if s and t are correct bracket sequences, then st (concatenation of s and t) is also a correct bracket sequence.
A string consisting of parentheses and question marks ('?') is called pretty if and only if there's a way to replace each question mark with either '(' or ')' such that the resulting string is a non-empty correct bracket sequence.
Will gave his mom a string s consisting of parentheses and question marks (using Morse code through the lights) and his coordinates are the number of pairs of integers (l, r) such that 1 ≤ l ≤ r ≤ |s| and the string slsl + 1... sr is pretty, where si is i-th character of s.
Joyce doesn't know anything about bracket sequences, so she asked for your help.
Input
The first and only line of input contains string s, consisting only of characters '(', ')' and '?' (2 ≤ |s| ≤ 5000).
Output
Print the answer to Will's puzzle in the first and only line of output.
Examples
Input
((?))
Output
4
Input
??()??
Output
7
Note
For the first sample testcase, the pretty substrings of s are:
1. "(?" which can be transformed to "()".
2. "?)" which can be transformed to "()".
3. "((?)" which can be transformed to "(())".
4. "(?))" which can be transformed to "(())".
For the second sample testcase, the pretty substrings of s are:
1. "??" which can be transformed to "()".
2. "()".
3. "??()" which can be transformed to "()()".
4. "?()?" which can be transformed to "(())".
5. "??" which can be transformed to "()".
6. "()??" which can be transformed to "()()".
7. "??()??" which can be transformed to "()()()".
Submitted Solution:
```
import sys, math, os.path
FILE_INPUT = "c.in"
DEBUG = os.path.isfile(FILE_INPUT)
if DEBUG:
sys.stdin = open(FILE_INPUT)
def ni():
return int(input())
def nvi():
return map(int, input().split(" "))
def nia():
return list(map(int,input().split()))
def log(x):
if (DEBUG):
print(x)
s = input()
ls = len(s)
count = 0
for i in range(0,ls):
low = 0
high = 0
if s[i] != ')':
log(i)
for j in range(i,ls):
if s[j] == '(':
low += 1
high += 1
elif s[j] == ')':
low -= 1
high -= 1
else:
low = max(0, low-1)
high += 1
log(str(i) + ": "+str(low) + "-"+str(high))
if (low > high):
break
if (low <= 0 and 0 <= high) and ((j - i) % 2 == 1):
log("count " + str(i) + " - " + str(j))
count += 1
print(count)
``` | instruction | 0 | 47,639 | 21 | 95,278 |
No | output | 1 | 47,639 | 21 | 95,279 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As Will is stuck in the Upside Down, he can still communicate with his mom, Joyce, through the Christmas lights (he can turn them on and off with his mind). He can't directly tell his mom where he is, because the monster that took him to the Upside Down will know and relocate him.
<image>
Thus, he came up with a puzzle to tell his mom his coordinates. His coordinates are the answer to the following problem.
A string consisting only of parentheses ('(' and ')') is called a bracket sequence. Some bracket sequence are called correct bracket sequences. More formally:
* Empty string is a correct bracket sequence.
* if s is a correct bracket sequence, then (s) is also a correct bracket sequence.
* if s and t are correct bracket sequences, then st (concatenation of s and t) is also a correct bracket sequence.
A string consisting of parentheses and question marks ('?') is called pretty if and only if there's a way to replace each question mark with either '(' or ')' such that the resulting string is a non-empty correct bracket sequence.
Will gave his mom a string s consisting of parentheses and question marks (using Morse code through the lights) and his coordinates are the number of pairs of integers (l, r) such that 1 ≤ l ≤ r ≤ |s| and the string slsl + 1... sr is pretty, where si is i-th character of s.
Joyce doesn't know anything about bracket sequences, so she asked for your help.
Input
The first and only line of input contains string s, consisting only of characters '(', ')' and '?' (2 ≤ |s| ≤ 5000).
Output
Print the answer to Will's puzzle in the first and only line of output.
Examples
Input
((?))
Output
4
Input
??()??
Output
7
Note
For the first sample testcase, the pretty substrings of s are:
1. "(?" which can be transformed to "()".
2. "?)" which can be transformed to "()".
3. "((?)" which can be transformed to "(())".
4. "(?))" which can be transformed to "(())".
For the second sample testcase, the pretty substrings of s are:
1. "??" which can be transformed to "()".
2. "()".
3. "??()" which can be transformed to "()()".
4. "?()?" which can be transformed to "(())".
5. "??" which can be transformed to "()".
6. "()??" which can be transformed to "()()".
7. "??()??" which can be transformed to "()()()".
Submitted Solution:
```
s = input()
ans = 0
ls = len(s)
for l in range(ls):
ct = 0
ctq = 0
for r in range(l,ls):
if s[r] == '(':
ct += 1
continue
elif s[r] == ')':
ct -= 1
if ct < 0:
if ctq > 0:
ctq -= 1
ct +=1
else:
break
elif s[r] == '?':
ctq += 1
if ctq-ct>=0 and (ctq-ct)%2==0:
ans +=1
print(ans)
``` | instruction | 0 | 47,640 | 21 | 95,280 |
No | output | 1 | 47,640 | 21 | 95,281 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As Will is stuck in the Upside Down, he can still communicate with his mom, Joyce, through the Christmas lights (he can turn them on and off with his mind). He can't directly tell his mom where he is, because the monster that took him to the Upside Down will know and relocate him.
<image>
Thus, he came up with a puzzle to tell his mom his coordinates. His coordinates are the answer to the following problem.
A string consisting only of parentheses ('(' and ')') is called a bracket sequence. Some bracket sequence are called correct bracket sequences. More formally:
* Empty string is a correct bracket sequence.
* if s is a correct bracket sequence, then (s) is also a correct bracket sequence.
* if s and t are correct bracket sequences, then st (concatenation of s and t) is also a correct bracket sequence.
A string consisting of parentheses and question marks ('?') is called pretty if and only if there's a way to replace each question mark with either '(' or ')' such that the resulting string is a non-empty correct bracket sequence.
Will gave his mom a string s consisting of parentheses and question marks (using Morse code through the lights) and his coordinates are the number of pairs of integers (l, r) such that 1 ≤ l ≤ r ≤ |s| and the string slsl + 1... sr is pretty, where si is i-th character of s.
Joyce doesn't know anything about bracket sequences, so she asked for your help.
Input
The first and only line of input contains string s, consisting only of characters '(', ')' and '?' (2 ≤ |s| ≤ 5000).
Output
Print the answer to Will's puzzle in the first and only line of output.
Examples
Input
((?))
Output
4
Input
??()??
Output
7
Note
For the first sample testcase, the pretty substrings of s are:
1. "(?" which can be transformed to "()".
2. "?)" which can be transformed to "()".
3. "((?)" which can be transformed to "(())".
4. "(?))" which can be transformed to "(())".
For the second sample testcase, the pretty substrings of s are:
1. "??" which can be transformed to "()".
2. "()".
3. "??()" which can be transformed to "()()".
4. "?()?" which can be transformed to "(())".
5. "??" which can be transformed to "()".
6. "()??" which can be transformed to "()()".
7. "??()??" which can be transformed to "()()()".
Submitted Solution:
```
import sys
import math
import bisect
from sys import stdin, stdout
from math import gcd, floor, sqrt, log
from collections import defaultdict as dd
from bisect import bisect_left as bl, bisect_right as br
from collections import Counter
#sys.setrecursionlimit(100000000)
inp = lambda: int(input())
strng = lambda: input().strip()
jn = lambda x, l: x.join(map(str, l))
strl = lambda: list(input().strip())
mul = lambda: map(int, input().strip().split())
mulf = lambda: map(float, input().strip().split())
seq = lambda: list(map(int, input().strip().split()))
ceil = lambda x: int(x) if (x == int(x)) else int(x) + 1
ceildiv = lambda x, d: x // d if (x % d == 0) else x // d + 1
flush = lambda: stdout.flush()
stdstr = lambda: stdin.readline()
stdint = lambda: int(stdin.readline())
stdpr = lambda x: stdout.write(str(x))
stdarr = lambda: map(int, stdstr().split())
mod = 1000000007
s = input()
res = 0
done = Counter()
for i in range(len(s)):
if(s[i] == ")"):
continue
c = Counter()
c[s[i]] += 1
for j in range(i+1, len(s)):
if(s[j] == "("):
c[s[j]] += 1
continue
else:
if (c[")"] >= c["("] + c["?"]):
c[s[j]] += 1
continue
c[s[j]] += 1
if(abs(c["("]-c[")"]) > c["?"]):
continue
else:
if((c["("]+c[")"]+c["?"])%2 == 0):
res += 1
else:
continue
print(res)
``` | instruction | 0 | 47,641 | 21 | 95,282 |
No | output | 1 | 47,641 | 21 | 95,283 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As Will is stuck in the Upside Down, he can still communicate with his mom, Joyce, through the Christmas lights (he can turn them on and off with his mind). He can't directly tell his mom where he is, because the monster that took him to the Upside Down will know and relocate him.
<image>
Thus, he came up with a puzzle to tell his mom his coordinates. His coordinates are the answer to the following problem.
A string consisting only of parentheses ('(' and ')') is called a bracket sequence. Some bracket sequence are called correct bracket sequences. More formally:
* Empty string is a correct bracket sequence.
* if s is a correct bracket sequence, then (s) is also a correct bracket sequence.
* if s and t are correct bracket sequences, then st (concatenation of s and t) is also a correct bracket sequence.
A string consisting of parentheses and question marks ('?') is called pretty if and only if there's a way to replace each question mark with either '(' or ')' such that the resulting string is a non-empty correct bracket sequence.
Will gave his mom a string s consisting of parentheses and question marks (using Morse code through the lights) and his coordinates are the number of pairs of integers (l, r) such that 1 ≤ l ≤ r ≤ |s| and the string slsl + 1... sr is pretty, where si is i-th character of s.
Joyce doesn't know anything about bracket sequences, so she asked for your help.
Input
The first and only line of input contains string s, consisting only of characters '(', ')' and '?' (2 ≤ |s| ≤ 5000).
Output
Print the answer to Will's puzzle in the first and only line of output.
Examples
Input
((?))
Output
4
Input
??()??
Output
7
Note
For the first sample testcase, the pretty substrings of s are:
1. "(?" which can be transformed to "()".
2. "?)" which can be transformed to "()".
3. "((?)" which can be transformed to "(())".
4. "(?))" which can be transformed to "(())".
For the second sample testcase, the pretty substrings of s are:
1. "??" which can be transformed to "()".
2. "()".
3. "??()" which can be transformed to "()()".
4. "?()?" which can be transformed to "(())".
5. "??" which can be transformed to "()".
6. "()??" which can be transformed to "()()".
7. "??()??" which can be transformed to "()()()".
Submitted Solution:
```
m=str(input())
p=[]
c=0
def ok(n):
zero=0
one=0
for i in range(len(n)):
if(n[i]==')'):
for t in range(i):
if(n[t]=='('):
zero=zero+1
elif(n[t]==')'):
one=one+1
if(zero<=one):
return False
return True
def isvalid(n):
zero=0
one=0
q=0
for i in range(len(n)):
if(n[i]=='('):
zero=zero+1
elif(n[i]==')'):
one=one+1
else:
q=q+1
if((q+zero+one )%2 !=0):
return False
if(zero==one):
return True
elif(zero+q==one):
for i in range(len(n)):
if(n[i]=='?'):
n=n[:i]+'('+n[i+1:]
if(ok(n)):
return True
elif(zero==one+q):
for i in range(len(n)):
if(n[i]=='?'):
n=n[:i]+')'+n[i+1:]
if(ok(n)):
return True
return False
for i in range(2,len(m),2):
for j in range(0,len(m)):
k=m[j:j+i]
if(isvalid(k)):
c=c+1
print(c)
``` | instruction | 0 | 47,642 | 21 | 95,284 |
No | output | 1 | 47,642 | 21 | 95,285 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As Will is stuck in the Upside Down, he can still communicate with his mom, Joyce, through the Christmas lights (he can turn them on and off with his mind). He can't directly tell his mom where he is, because the monster that took him to the Upside Down will know and relocate him.
<image>
Thus, he came up with a puzzle to tell his mom his coordinates. His coordinates are the answer to the following problem.
A string consisting only of parentheses ('(' and ')') is called a bracket sequence. Some bracket sequence are called correct bracket sequences. More formally:
* Empty string is a correct bracket sequence.
* if s is a correct bracket sequence, then (s) is also a correct bracket sequence.
* if s and t are correct bracket sequences, then st (concatenation of s and t) is also a correct bracket sequence.
A string consisting of parentheses and question marks ('?') is called pretty if and only if there's a way to replace each question mark with either '(' or ')' such that the resulting string is a non-empty correct bracket sequence.
Will gave his mom a string s consisting of parentheses and question marks (using Morse code through the lights) and his coordinates are the number of pairs of integers (l, r) such that 1 ≤ l ≤ r ≤ |s| and the string slsl + 1... sr is pretty, where si is i-th character of s.
Joyce doesn't know anything about bracket sequences, so she asked for your help.
Input
The first and only line of input contains string s, consisting only of characters '(', ')' and '?' (2 ≤ |s| ≤ 5000).
Output
Print the answer to Will's puzzle in the first and only line of output.
Examples
Input
((?))
Output
4
Input
??()??
Output
7
Note
For the first sample testcase, the pretty substrings of s are:
1. "(?" which can be transformed to "()".
2. "?)" which can be transformed to "()".
3. "((?)" which can be transformed to "(())".
4. "(?))" which can be transformed to "(())".
For the second sample testcase, the pretty substrings of s are:
1. "??" which can be transformed to "()".
2. "()".
3. "??()" which can be transformed to "()()".
4. "?()?" which can be transformed to "(())".
5. "??" which can be transformed to "()".
6. "()??" which can be transformed to "()()".
7. "??()??" which can be transformed to "()()()". | instruction | 0 | 51,032 | 21 | 102,064 |
Tags: dp, greedy, implementation, math
Correct Solution:
```
#!/user/bin/env python 3.5
# -*- coding: utf-8 -*-
s=input()
a=0
n=len(s)
for i in range(n):
l=0
k=0
for j in range(i,n):
l+=s[j]=='('
l-=s[j]==')'
k+=s[j]=='?'
if l+k<0:
break
if k>l:
l,k=k,l
if l==k:
a=a+1
print(a)
``` | output | 1 | 51,032 | 21 | 102,065 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As Will is stuck in the Upside Down, he can still communicate with his mom, Joyce, through the Christmas lights (he can turn them on and off with his mind). He can't directly tell his mom where he is, because the monster that took him to the Upside Down will know and relocate him.
