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 |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a bracket sequence s consisting of n opening '(' and closing ')' brackets.
A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example, bracket sequences "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.
You can change the type of some bracket s_i. It means that if s_i = ')' then you can change it to '(' and vice versa.
Your task is to calculate the number of positions i such that if you change the type of the i-th bracket, then the resulting bracket sequence becomes regular.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^6) — the length of the bracket sequence.
The second line of the input contains the string s consisting of n opening '(' and closing ')' brackets.
Output
Print one integer — the number of positions i such that if you change the type of the i-th bracket, then the resulting bracket sequence becomes regular.
Examples
Input
6
(((())
Output
3
Input
6
()()()
Output
0
Input
1
)
Output
0
Input
8
)))(((((
Output
0 | instruction | 0 | 45 | 21 | 90 |
Tags: implementation
Correct Solution:
```
from sys import stdin, stdout
import collections
import math
N = int(input())
s = input()
level = 0
L = [0]*N
for i in range(N):
if s[i]=='(':
level += 1
else:
level -= 1
L[i] = level
if level!=2 and level!=-2:
print(0)
quit()
mini = level
res = 0
if level==2:
last = [0]*N
front = [0]*N
for i in range(N-1,-1,-1):
mini = min(mini,L[i])
last[i] = mini
mini = L[0]
for i in range(N):
mini = min(mini,L[i])
front[i] = mini
for i in range(N):
if front[i]>=0 and last[i]>=2 and s[i]=='(':
res += 1
#print(front,last)
if level==-2:
last = [0]*N
front = [0]*N
for i in range(N-1,-1,-1):
mini = min(mini,L[i])
last[i] = mini
mini = 0
for i in range(N):
front[i] = mini
mini = min(mini,L[i])
for i in range(N):
if front[i]>=0 and last[i]>=-2 and s[i]==')':
res += 1
#print(front,last)
print(res)
``` | output | 1 | 45 | 21 | 91 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a bracket sequence s consisting of n opening '(' and closing ')' brackets.
A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example, bracket sequences "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.
You can change the type of some bracket s_i. It means that if s_i = ')' then you can change it to '(' and vice versa.
Your task is to calculate the number of positions i such that if you change the type of the i-th bracket, then the resulting bracket sequence becomes regular.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^6) — the length of the bracket sequence.
The second line of the input contains the string s consisting of n opening '(' and closing ')' brackets.
Output
Print one integer — the number of positions i such that if you change the type of the i-th bracket, then the resulting bracket sequence becomes regular.
Examples
Input
6
(((())
Output
3
Input
6
()()()
Output
0
Input
1
)
Output
0
Input
8
)))(((((
Output
0 | instruction | 0 | 46 | 21 | 92 |
Tags: implementation
Correct Solution:
```
n = int(input())
s = input().strip()
a = [0] * (n + 1)
m = [0] * (n + 1)
for i in range(n):
a[i] = a[i-1] + (1 if s[i] == "(" else -1)
m[i] = min(m[i-1], a[i])
ans = 0
mm = a[n - 1]
for j in range(n - 1, -1, -1):
mm = min(mm, a[j])
if s[j] == "(":
ans += a[n - 1] == 2 and mm == 2 and m[j - 1] >= 0
else:
ans += a[n - 1] == -2 and mm == -2 and m[j - 1] >= 0
print(ans)
``` | output | 1 | 46 | 21 | 93 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a bracket sequence s consisting of n opening '(' and closing ')' brackets.
A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example, bracket sequences "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.
You can change the type of some bracket s_i. It means that if s_i = ')' then you can change it to '(' and vice versa.
Your task is to calculate the number of positions i such that if you change the type of the i-th bracket, then the resulting bracket sequence becomes regular.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^6) — the length of the bracket sequence.
The second line of the input contains the string s consisting of n opening '(' and closing ')' brackets.
Output
Print one integer — the number of positions i such that if you change the type of the i-th bracket, then the resulting bracket sequence becomes regular.
Examples
Input
6
(((())
Output
3
Input
6
()()()
Output
0
Input
1
)
Output
0
Input
8
)))(((((
Output
0 | instruction | 0 | 47 | 21 | 94 |
Tags: implementation
Correct Solution:
```
n = int(input())
s = input()
diff, cura, curb = [], 0, 0
for item in s:
if item == '(':
cura += 1
else:
curb += 1
diff.append(cura - curb)
if diff[-1] == 2:
if min(diff) < 0:
print(0)
else:
ans = 0
for i in range(n-1, -1, -1):
if diff[i] < 2:
break
if s[i] == '(' and diff[i] >= 2:
ans += 1
print(ans)
elif diff[-1] == -2:
if min(diff) < -2:
print(0)
else:
ans = 0
for i in range(0, n-1):
if s[i] == ')' and diff[i] >= 0:
ans += 1
elif diff[i] < 0:
if s[i] == ')':
ans += 1
break
print(ans)
else:
print(0)
``` | output | 1 | 47 | 21 | 95 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a bracket sequence s consisting of n opening '(' and closing ')' brackets.
A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example, bracket sequences "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.
You can change the type of some bracket s_i. It means that if s_i = ')' then you can change it to '(' and vice versa.
Your task is to calculate the number of positions i such that if you change the type of the i-th bracket, then the resulting bracket sequence becomes regular.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^6) — the length of the bracket sequence.
The second line of the input contains the string s consisting of n opening '(' and closing ')' brackets.
Output
Print one integer — the number of positions i such that if you change the type of the i-th bracket, then the resulting bracket sequence becomes regular.
Examples
Input
6
(((())
Output
3
Input
6
()()()
Output
0
Input
1
)
Output
0
Input
8
)))(((((
Output
0 | instruction | 0 | 48 | 21 | 96 |
Tags: implementation
Correct Solution:
```
n = int(input())
s = input()
if n % 2 == 1:
print(0)
else:
a, b, diff = [], [], []
cura, curb = 0, 0
for item in s:
if item == '(':
cura += 1
else:
curb += 1
a.append(cura)
b.append(curb)
cur_diff = cura - curb
diff.append(cur_diff)
if a[-1] - b[-1] == 2:
if min(diff) < 0:
print(0)
else:
ans = 0
for i in range(n-1, -1, -1):
if diff[i] < 2:
break
if s[i] == '(' and diff[i] >= 2:
ans += 1
print(ans)
elif b[-1] - a[-1] == 2:
if min(diff) < -2:
print(0)
else:
ans = 0
for i in range(0, n-1):
if s[i] == ')' and diff[i] >= 0:
ans += 1
elif diff[i] < 0:
if s[i] == ')':
ans += 1
break
print(ans)
else:
print(0)
``` | output | 1 | 48 | 21 | 97 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a bracket sequence s consisting of n opening '(' and closing ')' brackets.
A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example, bracket sequences "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.
You can change the type of some bracket s_i. It means that if s_i = ')' then you can change it to '(' and vice versa.
Your task is to calculate the number of positions i such that if you change the type of the i-th bracket, then the resulting bracket sequence becomes regular.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^6) — the length of the bracket sequence.
The second line of the input contains the string s consisting of n opening '(' and closing ')' brackets.
Output
Print one integer — the number of positions i such that if you change the type of the i-th bracket, then the resulting bracket sequence becomes regular.
Examples
Input
6
(((())
Output
3
Input
6
()()()
Output
0
Input
1
)
Output
0
Input
8
)))(((((
Output
0 | instruction | 0 | 49 | 21 | 98 |
Tags: implementation
Correct Solution:
```
import sys
def main():
_ = sys.stdin.readline()
s = sys.stdin.readline().strip()
prefix = [0] * (len(s) + 1)
canPrefix = [True] * (len(s) + 1)
count = 0
for i, c in enumerate(s):
count = count+1 if c == '(' else count-1
isPossible = True if count >= 0 and canPrefix[i] else False
prefix[i+1] = count
canPrefix[i+1] = isPossible
sufix = [0] * (len(s) + 1)
canSuffix = [True] * (len(s) + 1)
count = 0
for i, c in enumerate(reversed(s)):
count = count + 1 if c == ')' else count - 1
isPossible = True if count >= 0 and canSuffix[len(s) - i] else False
sufix[len(s)-1-i] = count
canSuffix[len(s)-1-i] = isPossible
ans = 0
for i, c in enumerate(s):
if canPrefix[i] == False or canSuffix[i+1] == False:
continue
elif c == '(' and prefix[i] - 1 - sufix[i+1] == 0:
ans += 1
elif c == ')' and prefix[i] + 1 - sufix[i+1] == 0:
ans += 1
print(ans)
if __name__ == "__main__":
main()
``` | output | 1 | 49 | 21 | 99 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a bracket sequence s consisting of n opening '(' and closing ')' brackets.
A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example, bracket sequences "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.
You can change the type of some bracket s_i. It means that if s_i = ')' then you can change it to '(' and vice versa.
Your task is to calculate the number of positions i such that if you change the type of the i-th bracket, then the resulting bracket sequence becomes regular.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^6) — the length of the bracket sequence.
The second line of the input contains the string s consisting of n opening '(' and closing ')' brackets.
Output
Print one integer — the number of positions i such that if you change the type of the i-th bracket, then the resulting bracket sequence becomes regular.
Examples
Input
6
(((())
Output
3
Input
6
()()()
Output
0
Input
1
)
Output
0
Input
8
)))(((((
Output
0 | instruction | 0 | 50 | 21 | 100 |
Tags: implementation
Correct Solution:
```
def ii():
return int(input())
def mi():
return map(int, input().split())
def li():
return list(mi())
n = ii()
s = input().strip()
a = [0] * (n + 1)
m = [0] * (n + 1)
for i in range(n):
a[i] = a[i - 1] + (1 if s[i] == '(' else -1)
m[i] = min(m[i - 1], a[i])
ans = 0
mm = a[n - 1]
for j in range(n - 1, -1, -1):
mm = min(mm, a[j])
if s[j] == '(':
if a[n - 1] == 2 and mm == 2 and m[j - 1] >= 0:
ans += 1
else:
if a[n - 1] == -2 and mm == -2 and m[j - 1] >= 0:
ans += 1
print(ans)
``` | output | 1 | 50 | 21 | 101 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a bracket sequence s consisting of n opening '(' and closing ')' brackets.
A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example, bracket sequences "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.
You can change the type of some bracket s_i. It means that if s_i = ')' then you can change it to '(' and vice versa.
Your task is to calculate the number of positions i such that if you change the type of the i-th bracket, then the resulting bracket sequence becomes regular.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^6) — the length of the bracket sequence.
The second line of the input contains the string s consisting of n opening '(' and closing ')' brackets.
Output
Print one integer — the number of positions i such that if you change the type of the i-th bracket, then the resulting bracket sequence becomes regular.