<image>
Thus, he came up with a puzzle to tell his mom his coordinates. His coordinates are the answer to the following problem.
A string consisting only of parentheses ('(' and ')') is called a bracket sequence. Some bracket sequence are called correct bracket sequences. More formally:
* Empty string is a correct bracket sequence.
* if s is a correct bracket sequence, then (s) is also a correct bracket sequence.
* if s and t are correct bracket sequences, then st (concatenation of s and t) is also a correct bracket sequence.
A string consisting of parentheses and question marks ('?') is called pretty if and only if there's a way to replace each question mark with either '(' or ')' such that the resulting string is a non-empty correct bracket sequence.
Will gave his mom a string s consisting of parentheses and question marks (using Morse code through the lights) and his coordinates are the number of pairs of integers (l, r) such that 1 ≤ l ≤ r ≤ |s| and the string slsl + 1... sr is pretty, where si is i-th character of s.
Joyce doesn't know anything about bracket sequences, so she asked for your help.
Input
The first and only line of input contains string s, consisting only of characters '(', ')' and '?' (2 ≤ |s| ≤ 5000).
Output
Print the answer to Will's puzzle in the first and only line of output.
Examples
Input
((?))
Output
4
Input
??()??
Output
7
Note
For the first sample testcase, the pretty substrings of s are:
1. "(?" which can be transformed to "()".
2. "?)" which can be transformed to "()".
3. "((?)" which can be transformed to "(())".
4. "(?))" which can be transformed to "(())".
For the second sample testcase, the pretty substrings of s are:
1. "??" which can be transformed to "()".
2. "()".
3. "??()" which can be transformed to "()()".
4. "?()?" which can be transformed to "(())".
5. "??" which can be transformed to "()".
6. "()??" which can be transformed to "()()".
7. "??()??" which can be transformed to "()()()". | instruction | 0 | 51,033 | 21 | 102,066 |
Tags: dp, greedy, implementation, math
Correct Solution:
```
s = input()
l = len(s)
ans = 0
for i in range(0, l):
m = n = 0
for j in range(i, l):
m += s[j] == '('
m -= s[j] == ')'
n += s[j] == '?'
if m + n < 0:
break
if m < n:
m, n = n, m
ans += m == n
print (ans)
``` | output | 1 | 51,033 | 21 | 102,067 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As Will is stuck in the Upside Down, he can still communicate with his mom, Joyce, through the Christmas lights (he can turn them on and off with his mind). He can't directly tell his mom where he is, because the monster that took him to the Upside Down will know and relocate him.
<image>
Thus, he came up with a puzzle to tell his mom his coordinates. His coordinates are the answer to the following problem.
A string consisting only of parentheses ('(' and ')') is called a bracket sequence. Some bracket sequence are called correct bracket sequences. More formally:
* Empty string is a correct bracket sequence.
* if s is a correct bracket sequence, then (s) is also a correct bracket sequence.
* if s and t are correct bracket sequences, then st (concatenation of s and t) is also a correct bracket sequence.
A string consisting of parentheses and question marks ('?') is called pretty if and only if there's a way to replace each question mark with either '(' or ')' such that the resulting string is a non-empty correct bracket sequence.
Will gave his mom a string s consisting of parentheses and question marks (using Morse code through the lights) and his coordinates are the number of pairs of integers (l, r) such that 1 ≤ l ≤ r ≤ |s| and the string slsl + 1... sr is pretty, where si is i-th character of s.
Joyce doesn't know anything about bracket sequences, so she asked for your help.
Input
The first and only line of input contains string s, consisting only of characters '(', ')' and '?' (2 ≤ |s| ≤ 5000).
Output
Print the answer to Will's puzzle in the first and only line of output.
Examples
Input
((?))
Output
4
Input
??()??
Output
7
Note
For the first sample testcase, the pretty substrings of s are:
1. "(?" which can be transformed to "()".
2. "?)" which can be transformed to "()".
3. "((?)" which can be transformed to "(())".
4. "(?))" which can be transformed to "(())".
For the second sample testcase, the pretty substrings of s are:
1. "??" which can be transformed to "()".
2. "()".
3. "??()" which can be transformed to "()()".
4. "?()?" which can be transformed to "(())".
5. "??" which can be transformed to "()".
6. "()??" which can be transformed to "()()".
7. "??()??" which can be transformed to "()()()". | instruction | 0 | 51,034 | 21 | 102,068 |
Tags: dp, greedy, implementation, math
Correct Solution:
```
s = input()
res, n = 0, len(s)
for i in range(n-1):
j, c, q = i , 0, 0
while j < n and c + q >= 0:
if(s[j] == '('): c += 1
elif(s[j] == ')'): c -= 1
else: q += 1
if(c < q):
c, q = q, c
res += (c == q)
j += 1
print(res)
``` | output | 1 | 51,034 | 21 | 102,069 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As Will is stuck in the Upside Down, he can still communicate with his mom, Joyce, through the Christmas lights (he can turn them on and off with his mind). He can't directly tell his mom where he is, because the monster that took him to the Upside Down will know and relocate him.
<image>
Thus, he came up with a puzzle to tell his mom his coordinates. His coordinates are the answer to the following problem.
A string consisting only of parentheses ('(' and ')') is called a bracket sequence. Some bracket sequence are called correct bracket sequences. More formally:
* Empty string is a correct bracket sequence.
* if s is a correct bracket sequence, then (s) is also a correct bracket sequence.
* if s and t are correct bracket sequences, then st (concatenation of s and t) is also a correct bracket sequence.
A string consisting of parentheses and question marks ('?') is called pretty if and only if there's a way to replace each question mark with either '(' or ')' such that the resulting string is a non-empty correct bracket sequence.
Will gave his mom a string s consisting of parentheses and question marks (using Morse code through the lights) and his coordinates are the number of pairs of integers (l, r) such that 1 ≤ l ≤ r ≤ |s| and the string slsl + 1... sr is pretty, where si is i-th character of s.
Joyce doesn't know anything about bracket sequences, so she asked for your help.
Input
The first and only line of input contains string s, consisting only of characters '(', ')' and '?' (2 ≤ |s| ≤ 5000).
Output
Print the answer to Will's puzzle in the first and only line of output.
Examples
Input
((?))
Output
4
Input
??()??
Output
7
Note
For the first sample testcase, the pretty substrings of s are:
1. "(?" which can be transformed to "()".
2. "?)" which can be transformed to "()".
3. "((?)" which can be transformed to "(())".
4. "(?))" which can be transformed to "(())".
For the second sample testcase, the pretty substrings of s are:
1. "??" which can be transformed to "()".
2. "()".
3. "??()" which can be transformed to "()()".
4. "?()?" which can be transformed to "(())".
5. "??" which can be transformed to "()".
6. "()??" which can be transformed to "()()".
7. "??()??" which can be transformed to "()()()". | instruction | 0 | 51,035 | 21 | 102,070 |
Tags: dp, greedy, implementation, math
Correct Solution:
```
s=input().strip()
n=len(s)
ques=ans=ini=0
for i in range(n):
ques=ini=0
for j in range(i,n):
c=s[j]
if(c=="?"):ques+=1
elif(c=='('):ini+=1
else:ini-=1
if(ini<0):break
if(ques>ini):
ques-=1;ini+=1;
if((j-i+1)%2==0 and ques>=ini):ans+=1
print(ans)
``` | output | 1 | 51,035 | 21 | 102,071 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As Will is stuck in the Upside Down, he can still communicate with his mom, Joyce, through the Christmas lights (he can turn them on and off with his mind). He can't directly tell his mom where he is, because the monster that took him to the Upside Down will know and relocate him.
<image>
Thus, he came up with a puzzle to tell his mom his coordinates. His coordinates are the answer to the following problem.
A string consisting only of parentheses ('(' and ')') is called a bracket sequence. Some bracket sequence are called correct bracket sequences. More formally:
* Empty string is a correct bracket sequence.
* if s is a correct bracket sequence, then (s) is also a correct bracket sequence.
* if s and t are correct bracket sequences, then st (concatenation of s and t) is also a correct bracket sequence.
A string consisting of parentheses and question marks ('?') is called pretty if and only if there's a way to replace each question mark with either '(' or ')' such that the resulting string is a non-empty correct bracket sequence.
Will gave his mom a string s consisting of parentheses and question marks (using Morse code through the lights) and his coordinates are the number of pairs of integers (l, r) such that 1 ≤ l ≤ r ≤ |s| and the string slsl + 1... sr is pretty, where si is i-th character of s.
Joyce doesn't know anything about bracket sequences, so she asked for your help.
Input
The first and only line of input contains string s, consisting only of characters '(', ')' and '?' (2 ≤ |s| ≤ 5000).
Output
Print the answer to Will's puzzle in the first and only line of output.
Examples
Input
((?))
Output
4
Input
??()??
Output
7
Note
For the first sample testcase, the pretty substrings of s are:
1. "(?" which can be transformed to "()".
2. "?)" which can be transformed to "()".
3. "((?)" which can be transformed to "(())".
4. "(?))" which can be transformed to "(())".
For the second sample testcase, the pretty substrings of s are:
1. "??" which can be transformed to "()".
2. "()".
3. "??()" which can be transformed to "()()".
4. "?()?" which can be transformed to "(())".
5. "??" which can be transformed to "()".
6. "()??" which can be transformed to "()()".
7. "??()??" which can be transformed to "()()()". | instruction | 0 | 51,036 | 21 | 102,072 |
Tags: dp, greedy, implementation, math
Correct Solution:
```
s = str(input())
n = len(s)
ans = 0
dp = [[0 for _ in range(n)] for _ in range(2)]
for i in range(n - 1):
if s[i] == ')':
continue
dp[0][i] = 1
dp[1][i] = 1
for j in range(i + 1, n):
if s[j] == '(':
dp[0][j] = dp[0][j - 1] + 1
dp[1][j] = dp[1][j - 1] + 1
elif s[j] == '?':
dp[0][j] = dp[0][j - 1] + 1
dp[1][j] = max(dp[1][j - 1] - 1, 0)
elif s[j] == ')' and 0 < dp[0][j - 1]:
dp[0][j] = dp[0][j - 1] - 1
dp[1][j] = max(dp[1][j - 1] - 1, 0)
else:
break
if (j - i) % 2 == 1 and dp[1][j] == 0:
ans += 1
print(ans)
``` | output | 1 | 51,036 | 21 | 102,073 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As Will is stuck in the Upside Down, he can still communicate with his mom, Joyce, through the Christmas lights (he can turn them on and off with his mind). He can't directly tell his mom where he is, because the monster that took him to the Upside Down will know and relocate him.
<image>
Thus, he came up with a puzzle to tell his mom his coordinates. His coordinates are the answer to the following problem.
A string consisting only of parentheses ('(' and ')') is called a bracket sequence. Some bracket sequence are called correct bracket sequences. More formally:
* Empty string is a correct bracket sequence.
* if s is a correct bracket sequence, then (s) is also a correct bracket sequence.
* if s and t are correct bracket sequences, then st (concatenation of s and t) is also a correct bracket sequence.
A string consisting of parentheses and question marks ('?') is called pretty if and only if there's a way to replace each question mark with either '(' or ')' such that the resulting string is a non-empty correct bracket sequence.
Will gave his mom a string s consisting of parentheses and question marks (using Morse code through the lights) and his coordinates are the number of pairs of integers (l, r) such that 1 ≤ l ≤ r ≤ |s| and the string slsl + 1... sr is pretty, where si is i-th character of s.
Joyce doesn't know anything about bracket sequences, so she asked for your help.
Input
The first and only line of input contains string s, consisting only of characters '(', ')' and '?' (2 ≤ |s| ≤ 5000).
Output
Print the answer to Will's puzzle in the first and only line of output.
Examples
Input
((?))
Output
4
Input
??()??
Output
7
Note
For the first sample testcase, the pretty substrings of s are:
1. "(?" which can be transformed to "()".
2. "?)" which can be transformed to "()".
3. "((?)" which can be transformed to "(())".
4. "(?))" which can be transformed to "(())".
For the second sample testcase, the pretty substrings of s are:
1. "??" which can be transformed to "()".
2. "()".
3. "??()" which can be transformed to "()()".
4. "?()?" which can be transformed to "(())".
5. "??" which can be transformed to "()".
6. "()??" which can be transformed to "()()".
7. "??()??" which can be transformed to "()()()". | instruction | 0 | 51,037 | 21 | 102,074 |
Tags: dp, greedy, implementation, math
Correct Solution:
```
s = input()
l = len(s)
ans = 0
for i in range(0, l):
ln = n = 0
for j in range(i, l):
if s[j] == '(':
ln += 1
elif s[j] == ')':
ln -= 1
else:
n += 1
ln -= 1
if ln == 0:
ans += 1
elif ln < 0:
if n > 0:
ln += 2
n -= 1
else:
break
print (ans)
``` | output | 1 | 51,037 | 21 | 102,075 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As Will is stuck in the Upside Down, he can still communicate with his mom, Joyce, through the Christmas lights (he can turn them on and off with his mind). He can't directly tell his mom where he is, because the monster that took him to the Upside Down will know and relocate him.
<image>
Thus, he came up with a puzzle to tell his mom his coordinates. His coordinates are the answer to the following problem.
A string consisting only of parentheses ('(' and ')') is called a bracket sequence. Some bracket sequence are called correct bracket sequences. More formally:
* Empty string is a correct bracket sequence.
* if s is a correct bracket sequence, then (s) is also a correct bracket sequence.
* if s and t are correct bracket sequences, then st (concatenation of s and t) is also a correct bracket sequence.
A string consisting of parentheses and question marks ('?') is called pretty if and only if there's a way to replace each question mark with either '(' or ')' such that the resulting string is a non-empty correct bracket sequence.
Will gave his mom a string s consisting of parentheses and question marks (using Morse code through the lights) and his coordinates are the number of pairs of integers (l, r) such that 1 ≤ l ≤ r ≤ |s| and the string slsl + 1... sr is pretty, where si is i-th character of s.
Joyce doesn't know anything about bracket sequences, so she asked for your help.
Input
The first and only line of input contains string s, consisting only of characters '(', ')' and '?' (2 ≤ |s| ≤ 5000).
Output
Print the answer to Will's puzzle in the first and only line of output.
Examples
Input
((?))
Output
4
Input
??()??
Output
7
Note
For the first sample testcase, the pretty substrings of s are:
1. "(?" which can be transformed to "()".
2. "?)" which can be transformed to "()".
3. "((?)" which can be transformed to "(())".
4. "(?))" which can be transformed to "(())".
For the second sample testcase, the pretty substrings of s are:
1. "??" which can be transformed to "()".
2. "()".
3. "??()" which can be transformed to "()()".