Examples
Input
6
(((())
Output
3
Input
6
()()()
Output
0
Input
1
)
Output
0
Input
8
)))(((((
Output
0 | instruction | 0 | 51 | 21 | 102 |
Tags: implementation
Correct Solution:
```
# Created by nikita at 30/12/2018
n = int(input())
s = input()
prefBal = [0] * n
prefBal.append(0)
prefCan = [False] * n
prefCan.append(True)
suffBal = [0] * n
suffBal.append(0)
suffCan = [False] * n
suffCan.append(True)
currBal = 0
currCan = True
for i in range(n):
if s[i] == '(':
prefBal[i] = currBal + 1
else:
prefBal[i] = currBal - 1
currBal = prefBal[i]
prefCan[i] = currCan and (prefBal[i] >= 0)
currCan = prefCan[i]
currBal = 0
currCan = True
for i in range(n-1, -1,- 1):
if s[i] == ')':
suffBal[i] = currBal + 1
else:
suffBal[i] = currBal - 1
currBal = suffBal[i]
suffCan[i] = currCan and (suffBal[i] >= 0)
currCan = suffCan[i]
# print(prefBal)
# print(prefCan)
# print(suffBal)
# print(suffCan)
ans = 0
for i in range(n):
if s[i] == '(':
if prefCan[i-1] and suffCan[i+1] and prefBal[i-1] - 1 - suffBal[i+1] == 0:
ans += 1
if s[i] == ')':
if prefCan[i-1] and suffCan[i+1] and prefBal[i-1] + 1 - suffBal[i+1] == 0:
ans += 1
print(ans)
``` | output | 1 | 51 | 21 | 103 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a bracket sequence s consisting of n opening '(' and closing ')' brackets.
A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example, bracket sequences "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.
You can change the type of some bracket s_i. It means that if s_i = ')' then you can change it to '(' and vice versa.
Your task is to calculate the number of positions i such that if you change the type of the i-th bracket, then the resulting bracket sequence becomes regular.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^6) — the length of the bracket sequence.
The second line of the input contains the string s consisting of n opening '(' and closing ')' brackets.
Output
Print one integer — the number of positions i such that if you change the type of the i-th bracket, then the resulting bracket sequence becomes regular.
Examples
Input
6
(((())
Output
3
Input
6
()()()
Output
0
Input
1
)
Output
0
Input
8
)))(((((
Output
0 | instruction | 0 | 52 | 21 | 104 |
Tags: implementation
Correct Solution:
```
def main():
n = int(input())
s = ' ' + input()
a = [0] * (n + 1)
for i in range(1, n + 1):
if s[i] == '(':
a[i] = a[i - 1] + 1
else:
a[i] = a[i - 1] - 1
# print(a) # debug
if a[n] != 2 and a[n] != -2:
print(0)
return
if min(a) < -2:
print(0)
return
if a[n] == 2:
if min(a) < 0:
print(0)
return
for i in range(n, -1, -1):
if a[i] == 1:
print(s[(i + 1):].count('('))
break
else:
for i in range(n):
if a[i] == -1:
print(s[:i + 1].count(')'))
break
if __name__ == '__main__':
main()
``` | output | 1 | 52 | 21 | 105 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a bracket sequence s consisting of n opening '(' and closing ')' brackets.
A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example, bracket sequences "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.
You can change the type of some bracket s_i. It means that if s_i = ')' then you can change it to '(' and vice versa.
Your task is to calculate the number of positions i such that if you change the type of the i-th bracket, then the resulting bracket sequence becomes regular.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^6) — the length of the bracket sequence.
The second line of the input contains the string s consisting of n opening '(' and closing ')' brackets.
Output
Print one integer — the number of positions i such that if you change the type of the i-th bracket, then the resulting bracket sequence becomes regular.
Examples
Input
6
(((())
Output
3
Input
6
()()()
Output
0
Input
1
)
Output
0
Input
8
)))(((((
Output
0
Submitted Solution:
```
import sys,math
from collections import defaultdict
n = int(input())
#n,k = [int(__) for __ in input().split()]
#arr = [int(__) for __ in input().split()]
st = input()
#lines = sys.stdin.readlines()
if n%2:
print(0)
exit()
x = st.count('(')
if x not in (n // 2 - 1, n // 2 + 1):
print(0)
exit()
rem = ')'
if x > n // 2:
rem = '('
arrpr = [0] * (n+2)
arrpo = [0] * (n+2)
for i in range(n):
if st[i] == '(':
arrpr[i+1] += arrpr[i] + 1
else:
arrpr[i+1] = arrpr[i] - 1
if st[n-1-i] == ')':
arrpo[n-i] = arrpo[n+1-i] + 1
else:
arrpo[n-i] = arrpo[n+1-i] - 1
#print(arrpr)
#print(arrpo)
valpr = [False] * (n+2)
valpo = [False] * (n+2)
valpr[0] = valpo[-1] = True
for i in range(1,n+1):
valpr[i] = arrpr[i-1] >= 0 and valpr[i-1]
valpo[n+1-i] = arrpo[n+1-i] >= 0 and valpo[n-i+2]
res = 0
for i in range(1,n+1):
if valpr[i-1] == False or valpo[i+1] == False:
continue
if arrpr[i-1] - arrpo[i+1] == 1 and st[i-1] == '(' and arrpr[i-1] > 0:
#print(i)
res += 1
elif st[i-1] == ')' and arrpr[i-1] - arrpo[i+1] == -1 and arrpo[i+1] > 0:
res += 1
#print(valpr)
#print(valpo)
print(res)
``` | instruction | 0 | 53 | 21 | 106 |
Yes | output | 1 | 53 | 21 | 107 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a bracket sequence s consisting of n opening '(' and closing ')' brackets.
A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example, bracket sequences "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.
You can change the type of some bracket s_i. It means that if s_i = ')' then you can change it to '(' and vice versa.
Your task is to calculate the number of positions i such that if you change the type of the i-th bracket, then the resulting bracket sequence becomes regular.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^6) — the length of the bracket sequence.
The second line of the input contains the string s consisting of n opening '(' and closing ')' brackets.
Output
Print one integer — the number of positions i such that if you change the type of the i-th bracket, then the resulting bracket sequence becomes regular.
Examples
Input
6
(((())
Output
3
Input
6
()()()
Output
0
Input
1
)
Output
0
Input
8
)))(((((
Output
0
Submitted Solution:
```
def E():
n = int(input())
brackets = input()
if(n%2):
print(0)
return
cumm_brackets = []
for b in brackets:
if(not len(cumm_brackets)): cumm_brackets.append(1 if b=='(' else -1)
else:
cumm_brackets.append(cumm_brackets[-1]+1 if b=='(' else cumm_brackets[-1]-1)
if(abs(cumm_brackets[-1])!= 2):
print(0)
return
if(cumm_brackets[-1]==2):
if(-1 in cumm_brackets):
print(0)
return
last_under_2_index = 0
for i in range(n-1):
if cumm_brackets[i]<2:
last_under_2_index = i
answer = 0
for i in range(last_under_2_index+1,n):
if(brackets[i]=='('): answer+=1
if(cumm_brackets[-1]== -2):
if(min(cumm_brackets)<=-3):
print(0)
return
for first_under_minus_2_index in range(n):
if cumm_brackets[first_under_minus_2_index]==-1:
break
answer = 0
for i in range(first_under_minus_2_index+1):
if(brackets[i]==')'): answer+=1
print(answer)
E()
``` | instruction | 0 | 54 | 21 | 108 |
Yes | output | 1 | 54 | 21 | 109 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a bracket sequence s consisting of n opening '(' and closing ')' brackets.
A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example, bracket sequences "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.
You can change the type of some bracket s_i. It means that if s_i = ')' then you can change it to '(' and vice versa.
Your task is to calculate the number of positions i such that if you change the type of the i-th bracket, then the resulting bracket sequence becomes regular.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^6) — the length of the bracket sequence.
The second line of the input contains the string s consisting of n opening '(' and closing ')' brackets.
Output
Print one integer — the number of positions i such that if you change the type of the i-th bracket, then the resulting bracket sequence becomes regular.
Examples
Input
6
(((())
Output
3
Input
6
()()()
Output
0
Input
1
)
Output
0
Input
8
)))(((((
Output
0
Submitted Solution:
```
def read(type = 1):
if type:
file = open("input.dat", "r")
n = int(file.readline())
a = file.readline()
file.close()
else:
n = int(input().strip())
a = input().strip()
return n, a
def solve():
sol = 0
vs = []
v = 0
for i in range(n):
if a[i] == "(":
v += 1
else:
v -= 1
vs.append(v)
mins = [10000000 for i in range(n)]
last = n
for i in range(n):
if i:
mins[n-i-1] = min(vs[n-i-1], mins[n-i])
else:
mins[n-i-1] = vs[n-i-1]
if vs[n-i-1] < 0:
last = n-i-1
for i in range(n):
if a[i] == "(" and vs[n-1] == 2:
if i:
if mins[i] >= 2:
sol += 1
if a[i] == ")" and vs[n-1] == -2:
if i != n-1:
if mins[i] >= -2:
sol += 1
if i == last:
break
return sol
n, a = read(0)
sol = solve()
print(sol)
``` | instruction | 0 | 55 | 21 | 110 |
Yes | output | 1 | 55 | 21 | 111 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a bracket sequence s consisting of n opening '(' and closing ')' brackets.
A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example, bracket sequences "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.
You can change the type of some bracket s_i. It means that if s_i = ')' then you can change it to '(' and vice versa.
Your task is to calculate the number of positions i such that if you change the type of the i-th bracket, then the resulting bracket sequence becomes regular.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^6) — the length of the bracket sequence.
The second line of the input contains the string s consisting of n opening '(' and closing ')' brackets.
Output
Print one integer — the number of positions i such that if you change the type of the i-th bracket, then the resulting bracket sequence becomes regular.
Examples
Input
6
(((())
Output
3
Input
6
()()()
Output
0
Input
1
)
Output
0
Input
8
)))(((((
Output
0
Submitted Solution:
```
import sys
input = sys.stdin.readline
n = int(input())
s = list(input().rstrip())
c1, c2 = 0, 0
for i in s:
if i == "(":
c1 += 1
else:
c2 += 1
if abs(c1 - c2) ^ 2:
ans = 0
else:
x, now = [0] * n, 0
ans, m = 0, n
if c1 > c2:
for i in range(n):
now += 1 if s[i] == "(" else -1
x[i] = now
for i in range(n - 1, -1, -1):
if s[i] == ")":
m = min(m, x[i])
elif 1 < m and 1 < x[i]:
ans += 1
else:
for i in range(n - 1, -1, -1):
now += 1 if s[i] == ")" else -1
x[i] = now
for i in range(n):
if s[i] == "(":
m = min(m, x[i])
elif 1 < m and 1 < x[i]:
ans += 1
if min(x) < 0:
ans = 0
print(ans)
``` | instruction | 0 | 56 | 21 | 112 |
Yes | output | 1 | 56 | 21 | 113 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a bracket sequence s consisting of n opening '(' and closing ')' brackets.
A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example, bracket sequences "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.
You can change the type of some bracket s_i. It means that if s_i = ')' then you can change it to '(' and vice versa.
Your task is to calculate the number of positions i such that if you change the type of the i-th bracket, then the resulting bracket sequence becomes regular.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^6) — the length of the bracket sequence.
The second line of the input contains the string s consisting of n opening '(' and closing ')' brackets.
Output
Print one integer — the number of positions i such that if you change the type of the i-th bracket, then the resulting bracket sequence becomes regular.
Examples
Input
6
(((())
Output
3
Input
6
()()()
Output
0
Input
1
)
Output
0
Input
8
)))(((((
Output
0
Submitted Solution:
```
def isbalanced(ar,l,r):
if (l==r and ar[0]==1 and ar[len(ar)-1]==-1):
return True
else:
return False
n = int(input())
st = input()
l=r=0
if (n&1==0):
ar = [0]*n
c=0
for i in range(n):
if st[i] == "(":
ar[i] = 1
l += 1
else:
ar[i] = -1
r += 1
for i in range(n):
if (ar[i]==1):
ar[i]=-1
l-=1;r+=1
if (isbalanced(ar,l,r)):
c+=1
ar[i]=1
l+=1;r-=1
else:
ar[i]=1
l+=1;r-=1
if(isbalanced(ar,l,r)):
c+=1
ar[i]=-1
l-=1;r+=1
print(c)
else:
print(0)
``` | instruction | 0 | 57 | 21 | 114 |
No | output | 1 | 57 | 21 | 115 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a bracket sequence s consisting of n opening '(' and closing ')' brackets.
A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example, bracket sequences "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.
You can change the type of some bracket s_i. It means that if s_i = ')' then you can change it to '(' and vice versa.
Your task is to calculate the number of positions i such that if you change the type of the i-th bracket, then the resulting bracket sequence becomes regular.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^6) — the length of the bracket sequence.
The second line of the input contains the string s consisting of n opening '(' and closing ')' brackets.
Output
Print one integer — the number of positions i such that if you change the type of the i-th bracket, then the resulting bracket sequence becomes regular.
Examples
Input
6
(((())
Output
3
Input
6
()()()
Output
0
Input
1
)
Output
0
Input
8
)))(((((
Output
0
Submitted Solution:
```
def main():
n = int(input())
s = input()
dpleft = []
dpright = []
for i in range(n):
dpleft.append([0,0])
dpright.append([0,0])
left = 0
right = 0
for i in range(n):
if s[i] == '(':
left += 1
else:
if left > 0:
left -= 1
else:
right += 1
dpleft[i][0] = left
dpleft[i][1] = right
left = 0
right = 0
for i in range(n-1,-1,-1):
if s[i] == ')':
right += 1
else:
if right > 0:
right -= 1
else:
left += 1
dpright[i][0] = left
dpright[i][1] = right
ans = 0
#print(dpleft,dpright)
for i in range(1,n-1):
if dpleft[i-1][1] == 0 and dpright[i+1][0] == 0:
left = dpleft[i-1][0]
right = dpright[i+1][1]
if s[i] == ')':
left += 1
else:
right += 1
if left == right:
ans += 1
print(ans)
main()
``` | instruction | 0 | 58 | 21 | 116 |
No | output | 1 | 58 | 21 | 117 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a bracket sequence s consisting of n opening '(' and closing ')' brackets.
A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example, bracket sequences "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.
You can change the type of some bracket s_i. It means that if s_i = ')' then you can change it to '(' and vice versa.
Your task is to calculate the number of positions i such that if you change the type of the i-th bracket, then the resulting bracket sequence becomes regular.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^6) — the length of the bracket sequence.
The second line of the input contains the string s consisting of n opening '(' and closing ')' brackets.
Output
Print one integer — the number of positions i such that if you change the type of the i-th bracket, then the resulting bracket sequence becomes regular.
Examples
Input
6
(((())
Output
3
Input
6
()()()
Output
0
Input
1
)
Output
0
Input
8
)))(((((
Output
0
Submitted Solution:
```
n = int(input())
S = input()
counts = []
def solve(s):
openCount = 0
closeCount = 0
posCount = 0
if n % 2 == 1:
return posCount
for i in range(n):
if s[i] == '(':
openCount += 1
else:
closeCount += 1
if closeCount - openCount > 2:
return posCount
else:
counts.append([openCount, closeCount])
if abs(openCount-closeCount) == 2:
if openCount > closeCount:
for i in range(n):
if counts[i][0] - counts[i][1] >= 2 and S[i] == '(':
posCount += 1
return posCount
else:
for i in range(n):
if counts[i][1] - counts[i][0] < 2 and S[i] == ')':
posCount += 1
if counts[i][1] - counts[i][0] >= 1:
break
return posCount
else:
return posCount
print(solve(S))
``` | instruction | 0 | 59 | 21 | 118 |
No | output | 1 | 59 | 21 | 119 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a bracket sequence s consisting of n opening '(' and closing ')' brackets.
A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example, bracket sequences "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.
You can change the type of some bracket s_i. It means that if s_i = ')' then you can change it to '(' and vice versa.
Your task is to calculate the number of positions i such that if you change the type of the i-th bracket, then the resulting bracket sequence becomes regular.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^6) — the length of the bracket sequence.
The second line of the input contains the string s consisting of n opening '(' and closing ')' brackets.
Output
Print one integer — the number of positions i such that if you change the type of the i-th bracket, then the resulting bracket sequence becomes regular.
Examples
Input
6
(((())
Output
3
Input
6
()()()
Output
0
Input
1
)
Output
0
Input
8
)))(((((
Output
0
Submitted Solution:
```
from collections import deque
n = int(input())
s = input()
def sol(n, s):
times = 1
stack = deque([])
for i in range(n):
b = s[i]
if b=='(':
stack.append(i)
else:
if len(stack)==0 and times>0:
stack.append(i)
times -= 1
pos = i
elif len(stack) > 0 :
stack.popleft()
else:
return 0
if len(stack) == 2 and times>0:
return stack[-1]
elif times==0 and len(stack)==0:
return pos
else:
return 0
print(sol(n, s))
``` | instruction | 0 | 60 | 21 | 120 |
No | output | 1 | 60 | 21 | 121 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alex was programming while Valentina (his toddler daughter) got there and started asking many questions about the round brackets (or parenthesis) in the code. He explained her a bit and when she got it he gave her a task in order to finish his code on time.
For the purpose of this problem we consider only strings consisting of opening and closing round brackets, that is characters '(' and ')'.
The sequence of brackets is called correct if:
1. it's empty;
2. it's a correct sequence of brackets, enclosed in a pair of opening and closing brackets;
3. it's a concatenation of two correct sequences of brackets.
For example, the sequences "()()" and "((()))(())" are correct, while ")(()", "(((((" and "())" are not.
Alex took a piece of paper, wrote a string s consisting of brackets and asked Valentina to count the number of distinct non-empty substrings of s that are correct sequences of brackets. In other words, her task is to count the number of non-empty correct sequences of brackets that occur in a string s as a substring (don't mix up with subsequences).
When Valentina finished the task, Alex noticed he doesn't know the answer. Help him don't loose face in front of Valentina and solve the problem!
Input
The first line of the input contains an integer n (1 ≤ n ≤ 500 000) — the length of the string s.
The second line contains a string s of length n consisting of only '(' and ')'.
Output
Print the number of distinct non-empty correct sequences that occur in s as substring.
Examples
Input
10
()()()()()
Output
5
Input
7
)(())()
Output
3
Note
In the first sample, there are 5 distinct substrings we should count: "()", "()()", "()()()", "()()()()" and "()()()()()".
In the second sample, there are 3 distinct substrings we should count: "()", "(())" and "(())()".
Submitted Solution:
```
n = int(input())
s = input()
e = 0
z = 0
for c in s:
if c == "(":
e += 1
else:
e -= 1
z += 1
if e < 0:
z -= 1
e = 0
print(z)
``` | instruction | 0 | 442 | 21 | 884 |
No | output | 1 | 442 | 21 | 885 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alex was programming while Valentina (his toddler daughter) got there and started asking many questions about the round brackets (or parenthesis) in the code. He explained her a bit and when she got it he gave her a task in order to finish his code on time.
For the purpose of this problem we consider only strings consisting of opening and closing round brackets, that is characters '(' and ')'.
The sequence of brackets is called correct if:
1. it's empty;
2. it's a correct sequence of brackets, enclosed in a pair of opening and closing brackets;
3. it's a concatenation of two correct sequences of brackets.
For example, the sequences "()()" and "((()))(())" are correct, while ")(()", "(((((" and "())" are not.
Alex took a piece of paper, wrote a string s consisting of brackets and asked Valentina to count the number of distinct non-empty substrings of s that are correct sequences of brackets. In other words, her task is to count the number of non-empty correct sequences of brackets that occur in a string s as a substring (don't mix up with subsequences).
When Valentina finished the task, Alex noticed he doesn't know the answer. Help him don't loose face in front of Valentina and solve the problem!
Input
The first line of the input contains an integer n (1 ≤ n ≤ 500 000) — the length of the string s.
The second line contains a string s of length n consisting of only '(' and ')'.
Output
Print the number of distinct non-empty correct sequences that occur in s as substring.
Examples
Input
10
()()()()()
Output
5
Input
7
)(())()
Output
3
Note
In the first sample, there are 5 distinct substrings we should count: "()", "()()", "()()()", "()()()()" and "()()()()()".
In the second sample, there are 3 distinct substrings we should count: "()", "(())" and "(())()".
Submitted Solution:
```
# http://codeforces.com/problemset/problem/653/F
def BracketCounter(S):
counter = 0
length = int(S.split('\n')[0].strip())
bracket_string = S.split('\n')[1].strip()
#print('bracket string is: %s' %bracket_string)
if len(bracket_string) == 0:
#print('zero length string')
return 0
ind = bracket_string[0]
if ind == '(':
#print("first character is a open")
ind2 = bracket_string.find(')')
if ind2 == -1:
#print("no close brackets found")
return 0
else:
#print("close bracket at %s" %ind2)
L_bracket_string = list(bracket_string)
L_bracket_string.pop(ind2)
L_bracket_string.pop(0)
S_new = str(length-2) + '\n' + ''.join(L_bracket_string)
#print("This is the new string %s" %S_new)
ret_counter = BracketCounter(S_new)
counter += ret_counter + 1
# elif bracket_string[:] == ')':
# print("close bracket is at end")
else:
#print("first character is a close, disregard")
S_new = str(length-1) + '\n' + bracket_string[1:]
ret_counter = BracketCounter(S_new)
counter += ret_counter
return counter
if __name__ == '__main__':
S = str(input()) + '\n' + str(input())
counted = BracketCounter(S)
print(counted)
else:
print("I am being imported")
``` | instruction | 0 | 443 | 21 | 886 |
No | output | 1 | 443 | 21 | 887 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alex was programming while Valentina (his toddler daughter) got there and started asking many questions about the round brackets (or parenthesis) in the code. He explained her a bit and when she got it he gave her a task in order to finish his code on time.
For the purpose of this problem we consider only strings consisting of opening and closing round brackets, that is characters '(' and ')'.
The sequence of brackets is called correct if:
1. it's empty;
2. it's a correct sequence of brackets, enclosed in a pair of opening and closing brackets;
3. it's a concatenation of two correct sequences of brackets.
For example, the sequences "()()" and "((()))(())" are correct, while ")(()", "(((((" and "())" are not.
Alex took a piece of paper, wrote a string s consisting of brackets and asked Valentina to count the number of distinct non-empty substrings of s that are correct sequences of brackets. In other words, her task is to count the number of non-empty correct sequences of brackets that occur in a string s as a substring (don't mix up with subsequences).
When Valentina finished the task, Alex noticed he doesn't know the answer. Help him don't loose face in front of Valentina and solve the problem!
Input
The first line of the input contains an integer n (1 ≤ n ≤ 500 000) — the length of the string s.
The second line contains a string s of length n consisting of only '(' and ')'.
Output
Print the number of distinct non-empty correct sequences that occur in s as substring.
Examples
Input
10
()()()()()
Output
5
Input
7
)(())()
Output
3
Note
In the first sample, there are 5 distinct substrings we should count: "()", "()()", "()()()", "()()()()" and "()()()()()".