4. "?()?" which can be transformed to "(())".
5. "??" which can be transformed to "()".
6. "()??" which can be transformed to "()()".
7. "??()??" which can be transformed to "()()()". | instruction | 0 | 51,038 | 21 | 102,076 |
Tags: dp, greedy, implementation, math
Correct Solution:
```
import sys
li = lambda : [int(x) for x in sys.stdin.readline().strip().split()]
rw = lambda : sys.stdin.readline().strip().split()
ni = lambda : int(sys.stdin.readline().strip())
nsi = lambda : sys.stdin.readline().strip()
from collections import defaultdict as df
from math import *
s=nsi()
n=len(s)
ans=0
for i in range(n):
l=0
r=0
for j in range(i,n):
if(s[j]=='('):
l+=1
r+=1
elif(s[j]==')'):
l-=1
r-=1
else:
l-=1
r+=1
if(l<0):
l+=2
if(r<0):
break
if(l==0):
ans+=1
print(ans)
``` | output | 1 | 51,038 | 21 | 102,077 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As Will is stuck in the Upside Down, he can still communicate with his mom, Joyce, through the Christmas lights (he can turn them on and off with his mind). He can't directly tell his mom where he is, because the monster that took him to the Upside Down will know and relocate him.
<image>
Thus, he came up with a puzzle to tell his mom his coordinates. His coordinates are the answer to the following problem.
A string consisting only of parentheses ('(' and ')') is called a bracket sequence. Some bracket sequence are called correct bracket sequences. More formally:
* Empty string is a correct bracket sequence.
* if s is a correct bracket sequence, then (s) is also a correct bracket sequence.
* if s and t are correct bracket sequences, then st (concatenation of s and t) is also a correct bracket sequence.
A string consisting of parentheses and question marks ('?') is called pretty if and only if there's a way to replace each question mark with either '(' or ')' such that the resulting string is a non-empty correct bracket sequence.
Will gave his mom a string s consisting of parentheses and question marks (using Morse code through the lights) and his coordinates are the number of pairs of integers (l, r) such that 1 ≤ l ≤ r ≤ |s| and the string slsl + 1... sr is pretty, where si is i-th character of s.
Joyce doesn't know anything about bracket sequences, so she asked for your help.
Input
The first and only line of input contains string s, consisting only of characters '(', ')' and '?' (2 ≤ |s| ≤ 5000).
Output
Print the answer to Will's puzzle in the first and only line of output.
Examples
Input
((?))
Output
4
Input
??()??
Output
7
Note
For the first sample testcase, the pretty substrings of s are:
1. "(?" which can be transformed to "()".
2. "?)" which can be transformed to "()".
3. "((?)" which can be transformed to "(())".
4. "(?))" which can be transformed to "(())".
For the second sample testcase, the pretty substrings of s are:
1. "??" which can be transformed to "()".
2. "()".
3. "??()" which can be transformed to "()()".
4. "?()?" which can be transformed to "(())".
5. "??" which can be transformed to "()".
6. "()??" which can be transformed to "()()".
7. "??()??" which can be transformed to "()()()". | instruction | 0 | 51,039 | 21 | 102,078 |
Tags: dp, greedy, implementation, math
Correct Solution:
```
s = str(input())
n = len(s)
ans = 0
for l in range(n-1):
cnt = 0
qu = 0
chk = 0
if s[l] == '(':
cnt += 1
chk += 1
elif s[l] == '?':
qu += 1
chk = max(chk-1, 0)
else:
continue
for r in range(l+1, n):
if s[r] == '(':
cnt += 1
chk += 1
elif s[r] == '?':
qu += 1
chk = max(chk-1, 0)
else:
cnt -= 1
chk = max(chk-1, 0)
if cnt+qu < 0:
break
else:
if (qu-cnt)%2 == 0 and qu-cnt >= 0 and (qu-cnt)//2 <= qu and chk <= 0:
#print(l, r)
ans += 1
print(ans)
``` | output | 1 | 51,039 | 21 | 102,079 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As Will is stuck in the Upside Down, he can still communicate with his mom, Joyce, through the Christmas lights (he can turn them on and off with his mind). He can't directly tell his mom where he is, because the monster that took him to the Upside Down will know and relocate him.
<image>
Thus, he came up with a puzzle to tell his mom his coordinates. His coordinates are the answer to the following problem.
A string consisting only of parentheses ('(' and ')') is called a bracket sequence. Some bracket sequence are called correct bracket sequences. More formally:
* Empty string is a correct bracket sequence.
* if s is a correct bracket sequence, then (s) is also a correct bracket sequence.
* if s and t are correct bracket sequences, then st (concatenation of s and t) is also a correct bracket sequence.
A string consisting of parentheses and question marks ('?') is called pretty if and only if there's a way to replace each question mark with either '(' or ')' such that the resulting string is a non-empty correct bracket sequence.
Will gave his mom a string s consisting of parentheses and question marks (using Morse code through the lights) and his coordinates are the number of pairs of integers (l, r) such that 1 ≤ l ≤ r ≤ |s| and the string slsl + 1... sr is pretty, where si is i-th character of s.
Joyce doesn't know anything about bracket sequences, so she asked for your help.
Input
The first and only line of input contains string s, consisting only of characters '(', ')' and '?' (2 ≤ |s| ≤ 5000).
Output
Print the answer to Will's puzzle in the first and only line of output.
Examples
Input
((?))
Output
4
Input
??()??
Output
7
Note
For the first sample testcase, the pretty substrings of s are:
1. "(?" which can be transformed to "()".
2. "?)" which can be transformed to "()".
3. "((?)" which can be transformed to "(())".
4. "(?))" which can be transformed to "(())".
For the second sample testcase, the pretty substrings of s are:
1. "??" which can be transformed to "()".
2. "()".
3. "??()" which can be transformed to "()()".
4. "?()?" which can be transformed to "(())".
5. "??" which can be transformed to "()".
6. "()??" which can be transformed to "()()".
7. "??()??" which can be transformed to "()()()".
Submitted Solution:
```
s=input().strip();n=len(s)
ques=ans=ini=0
for i in range(n):
ques=ini=0
for j in range(i,n):
c=s[j]
if(c=="?"):ques+=1
elif(c=='('):ini+=1
else:ini-=1 #present situation dekh li
if(ini<0):break #invalid cheez h bhaaya
if(ques>ini): #agr mujhe wo question mark opening se replace krna jo ki krna hi h
ques-=1;ini+=1; #ye kr denge hum fir
if((j-i+1)%2==0 and ques==ini):ans+=1 #aur ques fir ini ke barabar hi hone chahiye kyu ki hum baa
print(ans) #baar baar quest mark ko ini jo ki >o h se replace kr rhe h
``` | instruction | 0 | 51,040 | 21 | 102,080 |
Yes | output | 1 | 51,040 | 21 | 102,081 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As Will is stuck in the Upside Down, he can still communicate with his mom, Joyce, through the Christmas lights (he can turn them on and off with his mind). He can't directly tell his mom where he is, because the monster that took him to the Upside Down will know and relocate him.
<image>
Thus, he came up with a puzzle to tell his mom his coordinates. His coordinates are the answer to the following problem.
A string consisting only of parentheses ('(' and ')') is called a bracket sequence. Some bracket sequence are called correct bracket sequences. More formally:
* Empty string is a correct bracket sequence.
* if s is a correct bracket sequence, then (s) is also a correct bracket sequence.
* if s and t are correct bracket sequences, then st (concatenation of s and t) is also a correct bracket sequence.
A string consisting of parentheses and question marks ('?') is called pretty if and only if there's a way to replace each question mark with either '(' or ')' such that the resulting string is a non-empty correct bracket sequence.
Will gave his mom a string s consisting of parentheses and question marks (using Morse code through the lights) and his coordinates are the number of pairs of integers (l, r) such that 1 ≤ l ≤ r ≤ |s| and the string slsl + 1... sr is pretty, where si is i-th character of s.
Joyce doesn't know anything about bracket sequences, so she asked for your help.
Input
The first and only line of input contains string s, consisting only of characters '(', ')' and '?' (2 ≤ |s| ≤ 5000).
Output
Print the answer to Will's puzzle in the first and only line of output.
Examples
Input
((?))
Output
4
Input
??()??
Output
7
Note
For the first sample testcase, the pretty substrings of s are:
1. "(?" which can be transformed to "()".
2. "?)" which can be transformed to "()".
3. "((?)" which can be transformed to "(())".
4. "(?))" which can be transformed to "(())".
For the second sample testcase, the pretty substrings of s are:
1. "??" which can be transformed to "()".
2. "()".
3. "??()" which can be transformed to "()()".
4. "?()?" which can be transformed to "(())".
5. "??" which can be transformed to "()".
6. "()??" which can be transformed to "()()".
7. "??()??" which can be transformed to "()()()".
Submitted Solution:
```
s=input().strip();n=len(s)
ques=ans=ini=0
for i in range(n):
ques=ini=0
for j in range(i,n):
c=s[j]
if(c=="?"):ques+=1
elif(c=='('):ini+=1
else:ini-=1
if(ini<0):break
if(ques>ini):
ques-=1;ini+=1;
if((j-i+1)%2==0 and ques==ini):ans+=1
print(ans)
``` | instruction | 0 | 51,041 | 21 | 102,082 |
Yes | output | 1 | 51,041 | 21 | 102,083 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As Will is stuck in the Upside Down, he can still communicate with his mom, Joyce, through the Christmas lights (he can turn them on and off with his mind). He can't directly tell his mom where he is, because the monster that took him to the Upside Down will know and relocate him.
<image>
Thus, he came up with a puzzle to tell his mom his coordinates. His coordinates are the answer to the following problem.
A string consisting only of parentheses ('(' and ')') is called a bracket sequence. Some bracket sequence are called correct bracket sequences. More formally:
* Empty string is a correct bracket sequence.
* if s is a correct bracket sequence, then (s) is also a correct bracket sequence.
* if s and t are correct bracket sequences, then st (concatenation of s and t) is also a correct bracket sequence.
A string consisting of parentheses and question marks ('?') is called pretty if and only if there's a way to replace each question mark with either '(' or ')' such that the resulting string is a non-empty correct bracket sequence.
Will gave his mom a string s consisting of parentheses and question marks (using Morse code through the lights) and his coordinates are the number of pairs of integers (l, r) such that 1 ≤ l ≤ r ≤ |s| and the string slsl + 1... sr is pretty, where si is i-th character of s.
Joyce doesn't know anything about bracket sequences, so she asked for your help.
Input
The first and only line of input contains string s, consisting only of characters '(', ')' and '?' (2 ≤ |s| ≤ 5000).
Output
Print the answer to Will's puzzle in the first and only line of output.
Examples
Input
((?))
Output
4
Input
??()??
Output
7
Note
For the first sample testcase, the pretty substrings of s are:
1. "(?" which can be transformed to "()".
2. "?)" which can be transformed to "()".
3. "((?)" which can be transformed to "(())".
4. "(?))" which can be transformed to "(())".
For the second sample testcase, the pretty substrings of s are:
1. "??" which can be transformed to "()".
2. "()".
3. "??()" which can be transformed to "()()".
4. "?()?" which can be transformed to "(())".
5. "??" which can be transformed to "()".
6. "()??" which can be transformed to "()()".
7. "??()??" which can be transformed to "()()()".
Submitted Solution:
```
S = input()
n = len(S)
Ans = 0
for l in range(0, n):
bs = 0
det = 0
for r in range(l, n):
#print(S[r])
if (S[r] == '?'):
if (bs > 0):
det = det + 1
bs = bs - 1
else:
bs = bs + 1
elif (S[r] == '('):
bs = bs + 1
else:
if (bs == 0 and det > 0):
bs = bs + 2
det = det - 1
bs = bs - 1
if (bs < 0):
break
if (bs > 0):
continue
if (bs % 2 == 0):
Ans = Ans + 1
#print(bs, det, S[l:r])
print(Ans)
``` | instruction | 0 | 51,042 | 21 | 102,084 |
Yes | output | 1 | 51,042 | 21 | 102,085 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a string $S$, which is balanced parentheses with a star symbol '*' inserted.
Any balanced parentheses can be constructed using the following rules:
* An empty string is balanced.
* Concatenation of two balanced parentheses is balanced.
* If $T$ is balanced parentheses, concatenation of '(', $T$, and ')' in this order is balanced.
For example, '()()' and '(()())' are balanced parentheses. ')(' and ')()(()' are not balanced parentheses.
Your task is to count how many matching pairs of parentheses surround the star.
Let $S_i$be the $i$-th character of a string $S$. The pair of $S_l$ and $S_r$ ($l < r$) is called a matching pair of parentheses if $S_l$ is '(', $S_r$ is ')' and the surrounded string by them is balanced when ignoring a star symbol.
Input
The input consists of a single test case formatted as follows.
$S$
$S$ is balanced parentheses with exactly one '*' inserted somewhere. The length of $S$ is between 1 and 100, inclusive.
Output
Print the answer in one line.
Examples
Input
((*)())
Output
2
Input
(*)
Output
1
Input
(()())*
Output
0
Input
()*()
Output
0
Input
((((((((((*))))))))))
Output
10
Input
*
Output
0 | instruction | 0 | 52,152 | 21 | 104,304 |
"Correct Solution:
```
line = input()
lebel = 0
for i in range(len(line)):
if line[i] == '*':
print(lebel)
break
if line[i] == '(':
lebel += 1
elif line[i] == ')':
lebel -= 1
``` | output | 1 | 52,152 | 21 | 104,305 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As Famil Door’s birthday is coming, some of his friends (like Gabi) decided to buy a present for him. His friends are going to buy a string consisted of round brackets since Famil Door loves string of brackets of length n more than any other strings!
The sequence of round brackets is called valid if and only if:
1. the total number of opening brackets is equal to the total number of closing brackets;
2. for any prefix of the sequence, the number of opening brackets is greater or equal than the number of closing brackets.
Gabi bought a string s of length m (m ≤ n) and want to complete it to obtain a valid sequence of brackets of length n. He is going to pick some strings p and q consisting of round brackets and merge them in a string p + s + q, that is add the string p at the beginning of the string s and string q at the end of the string s.
Now he wonders, how many pairs of strings p and q exists, such that the string p + s + q is a valid sequence of round brackets. As this number may be pretty large, he wants to calculate it modulo 109 + 7.
Input
First line contains n and m (1 ≤ m ≤ n ≤ 100 000, n - m ≤ 2000) — the desired length of the string and the length of the string bought by Gabi, respectively.
The second line contains string s of length m consisting of characters '(' and ')' only.
Output
Print the number of pairs of string p and q such that p + s + q is a valid sequence of round brackets modulo 109 + 7.