In the second sample, there are 3 distinct substrings we should count: "()", "(())" and "(())()".
Submitted Solution:
```
# http://codeforces.com/problemset/problem/653/F
import sys
def SubSequence(S_sub):
count = 0
#subsequence
open_count = 0
open_pos = []
close_pos = []
close_count = 0
for pos, char in enumerate(S_sub):
if char == '(':
open_count += 1
open_pos.append(pos)
elif char == ')':
close_count += 1
close_pos.append(pos)
if close_count > open_count:
#too many closed, can't be a sub sequence
string_rem = S_sub[pos+1:]
count += SubSequence(string_rem)
break
if open_count == close_count:
#found subsequence
count += BracketCounter(S_sub[0:pos+1])
string_rem = S_sub[pos+1:]
count += SubSequence(string_rem)
break
if open_count > close_count:
#loop ended and no close bracket
#don't need to do anything
count = 0
return count
def BracketCounter(S):
counter = 0
bracket_string = str(S)
print('bracket string is: %s' %bracket_string)
counter = len(S)/2
return counter
# if len(bracket_string) == 0:
# #print('zero length string')
# return 0
# ind = bracket_string[0]
# open_pos = []
# close_pos = []
# counter_bracket = -
# for pos, char in enumerate(bracket_string):
# if char == '(':
# open_pos.append(pos)
# if char == ')':
# close_pos.append(pos)
# for ix in len(open_pos):
# if ix+1 in open_pos:
# #there is another open directly after
# if ix+1 in close_pos:
# counter
# print(open_pos)
# print(close_pos)
# if ind == '(':
# #print("first character is a open")
# ind2 = bracket_string.find(')')
# if ind2 == -1:
# #print("no close brackets found")
# return 0
# else:
# #print("close bracket at %s" %ind2)
# L_bracket_string = list(bracket_string)
# L_bracket_string.pop(ind2)
# L_bracket_string.pop(0)
# S_new = str(length-2) + '\n' + ''.join(L_bracket_string)
# #print("This is the new string %s" %S_new)
# ret_counter = BracketCounter(S_new)
# counter += ret_counter + 1
# # elif bracket_string[:] == ')':
# # print("close bracket is at end")
# else:
# #print("first character is a close, disregard")
# S_new = str(length-1) + '\n' + bracket_string[1:]
# ret_counter = BracketCounter(S_new)
# counter += ret_counter
# return counter
if __name__ == '__main__':
NOPE = str(input())
S = str(input())
#S = '(())(()())(()()'
counted = SubSequence(S)
print(counted)
else:
print("I am being imported")
``` | instruction | 0 | 444 | 21 | 888 |
No | output | 1 | 444 | 21 | 889 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alex was programming while Valentina (his toddler daughter) got there and started asking many questions about the round brackets (or parenthesis) in the code. He explained her a bit and when she got it he gave her a task in order to finish his code on time.
For the purpose of this problem we consider only strings consisting of opening and closing round brackets, that is characters '(' and ')'.
The sequence of brackets is called correct if:
1. it's empty;
2. it's a correct sequence of brackets, enclosed in a pair of opening and closing brackets;
3. it's a concatenation of two correct sequences of brackets.
For example, the sequences "()()" and "((()))(())" are correct, while ")(()", "(((((" and "())" are not.
Alex took a piece of paper, wrote a string s consisting of brackets and asked Valentina to count the number of distinct non-empty substrings of s that are correct sequences of brackets. In other words, her task is to count the number of non-empty correct sequences of brackets that occur in a string s as a substring (don't mix up with subsequences).
When Valentina finished the task, Alex noticed he doesn't know the answer. Help him don't loose face in front of Valentina and solve the problem!
Input
The first line of the input contains an integer n (1 ≤ n ≤ 500 000) — the length of the string s.
The second line contains a string s of length n consisting of only '(' and ')'.
Output
Print the number of distinct non-empty correct sequences that occur in s as substring.
Examples
Input
10
()()()()()
Output
5
Input
7
)(())()
Output
3
Note
In the first sample, there are 5 distinct substrings we should count: "()", "()()", "()()()", "()()()()" and "()()()()()".
In the second sample, there are 3 distinct substrings we should count: "()", "(())" and "(())()".
Submitted Solution:
```
# http://codeforces.com/problemset/problem/653/F
def BracketCounter(S):
counter = 0
length = int(S.split('\n')[0].strip())
bracket_string = S.split('\n')[1].strip()
# print('bracket string is: %s' %bracket_string)
if len(bracket_string) == 0:
print('zero length string')
return 0
ind = bracket_string[0]
if ind == '(':
#print("first character is a open")
ind2 = bracket_string.find(')')
if ind2 == -1:
#print("no close brackets found")
return 0
else:
#print("close bracket at %s" %ind2)
L_bracket_string = list(bracket_string)
L_bracket_string.pop(ind2)
L_bracket_string.pop(0)
S_new = str(length-2) + '\n' + ''.join(L_bracket_string)
#print("This is the new string %s" %S_new)
ret_counter = BracketCounter(S_new)
counter += ret_counter + 1
# elif bracket_string[:] == ')':
# print("close bracket is at end")
else:
#print("first character is a close, disregard")
S_new = str(length-1) + '\n' + bracket_string[1:]
ret_counter = BracketCounter(S_new)
counter += ret_counter
return counter
if __name__ == '__main__':
print("I am being called directly")
# S = '10 \n ()()()()()'
# counted = BracketCounter(S)
# print(counted)
S = '7 \n )(())()'
counted = BracketCounter(S)
print(counted)
else:
print("I am being imported")
``` | instruction | 0 | 445 | 21 | 890 |
No | output | 1 | 445 | 21 | 891 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence of brackets is called balanced if one can turn it into a valid math expression by adding characters '+' and '1'. For example, sequences '(())()', '()', and '(()(()))' are balanced, while ')(', '(()', and '(()))(' are not.
You are given a binary string s of length n. Construct two balanced bracket sequences a and b of length n such that for all 1≤ i≤ n:
* if s_i=1, then a_i=b_i
* if s_i=0, then a_i≠ b_i
If it is impossible, you should report about it.
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (2≤ n≤ 2⋅ 10^5, n is even).
The next line contains a string s of length n, consisting of characters 0 and 1.
The sum of n across all test cases does not exceed 2⋅ 10^5.
Output
If such two balanced bracked sequences exist, output "YES" on the first line, otherwise output "NO". You can print each letter in any case (upper or lower).
If the answer is "YES", output the balanced bracket sequences a and b satisfying the conditions on the next two lines.
If there are multiple solutions, you may print any.
Example
Input
3
6
101101
10
1001101101
4
1100
Output
YES
()()()
((()))
YES
()()((()))
(())()()()
NO
Note
In the first test case, a="()()()" and b="((()))". The characters are equal in positions 1, 3, 4, and 6, which are the exact same positions where s_i=1.
In the second test case, a="()()((()))" and b="(())()()()". The characters are equal in positions 1, 4, 5, 7, 8, 10, which are the exact same positions where s_i=1.
In the third test case, there is no solution. | instruction | 0 | 1,083 | 21 | 2,166 |
Tags: constructive algorithms, greedy
Correct Solution:
```
from sys import stdin, stdout
t = int(stdin.readline())
for _ in range(t):
n = int(stdin.readline())
s = stdin.readline(n)
stdin.readline(1)
if s[0] == '0' or s[n-1] == '0':
stdout.write("NO\n")
continue
a = []
b = []
va, vb = 0, 0
for i in range(n):
if s[i] == '1':
if va + vb < 4:
va += 1
vb += 1
a.append('(')
b.append('(')
else:
va -= 1
vb -= 1
a.append(')')
b.append(')')
elif s[i] == '0':
if va <= 1:
va += 1
vb -= 1
a.append('(')
b.append(')')
else:
va -= 1
vb += 1
a.append(')')
b.append('(')
if not (va == vb == 2):
stdout.write("NO\n")
else:
a[-1] = ')'
b[-1] = ')'
stdout.write("YES\n")
stdout.write(''.join(a))
stdout.write("\n")
stdout.write(''.join(b))
stdout.write("\n")
``` | output | 1 | 1,083 | 21 | 2,167 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence of brackets is called balanced if one can turn it into a valid math expression by adding characters '+' and '1'. For example, sequences '(())()', '()', and '(()(()))' are balanced, while ')(', '(()', and '(()))(' are not.
You are given a binary string s of length n. Construct two balanced bracket sequences a and b of length n such that for all 1≤ i≤ n:
* if s_i=1, then a_i=b_i
* if s_i=0, then a_i≠ b_i
If it is impossible, you should report about it.
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (2≤ n≤ 2⋅ 10^5, n is even).
The next line contains a string s of length n, consisting of characters 0 and 1.
The sum of n across all test cases does not exceed 2⋅ 10^5.
Output
If such two balanced bracked sequences exist, output "YES" on the first line, otherwise output "NO". You can print each letter in any case (upper or lower).
If the answer is "YES", output the balanced bracket sequences a and b satisfying the conditions on the next two lines.
If there are multiple solutions, you may print any.
Example
Input
3
6
101101
10
1001101101
4
1100
Output
YES
()()()
((()))
YES
()()((()))
(())()()()
NO
Note
In the first test case, a="()()()" and b="((()))". The characters are equal in positions 1, 3, 4, and 6, which are the exact same positions where s_i=1.
In the second test case, a="()()((()))" and b="(())()()()". The characters are equal in positions 1, 4, 5, 7, 8, 10, which are the exact same positions where s_i=1.
In the third test case, there is no solution. | instruction | 0 | 1,084 | 21 | 2,168 |
Tags: constructive algorithms, greedy
Correct Solution:
```
import sys,functools,collections,bisect,math,heapq
input = sys.stdin.readline
#print = sys.stdout.write
sys.setrecursionlimit(200000)
mod = 10**9 + 7
t = int(input())
for _ in range(t):
n = int(input())
s = input().strip()
if s[0] == '0' or s[-1] == '0':
print('NO')
continue
d = collections.Counter(s)
if d['0']%2 or d['1']%2:
print('NO')
continue
res1 = [1]*n
res2 = [1]*n
x = d['1']//2
i = 0
while x:
if s[i] == '1':
res1[i] = '('
res2[i] = '('
x -= 1
i += 1
if x == 0:
while i < n:
if s[i] == '1':
res1[i] = ')'
res2[i] = ')'
i += 1
c = 0
for i in range(n):
if s[i] == '0':
if c%2:
res1[i] = ')'
res2[i] = '('
else:
res1[i] = '('
res2[i] = ')'
c += 1
print('YES')
print(''.join(res1))
print(''.join(res2))
``` | output | 1 | 1,084 | 21 | 2,169 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence of brackets is called balanced if one can turn it into a valid math expression by adding characters '+' and '1'. For example, sequences '(())()', '()', and '(()(()))' are balanced, while ')(', '(()', and '(()))(' are not.
You are given a binary string s of length n. Construct two balanced bracket sequences a and b of length n such that for all 1≤ i≤ n:
* if s_i=1, then a_i=b_i
* if s_i=0, then a_i≠ b_i
If it is impossible, you should report about it.
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (2≤ n≤ 2⋅ 10^5, n is even).
The next line contains a string s of length n, consisting of characters 0 and 1.
The sum of n across all test cases does not exceed 2⋅ 10^5.
Output
If such two balanced bracked sequences exist, output "YES" on the first line, otherwise output "NO". You can print each letter in any case (upper or lower).
If the answer is "YES", output the balanced bracket sequences a and b satisfying the conditions on the next two lines.
If there are multiple solutions, you may print any.
Example
Input
3
6
101101
10
1001101101
4
1100
Output
YES
()()()
((()))
YES
()()((()))
(())()()()
NO
Note
In the first test case, a="()()()" and b="((()))". The characters are equal in positions 1, 3, 4, and 6, which are the exact same positions where s_i=1.
In the second test case, a="()()((()))" and b="(())()()()". The characters are equal in positions 1, 4, 5, 7, 8, 10, which are the exact same positions where s_i=1.
In the third test case, there is no solution. | instruction | 0 | 1,085 | 21 | 2,170 |
Tags: constructive algorithms, greedy
Correct Solution:
```
import sys
input = sys.stdin.readline
def main():
n = int(input())
S = input().strip()
l = S.count("1")
o = S.count("0")
if l % 2 == 1:
print("NO")
return
if S[0] == "0" or S[-1] == "0":
print("NO")
return
print("YES")
cnt_o = 0
cnt_l = 0
T1 = []
T2 = []
for s in S:
if s == "1":
if cnt_l < l // 2:
T1.append("(")
T2.append("(")
else:
T1.append(")")
T2.append(")")
cnt_l += 1
else:
cnt_o += 1
if cnt_o % 2 == 0:
T1.append("(")
T2.append(")")
else:
T1.append(")")
T2.append("(")
print("".join(T1))
print("".join(T2))
for _ in range(int(input())):
main()
``` | output | 1 | 1,085 | 21 | 2,171 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence of brackets is called balanced if one can turn it into a valid math expression by adding characters '+' and '1'. For example, sequences '(())()', '()', and '(()(()))' are balanced, while ')(', '(()', and '(()))(' are not.
You are given a binary string s of length n. Construct two balanced bracket sequences a and b of length n such that for all 1≤ i≤ n:
* if s_i=1, then a_i=b_i
* if s_i=0, then a_i≠ b_i
If it is impossible, you should report about it.
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (2≤ n≤ 2⋅ 10^5, n is even).
The next line contains a string s of length n, consisting of characters 0 and 1.
The sum of n across all test cases does not exceed 2⋅ 10^5.
Output
If such two balanced bracked sequences exist, output "YES" on the first line, otherwise output "NO". You can print each letter in any case (upper or lower).
If the answer is "YES", output the balanced bracket sequences a and b satisfying the conditions on the next two lines.
If there are multiple solutions, you may print any.
Example
Input
3
6
101101
10
1001101101
4
1100
Output
YES
()()()
((()))
YES
()()((()))
(())()()()
NO
Note
In the first test case, a="()()()" and b="((()))". The characters are equal in positions 1, 3, 4, and 6, which are the exact same positions where s_i=1.
In the second test case, a="()()((()))" and b="(())()()()". The characters are equal in positions 1, 4, 5, 7, 8, 10, which are the exact same positions where s_i=1.
In the third test case, there is no solution. | instruction | 0 | 1,086 | 21 | 2,172 |
Tags: constructive algorithms, greedy
Correct Solution:
```
import sys
input=sys.stdin.readline
t=int(input())
for _ in range(t):
n=int(input())
s=input()
a=[]
b=[]
count1=0
for i in range(n):
if s[i]=='1':
count1+=1
count0=n-count1
if count0%2:
print('NO')
continue
leftbr0=count0//2
rightbr0=count0//2
leftbr1=n//2-leftbr0
score=0
stack=[]
for i in range(n):
if s[i]=='1':
if leftbr1:
a.append('(')
score+=1
leftbr1-=1
else:
a.append(')')
score-=1
if score<0 and stack:
rightbr0+=1
score+=2
k=stack.pop()
a[k]='('
else:
if rightbr0 and score>0:
a.append(')')
score-=1
rightbr0-=1
stack.append(i)
else:
a.append('(')
score+=1
b=[]
for i in range(n):
if s[i]=='1':
b.append(a[i])
else:
if a[i]=='(':
b.append(')')
else:
b.append('(')
score1=0
score2=0
flag=0
for i in range(n):
if a[i]=='(':
score1+=1
else:
score1-=1
if b[i]=='(':
score2+=1
else:
score2-=1
if score1<0 or score2<0:
flag=1
break
if flag or score1!=0 or score2!=0:
print('NO')
else:
print('YES')
print(''.join(a))
print(''.join(b))
``` | output | 1 | 1,086 | 21 | 2,173 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence of brackets is called balanced if one can turn it into a valid math expression by adding characters '+' and '1'. For example, sequences '(())()', '()', and '(()(()))' are balanced, while ')(', '(()', and '(()))(' are not.
You are given a binary string s of length n. Construct two balanced bracket sequences a and b of length n such that for all 1≤ i≤ n:
* if s_i=1, then a_i=b_i
* if s_i=0, then a_i≠ b_i
If it is impossible, you should report about it.
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (2≤ n≤ 2⋅ 10^5, n is even).
The next line contains a string s of length n, consisting of characters 0 and 1.
The sum of n across all test cases does not exceed 2⋅ 10^5.
Output
If such two balanced bracked sequences exist, output "YES" on the first line, otherwise output "NO". You can print each letter in any case (upper or lower).
If the answer is "YES", output the balanced bracket sequences a and b satisfying the conditions on the next two lines.
If there are multiple solutions, you may print any.
Example
Input
3
6
101101
10
1001101101
4
1100
Output
YES
()()()
((()))
YES
()()((()))
(())()()()
NO
Note
In the first test case, a="()()()" and b="((()))". The characters are equal in positions 1, 3, 4, and 6, which are the exact same positions where s_i=1.
In the second test case, a="()()((()))" and b="(())()()()". The characters are equal in positions 1, 4, 5, 7, 8, 10, which are the exact same positions where s_i=1.
In the third test case, there is no solution. | instruction | 0 | 1,087 | 21 | 2,174 |
Tags: constructive algorithms, greedy
Correct Solution:
```
import time
#start_time = time.time()
#def TIME_(): print(time.time()-start_time)
import os, sys
from io import BytesIO, IOBase
from types import GeneratorType
from bisect import bisect_left, bisect_right
from collections import defaultdict as dd, deque as dq, Counter as dc
import math, string, heapq as h
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
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 = self.os.read(self._fd, max(self.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 = self.os.read(self._fd, max(self.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:
self.os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def getInt(): return int(input())
def getStrs(): return input().split()
def getInts(): return list(map(int,input().split()))
def getStr(): return input()
def listStr(): return list(input())
def getMat(n): return [getInts() for _ in range(n)]
def getBin(): return list(map(int,list(input())))
def isInt(s): return '0' <= s[0] <= '9'
def ceil_(a,b): return a//b + (a%b > 0)
MOD = 10**9 + 7
"""
The ones must be split in half
Let's make the first half of the ones (, the second half )
"""
def solve():
N = getInt()
S = getBin()
ones = S.count(1)
if ones % 2:
print("NO")
return
ones //= 2
curr1 = curr2 = 0
A = []
B = []
one_count = 0
for i in range(N):
if S[i] == 1:
if one_count < ones:
A.append('(')
B.append('(')
curr1 += 1
curr2 += 1
one_count += 1
else:
A.append(')')
B.append(')')
curr1 -= 1
curr2 -= 1
else:
if curr1 < curr2:
A.append('(')
B.append(')')
curr1 += 1
curr2 -= 1
else:
A.append(')')
B.append('(')
curr1 -= 1
curr2 += 1
if min(curr1,curr2) < 0:
print("NO")
return
if curr1 != 0 or curr2 != 0:
print("NO")
return
print("YES")
print(''.join(A))
print(''.join(B))
return
for _ in range(getInt()):
#print(solve())
solve()
#TIME_()
``` | output | 1 | 1,087 | 21 | 2,175 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence of brackets is called balanced if one can turn it into a valid math expression by adding characters '+' and '1'. For example, sequences '(())()', '()', and '(()(()))' are balanced, while ')(', '(()', and '(()))(' are not.
You are given a binary string s of length n. Construct two balanced bracket sequences a and b of length n such that for all 1≤ i≤ n:
* if s_i=1, then a_i=b_i
* if s_i=0, then a_i≠ b_i
If it is impossible, you should report about it.
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (2≤ n≤ 2⋅ 10^5, n is even).
The next line contains a string s of length n, consisting of characters 0 and 1.
The sum of n across all test cases does not exceed 2⋅ 10^5.
Output
If such two balanced bracked sequences exist, output "YES" on the first line, otherwise output "NO". You can print each letter in any case (upper or lower).
If the answer is "YES", output the balanced bracket sequences a and b satisfying the conditions on the next two lines.
If there are multiple solutions, you may print any.
Example
Input
3
6
101101
10
1001101101
4
1100
Output
YES
()()()
((()))
YES
()()((()))
(())()()()
NO
Note
In the first test case, a="()()()" and b="((()))". The characters are equal in positions 1, 3, 4, and 6, which are the exact same positions where s_i=1.
In the second test case, a="()()((()))" and b="(())()()()". The characters are equal in positions 1, 4, 5, 7, 8, 10, which are the exact same positions where s_i=1.