Examples
Input
4 1
(
Output
4
Input
4 4
(())
Output
1
Input
4 3
(((
Output
0
Note
In the first sample there are four different valid pairs:
1. p = "(", q = "))"
2. p = "()", q = ")"
3. p = "", q = "())"
4. p = "", q = ")()"
In the second sample the only way to obtain a desired string is choose empty p and q.
In the third sample there is no way to get a valid sequence of brackets. | instruction | 0 | 52,627 | 21 | 105,254 |
Tags: dp, strings
Correct Solution:
```
def ij(s):
i = 0
j = 0
for c in s:
if c == ')':
if j > 0:
j -= 1
else:
i += 1
else:
j += 1
return i, j
K = 10**9 + 7
def ways(n, s):
I, J = ij(s)
f = n - len(s) - I - J
if f < 0 or f%2:
return 0
E = f//2
if E == 0:
return 1
C = [1]
for n in range(E+max(I,J)+1):
C.append(C[n] * 2 * (2*n + 1) // (n+2))
W = [[c % K for c in C]]
W.append(W[0][1:])
for i in range(2, E+max(I,J)+1):
W.append([(W[i-1][e+1]-W[i-2][e+1]) % K
for e in range(E+max(I,J)+1-i+1)])
result = sum((W[I+k][e-k] * W[J+k][E-e]) % K
for k in range(E+1)
for e in range(k, E+1))
return result
if __name__ == '__main__':
n, m = map(int, input().split())
s = input()
print(ways(n, s) % K)
``` | output | 1 | 52,627 | 21 | 105,255 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As Famil Door’s birthday is coming, some of his friends (like Gabi) decided to buy a present for him. His friends are going to buy a string consisted of round brackets since Famil Door loves string of brackets of length n more than any other strings!
The sequence of round brackets is called valid if and only if:
1. the total number of opening brackets is equal to the total number of closing brackets;
2. for any prefix of the sequence, the number of opening brackets is greater or equal than the number of closing brackets.
Gabi bought a string s of length m (m ≤ n) and want to complete it to obtain a valid sequence of brackets of length n. He is going to pick some strings p and q consisting of round brackets and merge them in a string p + s + q, that is add the string p at the beginning of the string s and string q at the end of the string s.
Now he wonders, how many pairs of strings p and q exists, such that the string p + s + q is a valid sequence of round brackets. As this number may be pretty large, he wants to calculate it modulo 109 + 7.
Input
First line contains n and m (1 ≤ m ≤ n ≤ 100 000, n - m ≤ 2000) — the desired length of the string and the length of the string bought by Gabi, respectively.
The second line contains string s of length m consisting of characters '(' and ')' only.
Output
Print the number of pairs of string p and q such that p + s + q is a valid sequence of round brackets modulo 109 + 7.
Examples
Input
4 1
(
Output
4
Input
4 4
(())
Output
1
Input
4 3
(((
Output
0
Note
In the first sample there are four different valid pairs:
1. p = "(", q = "))"
2. p = "()", q = ")"
3. p = "", q = "())"
4. p = "", q = ")()"
In the second sample the only way to obtain a desired string is choose empty p and q.
In the third sample there is no way to get a valid sequence of brackets. | instruction | 0 | 52,628 | 21 | 105,256 |
Tags: dp, strings
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
# import string
# characters = string.ascii_lowercase
# digits = string.digits
# sys.setrecursionlimit(int(1e6))
# dir = [-1,0,1,0,-1]
# moves = 'NESW'
inf = float('inf')
from functools import cmp_to_key
from collections import defaultdict as dd
from collections import Counter, deque
from heapq import *
import math
from math import floor, ceil, sqrt
def geti(): return map(int, input().strip().split())
def getl(): return list(map(int, input().strip().split()))
def getis(): return map(str, input().strip().split())
def getls(): return list(map(str, input().strip().split()))
def gets(): return input().strip()
def geta(): return int(input())
def print_s(s): stdout.write(s+'\n')
def solve():
n, m = geti()
diff = n - m
dp = [[0] * (diff+1) for _ in range(diff+1)]
dp[0][0] = 1
mod = int(1e9+7)
for i in range(diff):
for j in range(diff):
dp[i+1][j+1] += dp[i][j]
dp[i+1][j+1] %= mod
if j:
dp[i+1][j-1] += dp[i][j]
dp[i+1][j-1] %= mod
ans = 0
s = gets()
def calc(s, turn):
balance = 0
left = 0
for i in s:
if i == turn:
balance += 1
else:
balance -= 1
left = min(left, balance)
return abs(left)
left = calc(s, '(')
right = calc(s[::-1], ')')
# print(left, right)
for i in range(diff+1):
for l in range(left, diff+1):
r = l - left + right
if r > diff:
break
ans += dp[i][l] * dp[diff - i][r]
ans %= mod
print(ans)
# Fast IO region
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")
if __name__=='__main__':
solve()
``` | output | 1 | 52,628 | 21 | 105,257 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As Famil Door’s birthday is coming, some of his friends (like Gabi) decided to buy a present for him. His friends are going to buy a string consisted of round brackets since Famil Door loves string of brackets of length n more than any other strings!
The sequence of round brackets is called valid if and only if:
1. the total number of opening brackets is equal to the total number of closing brackets;
2. for any prefix of the sequence, the number of opening brackets is greater or equal than the number of closing brackets.
Gabi bought a string s of length m (m ≤ n) and want to complete it to obtain a valid sequence of brackets of length n. He is going to pick some strings p and q consisting of round brackets and merge them in a string p + s + q, that is add the string p at the beginning of the string s and string q at the end of the string s.
Now he wonders, how many pairs of strings p and q exists, such that the string p + s + q is a valid sequence of round brackets. As this number may be pretty large, he wants to calculate it modulo 109 + 7.
Input
First line contains n and m (1 ≤ m ≤ n ≤ 100 000, n - m ≤ 2000) — the desired length of the string and the length of the string bought by Gabi, respectively.
The second line contains string s of length m consisting of characters '(' and ')' only.
Output
Print the number of pairs of string p and q such that p + s + q is a valid sequence of round brackets modulo 109 + 7.
Examples
Input
4 1
(
Output
4
Input
4 4
(())
Output
1
Input
4 3
(((
Output
0
Note
In the first sample there are four different valid pairs:
1. p = "(", q = "))"
2. p = "()", q = ")"
3. p = "", q = "())"
4. p = "", q = ")()"
In the second sample the only way to obtain a desired string is choose empty p and q.
In the third sample there is no way to get a valid sequence of brackets. | instruction | 0 | 52,629 | 21 | 105,258 |
Tags: dp, strings
Correct Solution:
```
n, m = map(int, input().split())
s = input()
mod = 10 ** 9 + 7
c = b = 0
for x in s:
c += (x == '(') * 2 - 1
b = min(c, b)
d = [[1]]
for i in range(n - m):
nd = d[-1][1:] + [0] * 2
for j in range(1, i + 2):
nd[j] = (nd[j] + d[-1][j-1]) % mod
d.append(nd)
ans = 0
for i in range(n - m + 1):
l = n - m - i
for j in range(-b, min(l - c, i) + 1):
ans = (ans + d[i][j] * d[l][j + c]) % mod
print(ans)
``` | output | 1 | 52,629 | 21 | 105,259 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As Famil Door’s birthday is coming, some of his friends (like Gabi) decided to buy a present for him. His friends are going to buy a string consisted of round brackets since Famil Door loves string of brackets of length n more than any other strings!
The sequence of round brackets is called valid if and only if:
1. the total number of opening brackets is equal to the total number of closing brackets;
2. for any prefix of the sequence, the number of opening brackets is greater or equal than the number of closing brackets.
Gabi bought a string s of length m (m ≤ n) and want to complete it to obtain a valid sequence of brackets of length n. He is going to pick some strings p and q consisting of round brackets and merge them in a string p + s + q, that is add the string p at the beginning of the string s and string q at the end of the string s.
Now he wonders, how many pairs of strings p and q exists, such that the string p + s + q is a valid sequence of round brackets. As this number may be pretty large, he wants to calculate it modulo 109 + 7.
Input
First line contains n and m (1 ≤ m ≤ n ≤ 100 000, n - m ≤ 2000) — the desired length of the string and the length of the string bought by Gabi, respectively.
The second line contains string s of length m consisting of characters '(' and ')' only.
Output
Print the number of pairs of string p and q such that p + s + q is a valid sequence of round brackets modulo 109 + 7.
Examples
Input
4 1
(
Output
4
Input
4 4
(())
Output
1
Input
4 3
(((
Output
0
Note
In the first sample there are four different valid pairs:
1. p = "(", q = "))"
2. p = "()", q = ")"
3. p = "", q = "())"
4. p = "", q = ")()"
In the second sample the only way to obtain a desired string is choose empty p and q.
In the third sample there is no way to get a valid sequence of brackets. | instruction | 0 | 52,630 | 21 | 105,260 |
Tags: dp, strings
Correct Solution:
```
from sys import stdin
n,m=map(int, stdin.readline().strip().split())
s=input()
dp=[[0 for i in range(2004)] for j in range(2004)]
mod=10**9+7
d=n-m
ans=0
b=0
d1=0
for i in s:
if i=='(':
b+=1
else:
b-=1
if b<d1:
d1=b
d1=-d1
dp[0][0]=1
d+=1
try:
for i in range(1,d):
dp[i][0]=(dp[i-1][1]%mod)
for j in range(1,d):
dp[i][j]=(dp[i-1][j-1]%mod+dp[i-1][j+1]%mod)%mod
for i in range(d):
x=n-(i+m)
if x<0:
continue
for j in range(d1,d):
if b+j>2000:
break
ans=((ans%mod)+(dp[i][j]%mod*dp[x][b+j]%mod)%mod)%mod
print(ans)
except:
print(ans,i,j,x,b+j)
``` | output | 1 | 52,630 | 21 | 105,261 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As Famil Door’s birthday is coming, some of his friends (like Gabi) decided to buy a present for him. His friends are going to buy a string consisted of round brackets since Famil Door loves string of brackets of length n more than any other strings!
The sequence of round brackets is called valid if and only if:
1. the total number of opening brackets is equal to the total number of closing brackets;
2. for any prefix of the sequence, the number of opening brackets is greater or equal than the number of closing brackets.
Gabi bought a string s of length m (m ≤ n) and want to complete it to obtain a valid sequence of brackets of length n. He is going to pick some strings p and q consisting of round brackets and merge them in a string p + s + q, that is add the string p at the beginning of the string s and string q at the end of the string s.
Now he wonders, how many pairs of strings p and q exists, such that the string p + s + q is a valid sequence of round brackets. As this number may be pretty large, he wants to calculate it modulo 109 + 7.
Input
First line contains n and m (1 ≤ m ≤ n ≤ 100 000, n - m ≤ 2000) — the desired length of the string and the length of the string bought by Gabi, respectively.
The second line contains string s of length m consisting of characters '(' and ')' only.
Output
Print the number of pairs of string p and q such that p + s + q is a valid sequence of round brackets modulo 109 + 7.
Examples
Input
4 1
(
Output
4
Input
4 4
(())
Output
1
Input
4 3
(((
Output
0
Note
In the first sample there are four different valid pairs:
1. p = "(", q = "))"
2. p = "()", q = ")"
3. p = "", q = "())"
4. p = "", q = ")()"
In the second sample the only way to obtain a desired string is choose empty p and q.
In the third sample there is no way to get a valid sequence of brackets. | instruction | 0 | 52,631 | 21 | 105,262 |
Tags: dp, strings
Correct Solution:
```
n, m = map(int, input().split())
s = input()
mod = 10 ** 9 + 7
c, b, ans, d, k = 0, 0, 0, [[1]], n - m
for i in s:
c += (i == '(') * 2 - 1
b = min(c, b)
for i in range(n - m):
nd = d[-1][1:] + [0] * 2
for j in range(1, i + 2):
nd[j] = (nd[j] + d[-1][j - 1]) % mod
d.append(nd)
for i in range(k + 1):
for j in range(-b, min(k - i - c, i) + 1):
ans = (ans + d[i][j] * d[k - i][j + c]) % mod
print(ans)
``` | output | 1 | 52,631 | 21 | 105,263 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As Famil Door’s birthday is coming, some of his friends (like Gabi) decided to buy a present for him. His friends are going to buy a string consisted of round brackets since Famil Door loves string of brackets of length n more than any other strings!
The sequence of round brackets is called valid if and only if:
1. the total number of opening brackets is equal to the total number of closing brackets;
2. for any prefix of the sequence, the number of opening brackets is greater or equal than the number of closing brackets.
Gabi bought a string s of length m (m ≤ n) and want to complete it to obtain a valid sequence of brackets of length n. He is going to pick some strings p and q consisting of round brackets and merge them in a string p + s + q, that is add the string p at the beginning of the string s and string q at the end of the string s.
Now he wonders, how many pairs of strings p and q exists, such that the string p + s + q is a valid sequence of round brackets. As this number may be pretty large, he wants to calculate it modulo 109 + 7.
Input
First line contains n and m (1 ≤ m ≤ n ≤ 100 000, n - m ≤ 2000) — the desired length of the string and the length of the string bought by Gabi, respectively.
The second line contains string s of length m consisting of characters '(' and ')' only.
Output
Print the number of pairs of string p and q such that p + s + q is a valid sequence of round brackets modulo 109 + 7.
Examples
Input
4 1
(
Output
4
Input
4 4
(())
Output
1
Input
4 3
(((
Output
0
Note
In the first sample there are four different valid pairs:
1. p = "(", q = "))"
2. p = "()", q = ")"
3. p = "", q = "())"
4. p = "", q = ")()"
In the second sample the only way to obtain a desired string is choose empty p and q.
In the third sample there is no way to get a valid sequence of brackets.
Submitted Solution:
```
import sys
from functools import lru_cache
sys.setrecursionlimit(10000)
mn = input()
#print(mn)
m,n = [int(i) for i in mn.split()]
my_string = input()
leftfill = 0
c = 0
const = 1000000007
for char in my_string:
if char == ')':
c-=1
elif char == '(':
c+=1
leftfill = min(leftfill,c)
rightfill = max(c,0)
leftfill *= -1
@lru_cache(maxsize=None)
def leftdp(a,b):
#print(a,b)
#print('...')