In the third test case, there is no solution. | instruction | 0 | 1,088 | 21 | 2,176 |
Tags: constructive algorithms, greedy
Correct Solution:
```
#------------------Important Modules------------------#
from sys import stdin,stdout
from bisect import bisect_left as bl
from bisect import bisect_right as br
from heapq import *
from random import *
input=stdin.readline
prin=stdout.write
from random import sample
from collections import Counter,deque
from fractions import *
from math import sqrt,ceil,log2,gcd
#dist=[0]*(n+1)
mod=10**9+7
"""
class DisjSet:
def __init__(self, n):
self.rank = [1] * n
self.parent = [i for i in range(n)]
def find(self, x):
if (self.parent[x] != x):
self.parent[x] = self.find(self.parent[x])
return self.parent[x]
def union(self, x, y):
xset = self.find(x)
yset = self.find(y)
if xset == yset:
return
if self.rank[xset] < self.rank[yset]:
self.parent[xset] = yset
elif self.rank[xset] > self.rank[yset]:
self.parent[yset] = xset
else:
self.parent[yset] = xset
self.rank[xset] = self.rank[xset] + 1
"""
def f(arr,i,j,d,dist):
if i==j:
return
nn=max(arr[i:j])
for tl in range(i,j):
if arr[tl]==nn:
dist[tl]=d
#print(tl,dist[tl])
f(arr,i,tl,d+1,dist)
f(arr,tl+1,j,d+1,dist)
#return dist
def ps(n):
cp=0;lk=0;arr={}
#print(n)
#cc=0;
while n%2==0:
n=n//2
#arr.append(n);arr.append(2**(lk+1))
lk+=1
for ps in range(3,ceil(sqrt(n))+1,2):
lk=0
cc=0
while n%ps==0:
n=n//ps
#cc=1
#arr.append(n);arr.append(ps**(lk+1))
lk+=1
if n!=1:
#lk+=1
#arr[n]=1
#ans*=n;
lk+=1
return False
#return arr
#print(arr)
return True
#count=0
#dp=[[0 for i in range(m)] for j in range(n)]
#[int(x) for x in input().strip().split()]
def gcd(x, y):
while(y):
x, y = y, x % y
return x
# Driver Code
def factorials(n,r):
#This calculates ncr mod 10**9+7
slr=n;dpr=r
qlr=1;qs=1
mod=10**9+7
for ip in range(n-r+1,n+1):
qlr=(qlr*ip)%mod
for ij in range(1,r+1):
qs=(qs*ij)%mod
#print(qlr,qs)
ans=(qlr*modInverse(qs))%mod
return ans
def modInverse(b):
qr=10**9+7
return pow(b, qr - 2,qr)
#===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
def wrapped_func(*args, **kwargs):
if stack: return f(*args, **kwargs)
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
continue
stack.pop()
if not stack: break
to = stack[-1].send(to)
return to
return wrapped_func
def power(arr):
listrep = arr
subsets = []
for i in range(2**len(listrep)):
subset = []
for k in range(len(listrep)):
if i & 1<<k:
subset.append(listrep[k])
subsets.append(subset)
return subsets
def pda(n) :
list = []
for i in range(1, int(sqrt(n) + 1)) :
if (n % i == 0) :
if (n // i == i) :
list.append(i)
else :
list.append(n//i);list.append(i)
# The list will be printed in reverse
return list
def dis(xa,ya,xb,yb):
return sqrt((xa-xb)**2+(ya-yb)**2)
#### END ITERATE RECURSION ####
#===============================================================================================
#----------Input functions--------------------#
def ii():
return int(input())
def ilist():
return [int(x) for x in input().strip().split()]
def islist():
return list(map(str,input().split().rstrip()))
def inp():
return input().strip()
def google(test,ans):
return "Case #"+str(test)+": "+str(ans);
def overlap(x1,y1,x2,y2):
if x2>y1:
return y1-x2
if y1>y2:
return y2-x2
return y1-x2;
###-------------------------CODE STARTS HERE--------------------------------###########
def pal(s):
k=len(s)
n=len(s)//2
for i in range(n):
if s[i]==s[k-1-i]:
continue
else:
return 0
return 1
#########################################################################################
t=int(input())
#t=1
for p in range(t):
n=ii()
a=inp();b=a
xy=a.count('0');#yz=b.count('0')
if xy%2==1 or a[0]=='0' or a[-1]=='0':
print("No")
continue
ac=''
bc=''
az=a.count('1')//2
one=0
bit=0
for i in range(n):
if a[i]=='0':
if bit==0:
ac+='('
bc+=')'
else:
ac+=')'
bc+='('
bit^=1
else:
if one<az:
ac+='('
bc+='('
else:
ac+=')'
bc+=')'
one+=1
#if debug(ac) and debug(bc):
prin("YES"+'\n');prin(ac+'\n');prin(bc+'\n')
``` | output | 1 | 1,088 | 21 | 2,177 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence of brackets is called balanced if one can turn it into a valid math expression by adding characters '+' and '1'. For example, sequences '(())()', '()', and '(()(()))' are balanced, while ')(', '(()', and '(()))(' are not.
You are given a binary string s of length n. Construct two balanced bracket sequences a and b of length n such that for all 1≤ i≤ n:
* if s_i=1, then a_i=b_i
* if s_i=0, then a_i≠ b_i
If it is impossible, you should report about it.
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (2≤ n≤ 2⋅ 10^5, n is even).
The next line contains a string s of length n, consisting of characters 0 and 1.
The sum of n across all test cases does not exceed 2⋅ 10^5.
Output
If such two balanced bracked sequences exist, output "YES" on the first line, otherwise output "NO". You can print each letter in any case (upper or lower).
If the answer is "YES", output the balanced bracket sequences a and b satisfying the conditions on the next two lines.
If there are multiple solutions, you may print any.
Example
Input
3
6
101101
10
1001101101
4
1100
Output
YES
()()()
((()))
YES
()()((()))
(())()()()
NO
Note
In the first test case, a="()()()" and b="((()))". The characters are equal in positions 1, 3, 4, and 6, which are the exact same positions where s_i=1.
In the second test case, a="()()((()))" and b="(())()()()". The characters are equal in positions 1, 4, 5, 7, 8, 10, which are the exact same positions where s_i=1.
In the third test case, there is no solution. | instruction | 0 | 1,089 | 21 | 2,178 |
Tags: constructive algorithms, greedy
Correct Solution:
```
t=int(input())
for t1 in range(0,t):
n=int(input())
s=input()
if s[0]=='0' or s[-1]=='0':
print("NO")
else:
same=[]
diff=[]
for i in range(0,n):
if s[i]=='1':
same.append(i)
else:
diff.append(i)
ans1=list(s)
ans2=list(s)
if len(diff)%2==1:
print("NO")
else:
flag=1
for i in diff:
if flag==1:
ans1[i]=')'
ans2[i]='('
else:
ans1[i]='('
ans2[i]=')'
flag=(flag+1)%2
start=0
end=len(same)-1
while(start<end):
xx=same[start]
yy=same[end]
ans1[xx]='('
ans2[xx]='('
ans1[yy]=')'
ans2[yy]=')'
start+=1
end-=1
print("YES")
print(''.join(ans1))
print(''.join(ans2))
``` | output | 1 | 1,089 | 21 | 2,179 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence of brackets is called balanced if one can turn it into a valid math expression by adding characters '+' and '1'. For example, sequences '(())()', '()', and '(()(()))' are balanced, while ')(', '(()', and '(()))(' are not.
You are given a binary string s of length n. Construct two balanced bracket sequences a and b of length n such that for all 1≤ i≤ n:
* if s_i=1, then a_i=b_i
* if s_i=0, then a_i≠ b_i
If it is impossible, you should report about it.
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (2≤ n≤ 2⋅ 10^5, n is even).
The next line contains a string s of length n, consisting of characters 0 and 1.
The sum of n across all test cases does not exceed 2⋅ 10^5.
Output
If such two balanced bracked sequences exist, output "YES" on the first line, otherwise output "NO". You can print each letter in any case (upper or lower).
If the answer is "YES", output the balanced bracket sequences a and b satisfying the conditions on the next two lines.
If there are multiple solutions, you may print any.
Example
Input
3
6
101101
10
1001101101
4
1100
Output
YES
()()()
((()))
YES
()()((()))
(())()()()
NO
Note
In the first test case, a="()()()" and b="((()))". The characters are equal in positions 1, 3, 4, and 6, which are the exact same positions where s_i=1.
In the second test case, a="()()((()))" and b="(())()()()". The characters are equal in positions 1, 4, 5, 7, 8, 10, which are the exact same positions where s_i=1.
In the third test case, there is no solution. | instruction | 0 | 1,090 | 21 | 2,180 |
Tags: constructive algorithms, greedy
Correct Solution:
```
import sys
lines = sys.stdin.readlines()
lineidx = 0
def readLine():
global lineidx
s = lines[lineidx].strip()
lineidx += 1
return s
t = int(readLine())
for testcase in range(t):
n = int(readLine())
s = readLine()
cntone = s.count('1')
cntzero = s.count('0')
if (cntone % 2 != 0) or (cntzero % 2 != 0):
print("NO")
continue
cntone /= 2
a = ['x'] * n
b = ['x'] * n
bala = balb = 0
for i in range(n):
if s[i] == '1':
if cntone > 0:
cntone -= 1
a[i] = b[i] = '('
else:
a[i] = b[i] = ')'
else:
if cntzero % 2 == 0:
a[i] = '('
b[i] = ')'
else:
a[i] = ')'
b[i] = '('
cntzero -= 1
if a[i] == '(':
bala += 1
else:
bala -= 1
if b[i] == '(':
balb += 1
else:
balb -= 1
if bala < 0 or balb < 0:
break
if bala != 0 or balb != 0:
print("NO")
else:
print("YES")
print("".join(a))
print("".join(b))
``` | output | 1 | 1,090 | 21 | 2,181 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A sequence of brackets is called balanced if one can turn it into a valid math expression by adding characters '+' and '1'. For example, sequences '(())()', '()', and '(()(()))' are balanced, while ')(', '(()', and '(()))(' are not.
You are given a binary string s of length n. Construct two balanced bracket sequences a and b of length n such that for all 1≤ i≤ n:
* if s_i=1, then a_i=b_i
* if s_i=0, then a_i≠ b_i
If it is impossible, you should report about it.
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (2≤ n≤ 2⋅ 10^5, n is even).
The next line contains a string s of length n, consisting of characters 0 and 1.
The sum of n across all test cases does not exceed 2⋅ 10^5.
Output
If such two balanced bracked sequences exist, output "YES" on the first line, otherwise output "NO". You can print each letter in any case (upper or lower).
If the answer is "YES", output the balanced bracket sequences a and b satisfying the conditions on the next two lines.
If there are multiple solutions, you may print any.
Example
Input
3
6
101101
10
1001101101
4
1100
Output
YES
()()()
((()))
YES
()()((()))
(())()()()
NO
Note
In the first test case, a="()()()" and b="((()))". The characters are equal in positions 1, 3, 4, and 6, which are the exact same positions where s_i=1.
In the second test case, a="()()((()))" and b="(())()()()". The characters are equal in positions 1, 4, 5, 7, 8, 10, which are the exact same positions where s_i=1.
In the third test case, there is no solution.
Submitted Solution:
```
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
def main():
for _ in range(int(input())):
n = int(input())
a = input()
ans1 = ["*"] * n
ans2 = ["*"] * n
cnt1 = 0
for elem in a:
if elem == "1":
cnt1 += 1
if cnt1 % 2:
print("NO")
continue
cntPlus = cnt1 // 2
tmp = 0
for i in range(n):
if a[i] == "1":
if tmp < cntPlus:
ans1[i] = "("
ans2[i] = "("
tmp += 1
else:
ans1[i] = ")"
ans2[i] = ")"
count1 = 0
count2 = 0
flag = True
for i in range(n):
if a[i] == "1":
if ans1[i] == "(":
count1 += 1
count2 += 1
else:
count1 -= 1
count2 -= 1
if count1 < 0 or count2 < 0:
flag = False
break
else:
if count1 >= count2:
ans1[i] = ")"
ans2[i] = "("
count1 -= 1
count2 += 1
if count1 < 0:
flag = False
break
else:
ans1[i] = "("
ans2[i] = ")"
count1 += 1
count2 -= 1
if count2 < 0:
flag = False
break
if flag:
print("YES")
print("".join(ans1))
print("".join(ans2))
else:
print("NO")
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
``` | instruction | 0 | 1,091 | 21 | 2,182 |
Yes | output | 1 | 1,091 | 21 | 2,183 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A sequence of brackets is called balanced if one can turn it into a valid math expression by adding characters '+' and '1'. For example, sequences '(())()', '()', and '(()(()))' are balanced, while ')(', '(()', and '(()))(' are not.
You are given a binary string s of length n. Construct two balanced bracket sequences a and b of length n such that for all 1≤ i≤ n:
* if s_i=1, then a_i=b_i
* if s_i=0, then a_i≠ b_i
If it is impossible, you should report about it.
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (2≤ n≤ 2⋅ 10^5, n is even).
The next line contains a string s of length n, consisting of characters 0 and 1.
The sum of n across all test cases does not exceed 2⋅ 10^5.
Output
If such two balanced bracked sequences exist, output "YES" on the first line, otherwise output "NO". You can print each letter in any case (upper or lower).
If the answer is "YES", output the balanced bracket sequences a and b satisfying the conditions on the next two lines.
If there are multiple solutions, you may print any.
Example
Input
3
6
101101
10
1001101101
4
1100
Output
YES
()()()
((()))
YES
()()((()))
(())()()()
NO
Note
In the first test case, a="()()()" and b="((()))". The characters are equal in positions 1, 3, 4, and 6, which are the exact same positions where s_i=1.
In the second test case, a="()()((()))" and b="(())()()()". The characters are equal in positions 1, 4, 5, 7, 8, 10, which are the exact same positions where s_i=1.
In the third test case, there is no solution.
Submitted Solution:
```
# IMPORTS GO HERE
# *************************************************************************
# CLASSES YOU MAKE GO HERE
# *************************************************************************
# ALREADY MADE FUNCTIONS
# To initialize an array of n values with 0
def init_array(n):
arr1d = [0] * n
return arr1d
# To initialize a 2D array of rows * columns 0
def init_array2d(arr_rows, arr_columns):
arr2d = [[0 for a in range(arr_columns)] for b in range(arr_rows)]
return arr2d
# To print a 2d array => helpful in debugging
def print_array2d(arr2d):
# len(arr) = no of rows, len(arr[0]) = no of columns
for f in range(len(arr2d)):
for q in range(len(arr2d[0])):
print(arr2d[f][q])
# To get input of a 2D array given rows and columns
def get_array2d_input(arr_rows, arr_columns):
arr2d = [[0 for f in range(arr_columns)] for q in range(arr_rows)]
for f in range(arr_rows):
line = input()
words = line.split()
for q in range(arr_columns):
arr2d[f][q] = int(words[q])
return arr2d
# *************************************************************************
# FUNCTIONS YOU MAKE GO HERE
def solve():
dummy = 0 # dummy variable
# *************************************************************************
# ******************** MAIN ***********************************
testCases = int(input()) # get total test cases
for test_case in range(1, testCases + 1):
# ********************** SETTING INPUTS ********************************
line1 = input() # First line of a test case
word1 = line1.split() # Splitting line into list of words
N = int(word1[0]) # Getting the value of N
s = input() # Second line of a test case
# ********************** SOLVE HERE ************************************
cost = 0 # To calculate cost
brackets_a = ""
brackets_b = ""
stack_a = []
stack_b = []
possible = True
for i in range(N):
# remaining_str = s[i:]
# remaining_zeros = remaining_str.count('0')
# print("i: " + i)
#
# print("stack_a: " + str(stack_a))
# print("stack_b: " + str(stack_b))
if s[i] == '1':
if len(stack_a) == 0 or len(stack_b) == 0:
brackets_a += '('
brackets_b += '('
stack_a.append('(')
stack_b.append('(')
else:
if i < N - 1:
if len(stack_a) == 1 and len(stack_b) == 1:
if s[i + 1] == '0':
brackets_a += '('
brackets_b += '('
stack_a.append('(')
stack_b.append('(')
else:
if len(stack_a) == 0 or len(stack_b) == 0:
possible = False
break
brackets_a += ')'
brackets_b += ')'
stack_a.pop()
stack_b.pop()
else:
if len(stack_a) == 0 or len(stack_b) == 0:
possible = False
break
brackets_a += ')'
brackets_b += ')'
stack_a.pop()
stack_b.pop()
else:
if len(stack_a) == 0 or len(stack_b) == 0:
possible = False
break
brackets_a += ')'
brackets_b += ')'
stack_a.pop()
stack_b.pop()
else:
if len(stack_a) == 0 and len(stack_b) == 0:
possible = False
break
elif len(stack_a) >= len(stack_b):
if len(stack_a) == 0:
possible = False
break
brackets_a += ')'
brackets_b += '('
stack_a.pop()
stack_b.append('(')
else:
if len(stack_b) == 0:
possible = False
break
brackets_a += '('
brackets_b += ')'
stack_b.pop()
stack_a.append('(')
# print("a: " + brackets_a)
# print("b: " + brackets_b)
# print()
ans = ""
if possible:
if len(stack_a) == 0 and len(stack_b) == 0:
ans = "YES"
else:
ans = "NO"
else:
ans = "NO"
# **********************************************************************
# ************************** OUTPUT ************************************
if ans == "NO":
print("NO")
else:
print("YES")
print(brackets_a)
print(brackets_b)
# *************************************************************************
# ************************** USEFUL CODE SNIPPETS *************************
``` | instruction | 0 | 1,092 | 21 | 2,184 |
Yes | output | 1 | 1,092 | 21 | 2,185 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A sequence of brackets is called balanced if one can turn it into a valid math expression by adding characters '+' and '1'. For example, sequences '(())()', '()', and '(()(()))' are balanced, while ')(', '(()', and '(()))(' are not.
You are given a binary string s of length n. Construct two balanced bracket sequences a and b of length n such that for all 1≤ i≤ n:
* if s_i=1, then a_i=b_i
* if s_i=0, then a_i≠ b_i
If it is impossible, you should report about it.
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (2≤ n≤ 2⋅ 10^5, n is even).
The next line contains a string s of length n, consisting of characters 0 and 1.
The sum of n across all test cases does not exceed 2⋅ 10^5.
Output
If such two balanced bracked sequences exist, output "YES" on the first line, otherwise output "NO". You can print each letter in any case (upper or lower).
If the answer is "YES", output the balanced bracket sequences a and b satisfying the conditions on the next two lines.
If there are multiple solutions, you may print any.
Example
Input
3
6
101101
10
1001101101
4
1100
Output
YES
()()()
((()))
YES
()()((()))
(())()()()
NO
Note
In the first test case, a="()()()" and b="((()))". The characters are equal in positions 1, 3, 4, and 6, which are the exact same positions where s_i=1.
In the second test case, a="()()((()))" and b="(())()()()". The characters are equal in positions 1, 4, 5, 7, 8, 10, which are the exact same positions where s_i=1.
In the third test case, there is no solution.
Submitted Solution:
```
from copy import deepcopy
t = int(input())
for _ in range(t):
n = int(input())
s = input()
count_0 = s.count('0')
if count_0 % 2 != 0:
print("NO")
elif count_0 == 0:
print("YES")
print('(' * (n // 2) + ')' * (n // 2))
print('(' * (n // 2) + ')' * (n // 2))
elif s[0] == '0' or s[-1] == '0':
print("NO")
else:
count_1 = n - count_0
ans = ['0' for _ in range(n)]
counter = 0
for i in range(0, n):
if s[i] == '1':
if counter < count_1 // 2:
ans[i] = '('
else:
ans[i] = ')'
counter += 1
ans_2 = deepcopy(ans)
x,y='(',')'
for i in range(0,n):
if s[i]=='0':
ans[i]=x
ans_2[i]=y
x,y=y,x
print("YES")
print(''.join(ans))
print(''.join(ans_2))
``` | instruction | 0 | 1,093 | 21 | 2,186 |
Yes | output | 1 | 1,093 | 21 | 2,187 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A sequence of brackets is called balanced if one can turn it into a valid math expression by adding characters '+' and '1'. For example, sequences '(())()', '()', and '(()(()))' are balanced, while ')(', '(()', and '(()))(' are not.
You are given a binary string s of length n. Construct two balanced bracket sequences a and b of length n such that for all 1≤ i≤ n:
* if s_i=1, then a_i=b_i
* if s_i=0, then a_i≠ b_i
If it is impossible, you should report about it.
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (2≤ n≤ 2⋅ 10^5, n is even).
The next line contains a string s of length n, consisting of characters 0 and 1.
The sum of n across all test cases does not exceed 2⋅ 10^5.
Output
If such two balanced bracked sequences exist, output "YES" on the first line, otherwise output "NO". You can print each letter in any case (upper or lower).
If the answer is "YES", output the balanced bracket sequences a and b satisfying the conditions on the next two lines.
If there are multiple solutions, you may print any.
Example
Input
3
6
101101
10
1001101101
4
1100
Output
YES
()()()
((()))
YES
()()((()))
(())()()()
NO
Note
In the first test case, a="()()()" and b="((()))". The characters are equal in positions 1, 3, 4, and 6, which are the exact same positions where s_i=1.
In the second test case, a="()()((()))" and b="(())()()()". The characters are equal in positions 1, 4, 5, 7, 8, 10, which are the exact same positions where s_i=1.
In the third test case, there is no solution.
Submitted Solution:
```
import array
import math
def main() :
tc=1
tc=int(input())
while(tc) :
n=int(input())
s=input()
res=True
cnt=0
for i in range(n) :
if(s[i]=='0') :
if(i==0 or i==n-1) :
res=False
break;
cnt+=1
if(cnt%2==1) :
res=False
if(res) :
c, cnt1, c1, a, b = 0, n-cnt, 0, "", ""
for i in range(n) :
if(s[i]=='1') :
c1+=1
if(c1<=cnt1/2) :
a+="("
b+="("
else :
a+=")"
b+=")"
else :
c+=1
if(c%2) :
a+="("
b+=")"
else :
a+=")"
b+="("
print("YES")
print(a)
print(b)
else :
print("NO")
tc-=1
if __name__=="__main__" :
main()
##### ##### ##### ##### ##### ##### ##### # # # # ##### ######
# # # # # # # # # # # # ## # # # # # #
# # # # # ### ##### ##### ##### # # # ## ##### #
# # # # # # # # # # # # ## # # # # # #
##### ##### ##### ##### # # # # # # # # # # # ####
``` | instruction | 0 | 1,094 | 21 | 2,188 |
Yes | output | 1 | 1,094 | 21 | 2,189 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A sequence of brackets is called balanced if one can turn it into a valid math expression by adding characters '+' and '1'. For example, sequences '(())()', '()', and '(()(()))' are balanced, while ')(', '(()', and '(()))(' are not.
You are given a binary string s of length n. Construct two balanced bracket sequences a and b of length n such that for all 1≤ i≤ n:
* if s_i=1, then a_i=b_i
* if s_i=0, then a_i≠ b_i
If it is impossible, you should report about it.
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (2≤ n≤ 2⋅ 10^5, n is even).
The next line contains a string s of length n, consisting of characters 0 and 1.
The sum of n across all test cases does not exceed 2⋅ 10^5.
Output
If such two balanced bracked sequences exist, output "YES" on the first line, otherwise output "NO". You can print each letter in any case (upper or lower).
If the answer is "YES", output the balanced bracket sequences a and b satisfying the conditions on the next two lines.
If there are multiple solutions, you may print any.
Example
Input
3
6
101101
10
1001101101
4
1100
Output
YES
()()()
((()))
YES
()()((()))
(())()()()
NO
Note
In the first test case, a="()()()" and b="((()))". The characters are equal in positions 1, 3, 4, and 6, which are the exact same positions where s_i=1.
In the second test case, a="()()((()))" and b="(())()()()". The characters are equal in positions 1, 4, 5, 7, 8, 10, which are the exact same positions where s_i=1.
In the third test case, there is no solution.
Submitted Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
s = input()
s = list(s)
a = ''
b = ''
c1 = c2 = 0
c = 0
d = 0
for i in range(n):
if s[i]=='1':
c+=1
for i in range(n):
if s[0]=='0' or s[-1]=='0' or c%2==1 :
print('NO')
d = 1
break
if s[i]=='1':
c1+=1
if c1<n//2:
a=a+'('
b+='('
else:
a+=')'
b+=')'
else:
if c2==0:
a+='('
b+=')'
if c2==1:
a+=')'
b+='('
if c2==0:
c2=1
elif c2==1:
c2=0
if d==0:
print('YES')
print(a)
print(b)
``` | instruction | 0 | 1,095 | 21 | 2,190 |
No | output | 1 | 1,095 | 21 | 2,191 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A sequence of brackets is called balanced if one can turn it into a valid math expression by adding characters '+' and '1'. For example, sequences '(())()', '()', and '(()(()))' are balanced, while ')(', '(()', and '(()))(' are not.