#input('---')
if b==0:
return 1
if a==b:
# print(a,b,1,'...')
return 1
if a == 0:
ccc = leftdp(1,b-1) % const
# print(a,b,ccc,'...')
return ccc
if a >= 0 and b >= 0 and (a + b) % 2 == 0:
ccc = (leftdp(a-1,b-1) + leftdp(a+1,b-1)) % const
# print(a,b,ccc,'...')
return ccc
#print(a,b,0,'...')
return 0
@lru_cache(maxsize=None)
def rightdp(a,b):
# print(a,b)
# print('---')
#input('***')
if b==0:
return 1
if a==b:
# print(a,b,1,'***')
return 1
if a==0:
ccc = rightdp(1,b-1) % const
# print(a,b,ccc,'***')
return ccc
if a >= 0 and b >= 0 and (a + b) % 2 == 0:
ccc = (rightdp(a-1,b-1) + rightdp(a,b-2)) % const
# print(a,b,ccc,'***')
return ccc
#print(a,b,0,'***')
return 0
diff = m-n-leftfill - rightfill
sum_ = 0
for i in range(diff + 1):
ddd = min(diff-i,i)
for j in range(ddd + 1):
k1 = leftdp(leftfill + j,leftfill + i)
k2 = rightdp(rightfill + j, rightfill + diff - i)
#k1 = max(k1,1)
#k2 = max(k2,1)
sum_ += k1 * k2
print(sum_)
``` | instruction | 0 | 52,632 | 21 | 105,264 |
No | output | 1 | 52,632 | 21 | 105,265 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As Famil Door’s birthday is coming, some of his friends (like Gabi) decided to buy a present for him. His friends are going to buy a string consisted of round brackets since Famil Door loves string of brackets of length n more than any other strings!
The sequence of round brackets is called valid if and only if:
1. the total number of opening brackets is equal to the total number of closing brackets;
2. for any prefix of the sequence, the number of opening brackets is greater or equal than the number of closing brackets.
Gabi bought a string s of length m (m ≤ n) and want to complete it to obtain a valid sequence of brackets of length n. He is going to pick some strings p and q consisting of round brackets and merge them in a string p + s + q, that is add the string p at the beginning of the string s and string q at the end of the string s.
Now he wonders, how many pairs of strings p and q exists, such that the string p + s + q is a valid sequence of round brackets. As this number may be pretty large, he wants to calculate it modulo 109 + 7.
Input
First line contains n and m (1 ≤ m ≤ n ≤ 100 000, n - m ≤ 2000) — the desired length of the string and the length of the string bought by Gabi, respectively.
The second line contains string s of length m consisting of characters '(' and ')' only.
Output
Print the number of pairs of string p and q such that p + s + q is a valid sequence of round brackets modulo 109 + 7.
Examples
Input
4 1
(
Output
4
Input
4 4
(())
Output
1
Input
4 3
(((
Output
0
Note
In the first sample there are four different valid pairs:
1. p = "(", q = "))"
2. p = "()", q = ")"
3. p = "", q = "())"
4. p = "", q = ")()"
In the second sample the only way to obtain a desired string is choose empty p and q.
In the third sample there is no way to get a valid sequence of brackets.
Submitted Solution:
```
import sys
from functools import lru_cache
sys.setrecursionlimit(10000)
mn = input()
#print(mn)
m,n = [int(i) for i in mn.split()]
my_string = input()
leftfill = 0
c = 0
const = 1000000007
for char in my_string:
if char == ')':
c-=1
elif char == '(':
c+=1
leftfill = min(leftfill,c)
rightfill = max(c,0)
leftfill *= -1
#@lru_cache(maxsize=None)
#def gendp(nsteps,balance,maxdrawdown):
# if balance < maxdrawdown:
# return 0
# if abs(balance) == nsteps:
# return 1
# if nsteps == 0:
# return 0
@lru_cache(maxsize=None)
def leftdp(a,b):
#print(a,b)
#print('...')
#oinput('---')
if b==0:
return 0
if a==b:
# print(a,b,1,'...')
return 1
if a == 0:
ccc = leftdp(1,b-1) % const
# print(a,b,ccc,'...')
return ccc
if a >= 0 and b >= 0 and (a + b) % 2 == 0:
ccc = (leftdp(a-1,b-1) + leftdp(a+1,b-1)) % const
print(a,b,ccc,'...')
return ccc
#print(a,b,0,'...')
return 0
@lru_cache(maxsize=None)
def rightdp(a,b):
# print(a,b)
# print('---')
#input('***')
if b== 0:
return 0
if a==b:
# print(a,b,1,'***')
return 1
if a==0:
ccc = rightdp(1,b-1) % const
# print(a,b,ccc,'***')
return ccc
if a >= 0 and b >= 0 and (a + b) % 2 == 0:
ccc = (rightdp(a-1,b-1) + rightdp(a,b-2)) % const
print(a,b,ccc,'***')
return ccc
#print(a,b,0,'***')
return 0
diff = m-n-leftfill - rightfill
sum_ = 0
for i in range(diff + 1):
ddd = min(diff-i,i)
for j in range(ddd + 1):
k1p1 = leftfill + j
k1p2 = leftfill + i
k2p1 = rightfill + j
k2p2 = rightfill + diff - i
if k1p1 ==0 and k1p2 ==0:
k1 = 1
else:
k1 = leftdp(leftfill + j,leftfill + i)
if k2p1 == 0 and k2p2 == 0:
k2 == 1
else:
k2 = rightdp(rightfill + j, rightfill + diff - i)
#k1 = max(k1,1)
#k2 = max(k2,1)
sum_ += k1 * k2 % const
sum_ = sum_ % const
print(sum_)
``` | instruction | 0 | 52,633 | 21 | 105,266 |
No | output | 1 | 52,633 | 21 | 105,267 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As Famil Door’s birthday is coming, some of his friends (like Gabi) decided to buy a present for him. His friends are going to buy a string consisted of round brackets since Famil Door loves string of brackets of length n more than any other strings!
The sequence of round brackets is called valid if and only if:
1. the total number of opening brackets is equal to the total number of closing brackets;
2. for any prefix of the sequence, the number of opening brackets is greater or equal than the number of closing brackets.
Gabi bought a string s of length m (m ≤ n) and want to complete it to obtain a valid sequence of brackets of length n. He is going to pick some strings p and q consisting of round brackets and merge them in a string p + s + q, that is add the string p at the beginning of the string s and string q at the end of the string s.
Now he wonders, how many pairs of strings p and q exists, such that the string p + s + q is a valid sequence of round brackets. As this number may be pretty large, he wants to calculate it modulo 109 + 7.
Input
First line contains n and m (1 ≤ m ≤ n ≤ 100 000, n - m ≤ 2000) — the desired length of the string and the length of the string bought by Gabi, respectively.
The second line contains string s of length m consisting of characters '(' and ')' only.
Output
Print the number of pairs of string p and q such that p + s + q is a valid sequence of round brackets modulo 109 + 7.
Examples
Input
4 1
(
Output
4
Input
4 4
(())
Output
1
Input
4 3
(((
Output
0
Note
In the first sample there are four different valid pairs:
1. p = "(", q = "))"
2. p = "()", q = ")"
3. p = "", q = "())"
4. p = "", q = ")()"
In the second sample the only way to obtain a desired string is choose empty p and q.
In the third sample there is no way to get a valid sequence of brackets.
Submitted Solution:
```
def mp(): return map(int,input().split())
def lt(): return list(map(int,input().split()))
def pt(x): print(x)
def ip(): return input()
def it(): return int(input())
def sl(x): return [t for t in x]
def spl(x): return x.split()
def aj(liste, item): liste.append(item)
def bin(x): return "{0:b}".format(x)
def listring(l): return ' '.join([str(x) for x in l])
def printlist(l): print(' '.join([str(x) for x in l]))
n,m = mp()
s = ip()
dp = [[0 for j in range(n-m+1)]for i in range(n-m+1)]
dp[0][0] = 1
for i in range(1,n-m+1):
for j in range(n-m+1):
if j + 1 <= n-m and i - 1 >= 0:
dp[i][j] = dp[i-1][j+1] + dp[i-1][j-1]
elif i - 1 >= 0:
dp[i][j] = dp[i-1][j-1]
mini = 0
a = 0
for i in s:
if i == "(":
a += 1
else:
a -= 1
mini = min(mini,a)
result = 0
for c in range(n-m+1):
for d in range(n-m+1):
if -mini <= d:
result += dp[c][d]*(a+d <= n-m and dp[n-m-c][a+d])
print(result)
``` | instruction | 0 | 52,634 | 21 | 105,268 |
No | output | 1 | 52,634 | 21 | 105,269 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As Famil Door’s birthday is coming, some of his friends (like Gabi) decided to buy a present for him. His friends are going to buy a string consisted of round brackets since Famil Door loves string of brackets of length n more than any other strings!
The sequence of round brackets is called valid if and only if:
1. the total number of opening brackets is equal to the total number of closing brackets;
2. for any prefix of the sequence, the number of opening brackets is greater or equal than the number of closing brackets.
Gabi bought a string s of length m (m ≤ n) and want to complete it to obtain a valid sequence of brackets of length n. He is going to pick some strings p and q consisting of round brackets and merge them in a string p + s + q, that is add the string p at the beginning of the string s and string q at the end of the string s.
Now he wonders, how many pairs of strings p and q exists, such that the string p + s + q is a valid sequence of round brackets. As this number may be pretty large, he wants to calculate it modulo 109 + 7.
Input
First line contains n and m (1 ≤ m ≤ n ≤ 100 000, n - m ≤ 2000) — the desired length of the string and the length of the string bought by Gabi, respectively.
The second line contains string s of length m consisting of characters '(' and ')' only.
Output
Print the number of pairs of string p and q such that p + s + q is a valid sequence of round brackets modulo 109 + 7.
Examples
Input
4 1
(
Output
4
Input
4 4
(())
Output
1
Input
4 3
(((
Output
0
Note
In the first sample there are four different valid pairs:
1. p = "(", q = "))"
2. p = "()", q = ")"
3. p = "", q = "())"
4. p = "", q = ")()"
In the second sample the only way to obtain a desired string is choose empty p and q.
In the third sample there is no way to get a valid sequence of brackets.
Submitted Solution:
```
def ij(s):
i = 0
j = 0
for c in s:
if c == ')':
if j > 0:
j -= 1
else:
i += 1
else:
j += 1
return i, j
def ways(n, s):
I, J = ij(s)
f = n - len(s) - I - J
if f < 0 or f%2:
return 0
E = f//2
if E == 0:
return 1
C = [1]
for n in range(E+max(I,J)+1):
C.append(C[n] * 2 * (2*n + 1) // (n+2))
W = [C, C]
for i in range(2, E+max(I,J)+1):
W.append([W[i-1][e+1]-W[i-2][e+1] for e in range(E+max(I,J)+1-i+1)])
result = sum(W[I+k][e-k] * W[J+k][E-e]
for k in range(E+1)
for e in range(k, E+1))
return result
if __name__ == '__main__':
n, m = map(int, input().split())
s = input()
print(ways(n, s) % (10**9 + 7))
``` | instruction | 0 | 52,635 | 21 | 105,270 |
No | output | 1 | 52,635 | 21 | 105,271 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters + and 1 into this sequence. For example, sequences (())(), () and (()(())) are regular, while )(, (() and (()))( are not. Let's call a regular bracket sequence "RBS".
You are given a sequence s of n characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.
You have to replace every character ? with either ) or ( (different characters ? can be replaced with different brackets). You cannot reorder the characters, remove them, insert other characters, and each ? must be replaced.
Determine if it is possible to obtain an RBS after these replacements.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of one line containing s (2 ≤ |s| ≤ 100) — a sequence of characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.
Output
For each test case, print YES if it is possible to obtain a regular bracket sequence, or NO otherwise}.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
5
()
(?)
(??)
??()
)?(?
Output
YES
NO
YES
YES
NO
Note
In the first test case, the sequence is already an RBS.
In the third test case, you can obtain an RBS as follows: ()() or (()).
In the fourth test case, you can obtain an RBS as follows: ()(). | instruction | 0 | 55,796 | 21 | 111,592 |
Tags: constructive algorithms, greedy
Correct Solution:
```
t=int(input())
for o in range(t):
s=list(input())
c=0
if len(s)%2!=0:
print('NO')
elif s[0]==')':
print('NO')
elif s[len(s)-1]=='(':
print('NO')
else:
print('YES')
``` | output | 1 | 55,796 | 21 | 111,593 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters + and 1 into this sequence. For example, sequences (())(), () and (()(())) are regular, while )(, (() and (()))( are not. Let's call a regular bracket sequence "RBS".
You are given a sequence s of n characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.
You have to replace every character ? with either ) or ( (different characters ? can be replaced with different brackets). You cannot reorder the characters, remove them, insert other characters, and each ? must be replaced.
Determine if it is possible to obtain an RBS after these replacements.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of one line containing s (2 ≤ |s| ≤ 100) — a sequence of characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.
Output
For each test case, print YES if it is possible to obtain a regular bracket sequence, or NO otherwise}.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
5
()
(?)
(??)
??()
)?(?
Output
YES
NO
YES
YES
NO
Note
In the first test case, the sequence is already an RBS.
In the third test case, you can obtain an RBS as follows: ()() or (()).
In the fourth test case, you can obtain an RBS as follows: ()(). | instruction | 0 | 55,797 | 21 | 111,594 |
Tags: constructive algorithms, greedy
Correct Solution:
```
from math import *
from collections import deque
from copy import deepcopy
import sys
def inp(): return sys.stdin.readline().rstrip("\r\n") #for fast input
def multi(): return map(int,input().split())
def strmulti(): return map(str, inp().split())
def lis(): return list(map(int, inp().split()))
def lcm(a,b): return (a*b)//gcd(a,b)
def ncr(n,r): return factorial(n) // (factorial(r) * factorial(max(n - r, 1)))
def stringlis(): return list(map(str, inp().split()))
def out(var): sys.stdout.write(str(var)) #for fast output, always take string
def printlist(a) :
print(' '.join(str(a[i]) for i in range(len(a))))
def isPrime(n) :
if (n <= 1) : return False
if (n <= 3) : return True
if (n % 2 == 0 or n % 3 == 0) : return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
#copied functions end
#start coding
t=int(inp())
for _ in range(t):
s=list(inp())
count=0
if(s[0]==')' or s[-1]=='('):
print("NO")
continue
if(len(s)%2==0):
print("YES")
else:
print("NO")
# for i in s:
# if(i=='('):
# count+=1
# elif(i==')'):
# count-=1
# else:
# if(count<=0):
# count+=1
# else:
# count-=1
#
#
# if(count==0):
# print("YES")
# else:
# print("NO")
``` | output | 1 | 55,797 | 21 | 111,595 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters + and 1 into this sequence. For example, sequences (())(), () and (()(())) are regular, while )(, (() and (()))( are not. Let's call a regular bracket sequence "RBS".
You are given a sequence s of n characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.
You have to replace every character ? with either ) or ( (different characters ? can be replaced with different brackets). You cannot reorder the characters, remove them, insert other characters, and each ? must be replaced.
Determine if it is possible to obtain an RBS after these replacements.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of one line containing s (2 ≤ |s| ≤ 100) — a sequence of characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.
Output
For each test case, print YES if it is possible to obtain a regular bracket sequence, or NO otherwise}.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
5
()
(?)
(??)
??()
)?(?
Output
YES
NO
YES
YES
NO
Note
In the first test case, the sequence is already an RBS.
In the third test case, you can obtain an RBS as follows: ()() or (()).
In the fourth test case, you can obtain an RBS as follows: ()(). | instruction | 0 | 55,798 | 21 | 111,596 |
Tags: constructive algorithms, greedy
Correct Solution:
```
for _ in range(int(input())):
s = input()
if len(s) % 2 != 0:
print("NO")
elif s.startswith(')'):
print("NO")
elif s.endswith('('):
print("NO")
else:
lb = [s.count('('), s.count(')')]
qb = s.count('?')
if lb[1] > lb[0] + qb:
print("NO")
elif (lb[1] - lb[0] + qb) % 2 != 0:
print("NO")
else:
print("YES")
``` | output | 1 | 55,798 | 21 | 111,597 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters + and 1 into this sequence. For example, sequences (())(), () and (()(())) are regular, while )(, (() and (()))( are not. Let's call a regular bracket sequence "RBS".
You are given a sequence s of n characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.
You have to replace every character ? with either ) or ( (different characters ? can be replaced with different brackets). You cannot reorder the characters, remove them, insert other characters, and each ? must be replaced.
Determine if it is possible to obtain an RBS after these replacements.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of one line containing s (2 ≤ |s| ≤ 100) — a sequence of characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.
Output
For each test case, print YES if it is possible to obtain a regular bracket sequence, or NO otherwise}.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
5
()
(?)
(??)
??()
)?(?
Output
YES
NO
YES
YES
NO
Note
In the first test case, the sequence is already an RBS.
In the third test case, you can obtain an RBS as follows: ()() or (()).
In the fourth test case, you can obtain an RBS as follows: ()(). | instruction | 0 | 55,799 | 21 | 111,598 |
Tags: constructive algorithms, greedy
Correct Solution:
```
for _ in range(int(input())):
s=input()
if len(s)%2==1:
print("NO")
else:
if s[0]==')' or s[-1]=='(':
print("NO")
else:
print("YES")
``` | output | 1 | 55,799 | 21 | 111,599 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters + and 1 into this sequence. For example, sequences (())(), () and (()(())) are regular, while )(, (() and (()))( are not. Let's call a regular bracket sequence "RBS".
You are given a sequence s of n characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.
You have to replace every character ? with either ) or ( (different characters ? can be replaced with different brackets). You cannot reorder the characters, remove them, insert other characters, and each ? must be replaced.
Determine if it is possible to obtain an RBS after these replacements.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of one line containing s (2 ≤ |s| ≤ 100) — a sequence of characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.
Output
For each test case, print YES if it is possible to obtain a regular bracket sequence, or NO otherwise}.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
5
()
(?)
(??)
??()
)?(?
Output
YES
NO
YES
YES
NO
Note
In the first test case, the sequence is already an RBS.
In the third test case, you can obtain an RBS as follows: ()() or (()).
In the fourth test case, you can obtain an RBS as follows: ()(). | instruction | 0 | 55,800 | 21 | 111,600 |
Tags: constructive algorithms, greedy
Correct Solution:
```
t=int(input())
for _ in range(t):
s=input()
l=r=q=0
for c in s:
if c == '(': l += 1
if c == ')': r += 1
if c == '?': q += 1
if (l+r+q)%2:
print('NO')
continue
lq = (l+r+q)//2-l
if not 0 <= lq <= q:
print('NO')
continue
works = True
d = 0
for c in s:
if c == '(': d += 1
if c == ')': d -= 1
if c == '?':
if lq:
lq -= 1;d+=1
else:
d-=1
if d < 0:
print('NO')
break
else:
print('YES')
``` | output | 1 | 55,800 | 21 | 111,601 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters + and 1 into this sequence. For example, sequences (())(), () and (()(())) are regular, while )(, (() and (()))( are not. Let's call a regular bracket sequence "RBS".
You are given a sequence s of n characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.
You have to replace every character ? with either ) or ( (different characters ? can be replaced with different brackets). You cannot reorder the characters, remove them, insert other characters, and each ? must be replaced.
Determine if it is possible to obtain an RBS after these replacements.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of one line containing s (2 ≤ |s| ≤ 100) — a sequence of characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.
Output
For each test case, print YES if it is possible to obtain a regular bracket sequence, or NO otherwise}.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
5
()
(?)
(??)
??()
)?(?
Output
YES
NO
YES
YES
NO
Note
In the first test case, the sequence is already an RBS.
In the third test case, you can obtain an RBS as follows: ()() or (()).
In the fourth test case, you can obtain an RBS as follows: ()(). | instruction | 0 | 55,801 | 21 | 111,602 |
Tags: constructive algorithms, greedy
Correct Solution:
```
for _ in range(int(input())):
s = input()
if len(s) % 2 == 0 and s[0] != ')' and s[-1] != '(':
print('YES')
else:
print("NO")
``` | output | 1 | 55,801 | 21 | 111,603 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters + and 1 into this sequence. For example, sequences (())(), () and (()(())) are regular, while )(, (() and (()))( are not. Let's call a regular bracket sequence "RBS".
You are given a sequence s of n characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.
You have to replace every character ? with either ) or ( (different characters ? can be replaced with different brackets). You cannot reorder the characters, remove them, insert other characters, and each ? must be replaced.
Determine if it is possible to obtain an RBS after these replacements.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of one line containing s (2 ≤ |s| ≤ 100) — a sequence of characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.
Output
For each test case, print YES if it is possible to obtain a regular bracket sequence, or NO otherwise}.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
5
()
(?)
(??)
??()
)?(?
Output
YES
NO
YES
YES
NO
Note
In the first test case, the sequence is already an RBS.
In the third test case, you can obtain an RBS as follows: ()() or (()).
In the fourth test case, you can obtain an RBS as follows: ()(). | instruction | 0 | 55,802 | 21 | 111,604 |
Tags: constructive algorithms, greedy
Correct Solution:
```
import math
from sys import *
#input=stdin.readline
def solve():
#n=int(input())
s=input()
#n,k=map(int,input().split())
##m=int(input())
#l=list(map(int,input().split()))
#l1=list(map(int,input().split()))
#zero=s.count('0')
n=len(s)
if(s[n-1]=='(' or s[0]==')'):
print("NO")
else:
c1,c2,c3=0,0,0
for i in s:
if(i=='('):
c1+=1
elif(i==')'):
c2+=1
elif(i=='?'):
c3+=1
if(abs(c1-c2)==0 and c3%2!=0):
print("NO")
elif(abs(c1-c2)!=0 and abs(c1-c2)!=c3):
print("NO")
else:
print("YES")
#t=1
t=int(input())
while(t>0):
t-=1
solve()
``` | output | 1 | 55,802 | 21 | 111,605 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters + and 1 into this sequence. For example, sequences (())(), () and (()(())) are regular, while )(, (() and (()))( are not. Let's call a regular bracket sequence "RBS".
You are given a sequence s of n characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.
You have to replace every character ? with either ) or ( (different characters ? can be replaced with different brackets). You cannot reorder the characters, remove them, insert other characters, and each ? must be replaced.
Determine if it is possible to obtain an RBS after these replacements.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of one line containing s (2 ≤ |s| ≤ 100) — a sequence of characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.
Output
For each test case, print YES if it is possible to obtain a regular bracket sequence, or NO otherwise}.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
5
()
(?)
(??)
??()
)?(?
Output
YES
NO
YES
YES
NO
Note
In the first test case, the sequence is already an RBS.
In the third test case, you can obtain an RBS as follows: ()() or (()).
In the fourth test case, you can obtain an RBS as follows: ()(). | instruction | 0 | 55,803 | 21 | 111,606 |
Tags: constructive algorithms, greedy
Correct Solution:
```
def valid_brackets(s):
openBrackets = 0
for i in s:
if openBrackets < 0:
return False
if i == "(":
openBrackets += 1
elif i == ")":
openBrackets -= 1
if openBrackets != 0:
return False
return True
# #
# t = int(input())
# for i in range(t):
# s = input()
# if len(s) % 2 == 0 and s[0] != ')' and s[-1] != '(':
# print('YES')
# else:
# print('NO')
t = int(input())
for _ in range(t):
s = input()
a = list(s)
count = len(s)//2 -1
for i in range(len(s)):
if a[i] == '?':
if count > 0:
a[i] = '('
count -= 1
else:
a[i] = ')'
if valid_brackets(str(a)):
print("YES")
else:
print("NO")
``` | output | 1 | 55,803 | 21 | 111,607 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters + and 1 into this sequence. For example, sequences (())(), () and (()(())) are regular, while )(, (() and (()))( are not. Let's call a regular bracket sequence "RBS".
You are given a sequence s of n characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.
You have to replace every character ? with either ) or ( (different characters ? can be replaced with different brackets). You cannot reorder the characters, remove them, insert other characters, and each ? must be replaced.
Determine if it is possible to obtain an RBS after these replacements.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of one line containing s (2 ≤ |s| ≤ 100) — a sequence of characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.
Output
For each test case, print YES if it is possible to obtain a regular bracket sequence, or NO otherwise}.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
5
()
(?)
(??)
??()
)?(?
Output
YES
NO
YES
YES
NO
Note
In the first test case, the sequence is already an RBS.
In the third test case, you can obtain an RBS as follows: ()() or (()).
In the fourth test case, you can obtain an RBS as follows: ()().
Submitted Solution:
```
for _ in range(int(input())):
s = input()
a = s.find('(')
b = s.find(')')
if len(s)%2==1:
print("NO")
elif a<b:
print("YES")
else:
if b!=0 and a!=len(s)-1:
print("YES")
else:
print("NO")
``` | instruction | 0 | 55,804 | 21 | 111,608 |
Yes | output | 1 | 55,804 | 21 | 111,609 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters + and 1 into this sequence. For example, sequences (())(), () and (()(())) are regular, while )(, (() and (()))( are not. Let's call a regular bracket sequence "RBS".
You are given a sequence s of n characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.
You have to replace every character ? with either ) or ( (different characters ? can be replaced with different brackets). You cannot reorder the characters, remove them, insert other characters, and each ? must be replaced.
Determine if it is possible to obtain an RBS after these replacements.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of one line containing s (2 ≤ |s| ≤ 100) — a sequence of characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.
Output
For each test case, print YES if it is possible to obtain a regular bracket sequence, or NO otherwise}.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
5
()
(?)
(??)
??()
)?(?
Output
YES
NO
YES
YES
NO
Note
In the first test case, the sequence is already an RBS.
In the third test case, you can obtain an RBS as follows: ()() or (()).
In the fourth test case, you can obtain an RBS as follows: ()().
Submitted Solution:
```
from sys import stdin
input = stdin.readline
t = int(input())
for _ in range(t):
s = input().strip()
if len(s) % 2 > 0:
print("NO")
continue
if s[0] == ")" or s[-1] == "(":
print("NO")
continue
print("YES")
``` | instruction | 0 | 55,805 | 21 | 111,610 |
Yes | output | 1 | 55,805 | 21 | 111,611 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters + and 1 into this sequence. For example, sequences (())(), () and (()(())) are regular, while )(, (() and (()))( are not. Let's call a regular bracket sequence "RBS".
You are given a sequence s of n characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.
You have to replace every character ? with either ) or ( (different characters ? can be replaced with different brackets). You cannot reorder the characters, remove them, insert other characters, and each ? must be replaced.
Determine if it is possible to obtain an RBS after these replacements.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of one line containing s (2 ≤ |s| ≤ 100) — a sequence of characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.
Output
For each test case, print YES if it is possible to obtain a regular bracket sequence, or NO otherwise}.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
5
()
(?)
(??)
??()
)?(?
Output
YES
NO
YES
YES
NO
Note
In the first test case, the sequence is already an RBS.
In the third test case, you can obtain an RBS as follows: ()() or (()).
In the fourth test case, you can obtain an RBS as follows: ()().
Submitted Solution:
```
try:
test = int(input())
while test!=0:
string = input()
left = string.index("(")
right = string.index(")")
if left<right:
before = left - 0
mid = right-left-1
after = len(string)-1-right
if (mid+before+after)&1==0:
print("Yes")
else:
print("No")
else:
before = right-0
mid = left-right-1
after = len(string)-1-left
if (before-1)>=0 and (after-1)>=0 and (mid+(before-1)+(after-1))&1==0:
print("Yes")
else:
print("No")
test -= 1
except EOFError as e:
print("")
``` | instruction | 0 | 55,806 | 21 | 111,612 |
Yes | output | 1 | 55,806 | 21 | 111,613 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters + and 1 into this sequence. For example, sequences (())(), () and (()(())) are regular, while )(, (() and (()))( are not. Let's call a regular bracket sequence "RBS".
You are given a sequence s of n characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.
You have to replace every character ? with either ) or ( (different characters ? can be replaced with different brackets). You cannot reorder the characters, remove them, insert other characters, and each ? must be replaced.
Determine if it is possible to obtain an RBS after these replacements.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of one line containing s (2 ≤ |s| ≤ 100) — a sequence of characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.
Output
For each test case, print YES if it is possible to obtain a regular bracket sequence, or NO otherwise}.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
5
()
(?)
(??)
??()
)?(?
Output
YES
NO
YES
YES
NO
Note
In the first test case, the sequence is already an RBS.
In the third test case, you can obtain an RBS as follows: ()() or (()).
In the fourth test case, you can obtain an RBS as follows: ()().
Submitted Solution:
```
t=int(input())
for _ in range(t):
s=input().strip('\n')
cnt=0
cnt1=0
cnt2=0
f=0
if s[0] == ')' or s[-1] == '(':
#print('NO')
f=1
else:
for i in s:
if i == ")":
cnt2+=1
if i == "?":
cnt+=1
if i == "(":
cnt1+=1
d=abs(cnt1-cnt2)
if d==0 and cnt%2!=0:
#print('NO')
f=1
elif d!=0 and d!=cnt:
#print('NO')
f=1
if f:
print("NO")
else:
print("YES")
``` | instruction | 0 | 55,807 | 21 | 111,614 |
Yes | output | 1 | 55,807 | 21 | 111,615 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters + and 1 into this sequence. For example, sequences (())(), () and (()(())) are regular, while )(, (() and (()))( are not. Let's call a regular bracket sequence "RBS".
You are given a sequence s of n characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.
You have to replace every character ? with either ) or ( (different characters ? can be replaced with different brackets). You cannot reorder the characters, remove them, insert other characters, and each ? must be replaced.
Determine if it is possible to obtain an RBS after these replacements.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of one line containing s (2 ≤ |s| ≤ 100) — a sequence of characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.
Output
For each test case, print YES if it is possible to obtain a regular bracket sequence, or NO otherwise}.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
5
()
(?)
(??)
??()
)?(?
Output
YES
NO
YES
YES
NO
Note
In the first test case, the sequence is already an RBS.
In the third test case, you can obtain an RBS as follows: ()() or (()).
In the fourth test case, you can obtain an RBS as follows: ()().
Submitted Solution:
```
def bst(s):
if(len(s)%2!=0):
return "NO"
o,c=0,0
for i in s:
if(i=="(" and c==1):
return "NO"
elif(i=="(" and c==0):
o=1
elif(i==")"):
c=1
return "YES"
'''ob=[]
for i in s:
if(i=="?" and not ob):
ob.append("(")
elif(i=="?" or i==")"):
if(ob):
ob.pop()
else:
return "NO"
else:
ob.append(i)
if(not ob):
return "YES"
else:
return "NO"'''
n=int(input())
for _ in range(n):
s=input()
print(bst(s))
``` | instruction | 0 | 55,808 | 21 | 111,616 |
No | output | 1 | 55,808 | 21 | 111,617 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters + and 1 into this sequence. For example, sequences (())(), () and (()(())) are regular, while )(, (() and (()))( are not. Let's call a regular bracket sequence "RBS".
You are given a sequence s of n characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.
You have to replace every character ? with either ) or ( (different characters ? can be replaced with different brackets). You cannot reorder the characters, remove them, insert other characters, and each ? must be replaced.
Determine if it is possible to obtain an RBS after these replacements.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of one line containing s (2 ≤ |s| ≤ 100) — a sequence of characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.
Output
For each test case, print YES if it is possible to obtain a regular bracket sequence, or NO otherwise}.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
5
()
(?)
(??)
??()
)?(?
Output
YES
NO
YES
YES
NO
Note
In the first test case, the sequence is already an RBS.
In the third test case, you can obtain an RBS as follows: ()() or (()).
In the fourth test case, you can obtain an RBS as follows: ()().
Submitted Solution:
```
n = int(input())
res = []
for _ in range(n):
arr = input()
openb = 0
closeb = 0
w = 1
for i in arr:
if i == '(':
openb+=1
elif i == ')':
if openb>0:
openb-=1
else:
w = 0
res.append('NO')
break
elif openb>0:
openb-=1
elif openb == 0:
openb+=1
if w == 1:
res.append('YES')
for i in res:
print(i)
``` | instruction | 0 | 55,809 | 21 | 111,618 |
No | output | 1 | 55,809 | 21 | 111,619 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters + and 1 into this sequence. For example, sequences (())(), () and (()(())) are regular, while )(, (() and (()))( are not. Let's call a regular bracket sequence "RBS".
You are given a sequence s of n characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.
You have to replace every character ? with either ) or ( (different characters ? can be replaced with different brackets). You cannot reorder the characters, remove them, insert other characters, and each ? must be replaced.
Determine if it is possible to obtain an RBS after these replacements.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of one line containing s (2 ≤ |s| ≤ 100) — a sequence of characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.
Output
For each test case, print YES if it is possible to obtain a regular bracket sequence, or NO otherwise}.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
5
()
(?)
(??)
??()
)?(?
Output
YES
NO
YES
YES
NO
Note
In the first test case, the sequence is already an RBS.
In the third test case, you can obtain an RBS as follows: ()() or (()).
In the fourth test case, you can obtain an RBS as follows: ()().
Submitted Solution:
```
t = int(input())
for i in range(t):
sequence = input()
ans = 'YES'
stack = []
for c in sequence:
if c == ')':
if len(stack) != 0 and stack[-1] != ')':
stack.pop()
else:
ans = 'NO'
break
else:
stack.append(c)
open_minus_close = 0
for c in stack:
if c == '(':
open_minus_close += 1
else:
open_minus_close += 1 if open_minus_close == 0 else -1
print(ans if open_minus_close == 0 else 'NO')
``` | instruction | 0 | 55,810 | 21 | 111,620 |
No | output | 1 | 55,810 | 21 | 111,621 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters + and 1 into this sequence. For example, sequences (())(), () and (()(())) are regular, while )(, (() and (()))( are not. Let's call a regular bracket sequence "RBS".
You are given a sequence s of n characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.
You have to replace every character ? with either ) or ( (different characters ? can be replaced with different brackets). You cannot reorder the characters, remove them, insert other characters, and each ? must be replaced.
Determine if it is possible to obtain an RBS after these replacements.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each test case consists of one line containing s (2 ≤ |s| ≤ 100) — a sequence of characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.
Output
For each test case, print YES if it is possible to obtain a regular bracket sequence, or NO otherwise}.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
5
()
(?)
(??)
??()
)?(?
Output
YES
NO
YES
YES
NO
Note
In the first test case, the sequence is already an RBS.
In the third test case, you can obtain an RBS as follows: ()() or (()).
In the fourth test case, you can obtain an RBS as follows: ()().
Submitted Solution:
```
for _ in range(int(input())):
#n=int(input())
l=list(input())
s=list()
s.append(l[0])
n=len(l)
for i in range(1,n):
if l[i]=='(':
s.append(l[i])
elif l[i]==')':
if len(s)!=0:
if s[-1]=='(' or s[-1]=='?':
s.pop(-1)
else:
s.append(l[i])
elif l[i]=='?':
if len(s)!=0:
if s[-1]=='(' or s[-1]=='?':
s.pop(-1)
else:
s.append(l[i])
#print(s)
if s:
print('NO')
else:
print('YES')
``` | instruction | 0 | 55,811 | 21 | 111,622 |
No | output | 1 | 55,811 | 21 | 111,623 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A string is called bracket sequence if it does not contain any characters other than "(" and ")". A bracket sequence is called regular (shortly, RBS) if it is possible to obtain correct arithmetic expression by inserting characters "+" and "1" into this sequence. For example, "", "(())" and "()()" are RBS and ")(" and "(()" are not.
We can see that each opening bracket in RBS is paired with some closing bracket, and, using this fact, we can define nesting depth of the RBS as maximum number of bracket pairs, such that the 2-nd pair lies inside the 1-st one, the 3-rd one — inside the 2-nd one and so on. For example, nesting depth of "" is 0, "()()()" is 1 and "()((())())" is 3.
Now, you are given RBS s of even length n. You should color each bracket of s into one of two colors: red or blue. Bracket sequence r, consisting only of red brackets, should be RBS, and bracket sequence, consisting only of blue brackets b, should be RBS. Any of them can be empty. You are not allowed to reorder characters in s, r or b. No brackets can be left uncolored.
Among all possible variants you should choose one that minimizes maximum of r's and b's nesting depth. If there are multiple solutions you can print any of them.
Input
The first line contains an even integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of RBS s.
The second line contains regular bracket sequence s (|s| = n, s_i ∈ \{"(", ")"\}).
Output
Print single string t of length n consisting of "0"-s and "1"-s. If t_i is equal to 0 then character s_i belongs to RBS r, otherwise s_i belongs to b.
Examples
Input
2
()
Output
11
Input
4
(())
Output
0101
Input
10
((()())())
Output
0110001111
Note
In the first example one of optimal solutions is s = "\color{blue}{()}". r is empty and b = "()". The answer is max(0, 1) = 1.
In the second example it's optimal to make s = "\color{red}{(}\color{blue}{(}\color{red}{)}\color{blue}{)}". r = b = "()" and the answer is 1.
In the third example we can make s = "\color{red}{(}\color{blue}{((}\color{red}{)()}\color{blue}{)())}". r = "()()" and b = "(()())" and the answer is 2. | instruction | 0 | 56,442 | 21 | 112,884 |
Tags: constructive algorithms, greedy
Correct Solution:
```
input()
brackets = list(input())
a = 0
b = 0
result =[]
for bracket in brackets:
if bracket == '(':
if a == b:
a+=1
result.append("1")
else:
b+=1
result.append("0")
else:
if a>b:
a-=1
result.append("1")
else:
b-=1
result.append("0")
print("".join(result))
``` | output | 1 | 56,442 | 21 | 112,885 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A string is called bracket sequence if it does not contain any characters other than "(" and ")". A bracket sequence is called regular (shortly, RBS) if it is possible to obtain correct arithmetic expression by inserting characters "+" and "1" into this sequence. For example, "", "(())" and "()()" are RBS and ")(" and "(()" are not.
We can see that each opening bracket in RBS is paired with some closing bracket, and, using this fact, we can define nesting depth of the RBS as maximum number of bracket pairs, such that the 2-nd pair lies inside the 1-st one, the 3-rd one — inside the 2-nd one and so on. For example, nesting depth of "" is 0, "()()()" is 1 and "()((())())" is 3.
Now, you are given RBS s of even length n. You should color each bracket of s into one of two colors: red or blue. Bracket sequence r, consisting only of red brackets, should be RBS, and bracket sequence, consisting only of blue brackets b, should be RBS. Any of them can be empty. You are not allowed to reorder characters in s, r or b. No brackets can be left uncolored.
Among all possible variants you should choose one that minimizes maximum of r's and b's nesting depth. If there are multiple solutions you can print any of them.
Input
The first line contains an even integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of RBS s.
The second line contains regular bracket sequence s (|s| = n, s_i ∈ \{"(", ")"\}).
Output
Print single string t of length n consisting of "0"-s and "1"-s. If t_i is equal to 0 then character s_i belongs to RBS r, otherwise s_i belongs to b.
Examples
Input
2
()
Output
11
Input
4
(())
Output
0101
Input
10
((()())())
Output
0110001111
Note
In the first example one of optimal solutions is s = "\color{blue}{()}". r is empty and b = "()". The answer is max(0, 1) = 1.
In the second example it's optimal to make s = "\color{red}{(}\color{blue}{(}\color{red}{)}\color{blue}{)}". r = b = "()" and the answer is 1.
In the third example we can make s = "\color{red}{(}\color{blue}{((}\color{red}{)()}\color{blue}{)())}". r = "()()" and b = "(()())" and the answer is 2. | instruction | 0 | 56,443 | 21 | 112,886 |
Tags: constructive algorithms, greedy
Correct Solution:
```
from collections import deque
n = int(input())
s = input()
left = deque([])
ans = [-1]*n
flag = True
for i in range(n):
if s[i]=='(':
left.append(i)
else:
if flag:
l_idx = left.popleft()
ans[l_idx] = 0
ans[i] = 0
flag = False
else:
l_idx = left.popleft()
ans[l_idx] = 1
ans[i] = 1
flag = True
ans = list(map(str, ans))
print(''.join(ans))
``` | output | 1 | 56,443 | 21 | 112,887 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A string is called bracket sequence if it does not contain any characters other than "(" and ")". A bracket sequence is called regular (shortly, RBS) if it is possible to obtain correct arithmetic expression by inserting characters "+" and "1" into this sequence. For example, "", "(())" and "()()" are RBS and ")(" and "(()" are not.
We can see that each opening bracket in RBS is paired with some closing bracket, and, using this fact, we can define nesting depth of the RBS as maximum number of bracket pairs, such that the 2-nd pair lies inside the 1-st one, the 3-rd one — inside the 2-nd one and so on. For example, nesting depth of "" is 0, "()()()" is 1 and "()((())())" is 3.
Now, you are given RBS s of even length n. You should color each bracket of s into one of two colors: red or blue. Bracket sequence r, consisting only of red brackets, should be RBS, and bracket sequence, consisting only of blue brackets b, should be RBS. Any of them can be empty. You are not allowed to reorder characters in s, r or b. No brackets can be left uncolored.
Among all possible variants you should choose one that minimizes maximum of r's and b's nesting depth. If there are multiple solutions you can print any of them.
Input
The first line contains an even integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of RBS s.
The second line contains regular bracket sequence s (|s| = n, s_i ∈ \{"(", ")"\}).
Output
Print single string t of length n consisting of "0"-s and "1"-s. If t_i is equal to 0 then character s_i belongs to RBS r, otherwise s_i belongs to b.
Examples
Input
2
()
Output
11
Input
4
(())
Output
0101
Input
10
((()())())
Output
0110001111
Note
In the first example one of optimal solutions is s = "\color{blue}{()}". r is empty and b = "()". The answer is max(0, 1) = 1.
In the second example it's optimal to make s = "\color{red}{(}\color{blue}{(}\color{red}{)}\color{blue}{)}". r = b = "()" and the answer is 1.
In the third example we can make s = "\color{red}{(}\color{blue}{((}\color{red}{)()}\color{blue}{)())}". r = "()()" and b = "(()())" and the answer is 2. | instruction | 0 | 56,444 | 21 | 112,888 |
Tags: constructive algorithms, greedy
Correct Solution:
```
n = int(input())
s = str(input())
cum = [0]*(n+1)
depth = [0]*(n+1)
for i in range(n):
if s[i] == '(':
cum[i+1] = cum[i]+1
depth[i+1] = cum[i+1]
else:
depth[i+1] = cum[i]
cum[i+1] = cum[i]-1
#print(cum)
#print(depth)
ans = ['0']*n
for i in range(n):
if depth[i+1]%2 == 0:
ans[i] = '0'
else:
ans[i] = '1'
print(''.join(ans))
``` | output | 1 | 56,444 | 21 | 112,889 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A string is called bracket sequence if it does not contain any characters other than "(" and ")". A bracket sequence is called regular (shortly, RBS) if it is possible to obtain correct arithmetic expression by inserting characters "+" and "1" into this sequence. For example, "", "(())" and "()()" are RBS and ")(" and "(()" are not.
We can see that each opening bracket in RBS is paired with some closing bracket, and, using this fact, we can define nesting depth of the RBS as maximum number of bracket pairs, such that the 2-nd pair lies inside the 1-st one, the 3-rd one — inside the 2-nd one and so on. For example, nesting depth of "" is 0, "()()()" is 1 and "()((())())" is 3.
Now, you are given RBS s of even length n. You should color each bracket of s into one of two colors: red or blue. Bracket sequence r, consisting only of red brackets, should be RBS, and bracket sequence, consisting only of blue brackets b, should be RBS. Any of them can be empty. You are not allowed to reorder characters in s, r or b. No brackets can be left uncolored.
Among all possible variants you should choose one that minimizes maximum of r's and b's nesting depth. If there are multiple solutions you can print any of them.
Input
The first line contains an even integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of RBS s.
The second line contains regular bracket sequence s (|s| = n, s_i ∈ \{"(", ")"\}).
Output
Print single string t of length n consisting of "0"-s and "1"-s. If t_i is equal to 0 then character s_i belongs to RBS r, otherwise s_i belongs to b.
Examples
Input
2
()
Output
11
Input
4
(())
Output
0101
Input
10
((()())())
Output
0110001111
Note
In the first example one of optimal solutions is s = "\color{blue}{()}". r is empty and b = "()". The answer is max(0, 1) = 1.
In the second example it's optimal to make s = "\color{red}{(}\color{blue}{(}\color{red}{)}\color{blue}{)}". r = b = "()" and the answer is 1.
In the third example we can make s = "\color{red}{(}\color{blue}{((}\color{red}{)()}\color{blue}{)())}". r = "()()" and b = "(()())" and the answer is 2. | instruction | 0 | 56,445 | 21 | 112,890 |
Tags: constructive algorithms, greedy
Correct Solution:
```
n = int(input())
s = input()
ans = []
l = 0
for i in range(n):
if s[i] == '(':
l += 1
ans.append(l % 2)
if s[i] == ')':
ans.append(l% 2)
l -=1
print(''.join([str(i) for i in ans]))
``` | output | 1 | 56,445 | 21 | 112,891 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A string is called bracket sequence if it does not contain any characters other than "(" and ")". A bracket sequence is called regular (shortly, RBS) if it is possible to obtain correct arithmetic expression by inserting characters "+" and "1" into this sequence. For example, "", "(())" and "()()" are RBS and ")(" and "(()" are not.
We can see that each opening bracket in RBS is paired with some closing bracket, and, using this fact, we can define nesting depth of the RBS as maximum number of bracket pairs, such that the 2-nd pair lies inside the 1-st one, the 3-rd one — inside the 2-nd one and so on. For example, nesting depth of "" is 0, "()()()" is 1 and "()((())())" is 3.
Now, you are given RBS s of even length n. You should color each bracket of s into one of two colors: red or blue. Bracket sequence r, consisting only of red brackets, should be RBS, and bracket sequence, consisting only of blue brackets b, should be RBS. Any of them can be empty. You are not allowed to reorder characters in s, r or b. No brackets can be left uncolored.
Among all possible variants you should choose one that minimizes maximum of r's and b's nesting depth. If there are multiple solutions you can print any of them.
Input
The first line contains an even integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of RBS s.
The second line contains regular bracket sequence s (|s| = n, s_i ∈ \{"(", ")"\}).
Output
Print single string t of length n consisting of "0"-s and "1"-s. If t_i is equal to 0 then character s_i belongs to RBS r, otherwise s_i belongs to b.
Examples
Input
2
()
Output
11
Input
4
(())
Output
0101
Input
10
((()())())
Output
0110001111
Note
In the first example one of optimal solutions is s = "\color{blue}{()}". r is empty and b = "()". The answer is max(0, 1) = 1.
In the second example it's optimal to make s = "\color{red}{(}\color{blue}{(}\color{red}{)}\color{blue}{)}". r = b = "()" and the answer is 1.
In the third example we can make s = "\color{red}{(}\color{blue}{((}\color{red}{)()}\color{blue}{)())}". r = "()()" and b = "(()())" and the answer is 2. | instruction | 0 | 56,446 | 21 | 112,892 |
Tags: constructive algorithms, greedy
Correct Solution:
```
n = int(input())
A = input()
stack = []
for i in A:
if i == '(':
if len(stack) == 0 or stack[-1] == 1:
stack.append(0)
print(0, end='')
else:
stack.append(1)
print(1, end='')
else:
print(stack[-1], end='')
stack.pop()
``` | output | 1 | 56,446 | 21 | 112,893 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A string is called bracket sequence if it does not contain any characters other than "(" and ")". A bracket sequence is called regular (shortly, RBS) if it is possible to obtain correct arithmetic expression by inserting characters "+" and "1" into this sequence. For example, "", "(())" and "()()" are RBS and ")(" and "(()" are not.
We can see that each opening bracket in RBS is paired with some closing bracket, and, using this fact, we can define nesting depth of the RBS as maximum number of bracket pairs, such that the 2-nd pair lies inside the 1-st one, the 3-rd one — inside the 2-nd one and so on. For example, nesting depth of "" is 0, "()()()" is 1 and "()((())())" is 3.
Now, you are given RBS s of even length n. You should color each bracket of s into one of two colors: red or blue. Bracket sequence r, consisting only of red brackets, should be RBS, and bracket sequence, consisting only of blue brackets b, should be RBS. Any of them can be empty. You are not allowed to reorder characters in s, r or b. No brackets can be left uncolored.
Among all possible variants you should choose one that minimizes maximum of r's and b's nesting depth. If there are multiple solutions you can print any of them.
Input
The first line contains an even integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of RBS s.
The second line contains regular bracket sequence s (|s| = n, s_i ∈ \{"(", ")"\}).
Output
Print single string t of length n consisting of "0"-s and "1"-s. If t_i is equal to 0 then character s_i belongs to RBS r, otherwise s_i belongs to b.
Examples
Input
2
()
Output
11
Input
4
(())
Output
0101
Input
10
((()())())
Output
0110001111
Note
In the first example one of optimal solutions is s = "\color{blue}{()}". r is empty and b = "()". The answer is max(0, 1) = 1.
In the second example it's optimal to make s = "\color{red}{(}\color{blue}{(}\color{red}{)}\color{blue}{)}". r = b = "()" and the answer is 1.
In the third example we can make s = "\color{red}{(}\color{blue}{((}\color{red}{)()}\color{blue}{)())}". r = "()()" and b = "(()())" and the answer is 2. | instruction | 0 | 56,447 | 21 | 112,894 |
Tags: constructive algorithms, greedy
Correct Solution:
```
a=int(input())
red=0
blue=0
li=[]
if a==2:
input()
print(11)
else:
Evenprnths=input()
for elem in Evenprnths:
if elem =="(":
#todo add this to the one having least open parantheses of if draw then red
if red>blue:
blue+=1
print(0,end="")
else:
red+=1
print(1,end="")
elif elem==")":
if blue>red:
blue-=1
print(0,end="")
else:
red-=1
print(1,end="")
print()
#todo remove from the one having the most open parantehess
``` | output | 1 | 56,447 | 21 | 112,895 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A string is called bracket sequence if it does not contain any characters other than "(" and ")". A bracket sequence is called regular (shortly, RBS) if it is possible to obtain correct arithmetic expression by inserting characters "+" and "1" into this sequence. For example, "", "(())" and "()()" are RBS and ")(" and "(()" are not.
We can see that each opening bracket in RBS is paired with some closing bracket, and, using this fact, we can define nesting depth of the RBS as maximum number of bracket pairs, such that the 2-nd pair lies inside the 1-st one, the 3-rd one — inside the 2-nd one and so on. For example, nesting depth of "" is 0, "()()()" is 1 and "()((())())" is 3.
Now, you are given RBS s of even length n. You should color each bracket of s into one of two colors: red or blue. Bracket sequence r, consisting only of red brackets, should be RBS, and bracket sequence, consisting only of blue brackets b, should be RBS. Any of them can be empty. You are not allowed to reorder characters in s, r or b. No brackets can be left uncolored.
Among all possible variants you should choose one that minimizes maximum of r's and b's nesting depth. If there are multiple solutions you can print any of them.
Input
The first line contains an even integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of RBS s.
The second line contains regular bracket sequence s (|s| = n, s_i ∈ \{"(", ")"\}).
Output
Print single string t of length n consisting of "0"-s and "1"-s. If t_i is equal to 0 then character s_i belongs to RBS r, otherwise s_i belongs to b.
Examples
Input
2
()
Output
11
Input
4
(())
Output
0101
Input
10
((()())())
Output
0110001111
Note
In the first example one of optimal solutions is s = "\color{blue}{()}". r is empty and b = "()". The answer is max(0, 1) = 1.
In the second example it's optimal to make s = "\color{red}{(}\color{blue}{(}\color{red}{)}\color{blue}{)}". r = b = "()" and the answer is 1.
In the third example we can make s = "\color{red}{(}\color{blue}{((}\color{red}{)()}\color{blue}{)())}". r = "()()" and b = "(()())" and the answer is 2. | instruction | 0 | 56,448 | 21 | 112,896 |
Tags: constructive algorithms, greedy
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#######################################
from collections import *
from operator import itemgetter , attrgetter
from decimal import *
import bisect
import math
import heapq as hq
#import sympy
MOD=10**9 +7
def is_prime(n):
if n == 2 or n == 3: return True
if n < 2 or n%2 == 0: return False
if n < 9: return True
if n%3 == 0: return False
r = int(n**0.5)
# since all primes > 3 are of the form 6n ± 1
# start with f=5 (which is prime)
# and test f, f+2 for being prime
# then loop by 6.
f = 5
while f <= r:
if n % f == 0: return False
if n % (f+2) == 0: return False
f += 6
return True
def pow(a,b,m):
ans=1
while b:
if b&1:
ans=(ans*a)%m
b//=2
a=(a*a)%m
return ans
vis=[]
graph=[]
def dfs(v):
if vis[v]: return 0
vis[v]=True
temp=0
for vv in graph[v]:
temp+=dfs(vv)
return 1+temp
def ispalindrome(s):
if s[:]==s[::-1]:
return 1
return 0
ans=[]
n=int(input())
s=input()
open=[]
for i in s:
if len(open)==0 and i=="(":
open.append(0)
ans.append("0")
continue
if i=="(":
if open[-1]==0:
open.append(1)
ans.append("1")
else:
open.append(0)
ans.append("0")
continue
ans.append(str(open[-1]))
open.pop()
print("".join(ans))
``` | output | 1 | 56,448 | 21 | 112,897 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A string is called bracket sequence if it does not contain any characters other than "(" and ")". A bracket sequence is called regular (shortly, RBS) if it is possible to obtain correct arithmetic expression by inserting characters "+" and "1" into this sequence. For example, "", "(())" and "()()" are RBS and ")(" and "(()" are not.
We can see that each opening bracket in RBS is paired with some closing bracket, and, using this fact, we can define nesting depth of the RBS as maximum number of bracket pairs, such that the 2-nd pair lies inside the 1-st one, the 3-rd one — inside the 2-nd one and so on. For example, nesting depth of "" is 0, "()()()" is 1 and "()((())())" is 3.
Now, you are given RBS s of even length n. You should color each bracket of s into one of two colors: red or blue. Bracket sequence r, consisting only of red brackets, should be RBS, and bracket sequence, consisting only of blue brackets b, should be RBS. Any of them can be empty. You are not allowed to reorder characters in s, r or b. No brackets can be left uncolored.
Among all possible variants you should choose one that minimizes maximum of r's and b's nesting depth. If there are multiple solutions you can print any of them.
Input
The first line contains an even integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of RBS s.
The second line contains regular bracket sequence s (|s| = n, s_i ∈ \{"(", ")"\}).
Output
Print single string t of length n consisting of "0"-s and "1"-s. If t_i is equal to 0 then character s_i belongs to RBS r, otherwise s_i belongs to b.
Examples
Input
2
()
Output
11
Input
4
(())
Output
0101
Input
10
((()())())
Output
0110001111
Note
In the first example one of optimal solutions is s = "\color{blue}{()}". r is empty and b = "()". The answer is max(0, 1) = 1.
In the second example it's optimal to make s = "\color{red}{(}\color{blue}{(}\color{red}{)}\color{blue}{)}". r = b = "()" and the answer is 1.
In the third example we can make s = "\color{red}{(}\color{blue}{((}\color{red}{)()}\color{blue}{)())}". r = "()()" and b = "(()())" and the answer is 2. | instruction | 0 | 56,449 | 21 | 112,898 |
Tags: constructive algorithms, greedy
Correct Solution:
```
n = int(input())
l = input()
r = 0
b = 0
maxb = 0
for j in range(n):
if(l[j]=='('):
if(r>b):
b+=1
print("1",end="")
else:
r+=1
print("0",end="")
else:
if(r<b):
b-=1
print("1",end="")
else:
r-=1
print("0",end="")
print()
``` | output | 1 | 56,449 | 21 | 112,899 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string is called bracket sequence if it does not contain any characters other than "(" and ")". A bracket sequence is called regular (shortly, RBS) if it is possible to obtain correct arithmetic expression by inserting characters "+" and "1" into this sequence. For example, "", "(())" and "()()" are RBS and ")(" and "(()" are not.
We can see that each opening bracket in RBS is paired with some closing bracket, and, using this fact, we can define nesting depth of the RBS as maximum number of bracket pairs, such that the 2-nd pair lies inside the 1-st one, the 3-rd one — inside the 2-nd one and so on. For example, nesting depth of "" is 0, "()()()" is 1 and "()((())())" is 3.
Now, you are given RBS s of even length n. You should color each bracket of s into one of two colors: red or blue. Bracket sequence r, consisting only of red brackets, should be RBS, and bracket sequence, consisting only of blue brackets b, should be RBS. Any of them can be empty. You are not allowed to reorder characters in s, r or b. No brackets can be left uncolored.
Among all possible variants you should choose one that minimizes maximum of r's and b's nesting depth. If there are multiple solutions you can print any of them.
Input
The first line contains an even integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of RBS s.
The second line contains regular bracket sequence s (|s| = n, s_i ∈ \{"(", ")"\}).
Output
Print single string t of length n consisting of "0"-s and "1"-s. If t_i is equal to 0 then character s_i belongs to RBS r, otherwise s_i belongs to b.
Examples
Input
2
()
Output
11
Input
4
(())
Output
0101
Input
10
((()())())
Output
0110001111
Note
In the first example one of optimal solutions is s = "\color{blue}{()}". r is empty and b = "()". The answer is max(0, 1) = 1.
In the second example it's optimal to make s = "\color{red}{(}\color{blue}{(}\color{red}{)}\color{blue}{)}". r = b = "()" and the answer is 1.
In the third example we can make s = "\color{red}{(}\color{blue}{((}\color{red}{)()}\color{blue}{)())}". r = "()()" and b = "(()())" and the answer is 2.
Submitted Solution:
```
n = int(input())
s = input()
bal = 0
#ans = ''
for c in s:
if c == ')':
bal -= 1
#print(bal)
print(bal & 1, end = '')
if c == '(':
bal += 1
#print(ans)
``` | instruction | 0 | 56,450 | 21 | 112,900 |
Yes | output | 1 | 56,450 | 21 | 112,901 |
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