You are given a binary string s of length n. Construct two balanced bracket sequences a and b of length n such that for all 1≤ i≤ n:
* if s_i=1, then a_i=b_i
* if s_i=0, then a_i≠ b_i
If it is impossible, you should report about it.
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (2≤ n≤ 2⋅ 10^5, n is even).
The next line contains a string s of length n, consisting of characters 0 and 1.
The sum of n across all test cases does not exceed 2⋅ 10^5.
Output
If such two balanced bracked sequences exist, output "YES" on the first line, otherwise output "NO". You can print each letter in any case (upper or lower).
If the answer is "YES", output the balanced bracket sequences a and b satisfying the conditions on the next two lines.
If there are multiple solutions, you may print any.
Example
Input
3
6
101101
10
1001101101
4
1100
Output
YES
()()()
((()))
YES
()()((()))
(())()()()
NO
Note
In the first test case, a="()()()" and b="((()))". The characters are equal in positions 1, 3, 4, and 6, which are the exact same positions where s_i=1.
In the second test case, a="()()((()))" and b="(())()()()". The characters are equal in positions 1, 4, 5, 7, 8, 10, which are the exact same positions where s_i=1.
In the third test case, there is no solution.
Submitted Solution:
```
t=int(input())
for _ in range(t):
n=int(input())
st=input()
if(st[0]=="0" or st[-1]=="0"):
print("NO")
else:
s1=[]
s2=[]
o1,o2=0,0
f=1
for i in range(len(st)):
if(st[i]=="1"):
if(o1>0 and o2>0):
o1-=1
o2-=1
s1.append(")")
s2.append(")")
elif(o1==0 or o2==0):
o1+=1
o2+=1
s1.append("(")
s2.append("(")
elif(o1<0 or o2<0):
print("NO")
f=0
break
else:
if(o1>0 and o2>0):
if(o1>=o2):
o1-=1
o2+=1
s1.append(")")
s2.append("(")
else:
o2-=1
o1+=1
s1.append(")")
s2.append("(")
elif(o1>0 and o2==0):
o1-=1
o2+=1
s1.append(")")
s2.append("(")
elif(o2>0 and o1==0):
o1+=1
o2-=1
s1.append("(")
s2.append(")")
elif((o2< 0 or o1<0) or (o1==0 and o2==0)):
print("NO")
f=0
break
if(f==1 and o1==0 and o2==0):
print("YES")
print("".join(s1))
print("".join(s2))
``` | instruction | 0 | 1,096 | 21 | 2,192 |
No | output | 1 | 1,096 | 21 | 2,193 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A sequence of brackets is called balanced if one can turn it into a valid math expression by adding characters '+' and '1'. For example, sequences '(())()', '()', and '(()(()))' are balanced, while ')(', '(()', and '(()))(' are not.
You are given a binary string s of length n. Construct two balanced bracket sequences a and b of length n such that for all 1≤ i≤ n:
* if s_i=1, then a_i=b_i
* if s_i=0, then a_i≠ b_i
If it is impossible, you should report about it.
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (2≤ n≤ 2⋅ 10^5, n is even).
The next line contains a string s of length n, consisting of characters 0 and 1.
The sum of n across all test cases does not exceed 2⋅ 10^5.
Output
If such two balanced bracked sequences exist, output "YES" on the first line, otherwise output "NO". You can print each letter in any case (upper or lower).
If the answer is "YES", output the balanced bracket sequences a and b satisfying the conditions on the next two lines.
If there are multiple solutions, you may print any.
Example
Input
3
6
101101
10
1001101101
4
1100
Output
YES
()()()
((()))
YES
()()((()))
(())()()()
NO
Note
In the first test case, a="()()()" and b="((()))". The characters are equal in positions 1, 3, 4, and 6, which are the exact same positions where s_i=1.
In the second test case, a="()()((()))" and b="(())()()()". The characters are equal in positions 1, 4, 5, 7, 8, 10, which are the exact same positions where s_i=1.
In the third test case, there is no solution.
Submitted Solution:
```
t = int(input())
while(t>0):
n = int(input())
s = input()
a = ""
b = ""
one = 0
zero = 0
for i in s:
if i is "1":
one+=1
elif i is "0":
zero+=1
if(zero%2==1 or one%2==1 or s[0]=="0" or s[-1] =="0"):
print("NO")
else:
for i in s:
if(i=="1"):
if(one):
a+="("
b+="("
one-=2
else:
a+=")"
b+=")"
else:
if(zero%2==0):
a+=")"
b+="("
zero-=1
else:
a+="("
b+=")"
print("YES")
print(a)
print(b)
t-=1
``` | instruction | 0 | 1,097 | 21 | 2,194 |
No | output | 1 | 1,097 | 21 | 2,195 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A sequence of brackets is called balanced if one can turn it into a valid math expression by adding characters '+' and '1'. For example, sequences '(())()', '()', and '(()(()))' are balanced, while ')(', '(()', and '(()))(' are not.
You are given a binary string s of length n. Construct two balanced bracket sequences a and b of length n such that for all 1≤ i≤ n:
* if s_i=1, then a_i=b_i
* if s_i=0, then a_i≠ b_i
If it is impossible, you should report about it.
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (2≤ n≤ 2⋅ 10^5, n is even).
The next line contains a string s of length n, consisting of characters 0 and 1.
The sum of n across all test cases does not exceed 2⋅ 10^5.
Output
If such two balanced bracked sequences exist, output "YES" on the first line, otherwise output "NO". You can print each letter in any case (upper or lower).
If the answer is "YES", output the balanced bracket sequences a and b satisfying the conditions on the next two lines.
If there are multiple solutions, you may print any.
Example
Input
3
6
101101
10
1001101101
4
1100
Output
YES
()()()
((()))
YES
()()((()))
(())()()()
NO
Note
In the first test case, a="()()()" and b="((()))". The characters are equal in positions 1, 3, 4, and 6, which are the exact same positions where s_i=1.
In the second test case, a="()()((()))" and b="(())()()()". The characters are equal in positions 1, 4, 5, 7, 8, 10, which are the exact same positions where s_i=1.
In the third test case, there is no solution.
Submitted Solution:
```
def TestLine(sLine, nLen):
x = 0
for i in range(nLen):
if sLine[i] == "(":
x = x + 1
else:
x = x - 1
if x < 0:
return False
if x != 0:
return False
return True
nTests = int(input())
for i in range(nTests):
nLen = int(input())
if nLen//2*2 != nLen:
print("NO")
break
sMask = input()
a = ""
for j in range(nLen//2):
a = a + "("
for j in range(nLen//2):
a = a + ")"
b = ""
for j in range(nLen):
if sMask[j] == "1":
b = b + a[j]
elif a[j] == "(":
b = b + ")"
else:
b = b + "("
if TestLine(b, nLen):
print("YES")
print(a)
print(b)
else:
print("NO")
# print(a)
# print(b)
``` | instruction | 0 | 1,098 | 21 | 2,196 |
No | output | 1 | 1,098 | 21 | 2,197 |
Provide a correct Python 3 solution for this coding contest problem.
A bracket sequence is a string that is one of the following:
1. An empty string;
2. The concatenation of `(`, A, and `)` in this order, for some bracket sequence A ;
3. The concatenation of A and B in this order, for some non-empty bracket sequences A and B /
Given are N strings S_i. Can a bracket sequence be formed by concatenating all the N strings in some order?
Constraints
* 1 \leq N \leq 10^6
* The total length of the strings S_i is at most 10^6.
* S_i is a non-empty string consisting of `(` and `)`.
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If a bracket sequence can be formed by concatenating all the N strings in some order, print `Yes`; otherwise, print `No`.
Examples
Input
2
)
(()
Output
Yes
Input
2
)(
()
Output
No
Input
4
((()))
((((((
))))))
()()()
Output
Yes
Input
3
(((
)
)
Output
No | instruction | 0 | 1,444 | 21 | 2,888 |
"Correct Solution:
```
def f():
N = int(input())
UP = []
DOWN = []
for _ in range(N):
S = input()
c = 0
minC = 0
for s in S:
if s == '(':
c += 1
else:
c -= 1
minC = min(minC, c)
if c >= 0:
UP.append((minC, c))
else:
DOWN.append((c - minC, c))
c = 0
for up in sorted(UP, reverse=True):
if c + up[0] < 0:
return False
c += up[1]
for down in sorted(DOWN, reverse=True):
if c + down[1] - down[0] < 0:
return False
c += down[1]
if c != 0:
return False
return True
if f():
print('Yes')
else:
print('No')
``` | output | 1 | 1,444 | 21 | 2,889 |
Provide a correct Python 3 solution for this coding contest problem.
A bracket sequence is a string that is one of the following:
1. An empty string;
2. The concatenation of `(`, A, and `)` in this order, for some bracket sequence A ;
3. The concatenation of A and B in this order, for some non-empty bracket sequences A and B /
Given are N strings S_i. Can a bracket sequence be formed by concatenating all the N strings in some order?
Constraints
* 1 \leq N \leq 10^6
* The total length of the strings S_i is at most 10^6.
* S_i is a non-empty string consisting of `(` and `)`.
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If a bracket sequence can be formed by concatenating all the N strings in some order, print `Yes`; otherwise, print `No`.
Examples
Input
2
)
(()
Output
Yes
Input
2
)(
()
Output
No
Input
4
((()))
((((((
))))))
()()()
Output
Yes
Input
3
(((
)
)
Output
No | instruction | 0 | 1,445 | 21 | 2,890 |
"Correct Solution:
```
import sys
input = sys.stdin.readline
n = int(input())
br = [input().rstrip() for i in range(n)]
ls = []
numsum = 0
charge = 0
for i in range(n):
s = br[i]
need = 0
num = 0
for t in s:
if t == "(":
num += 1
else:
num -= 1
need = max(need,-num)
numsum += num
if need == 0:
charge += num
else:
ls.append([need,num])
if numsum != 0:
print("No")
exit()
ls.sort()
l = len(ls)
vis = [0]*l
for i,x in enumerate(ls):
need,num = x
if need <= charge and num >= 0:
charge += num
ls[i][0] = -1
ls.sort(key = lambda x:x[0]+x[1],reverse=True)
for i in range(l):
need,num = ls[i]
if need == -1:
continue
if need > charge:
print("No")
exit()
charge += num
print("Yes")
``` | output | 1 | 1,445 | 21 | 2,891 |
Provide a correct Python 3 solution for this coding contest problem.
A bracket sequence is a string that is one of the following:
1. An empty string;
2. The concatenation of `(`, A, and `)` in this order, for some bracket sequence A ;
3. The concatenation of A and B in this order, for some non-empty bracket sequences A and B /
Given are N strings S_i. Can a bracket sequence be formed by concatenating all the N strings in some order?
Constraints
* 1 \leq N \leq 10^6
* The total length of the strings S_i is at most 10^6.
* S_i is a non-empty string consisting of `(` and `)`.
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If a bracket sequence can be formed by concatenating all the N strings in some order, print `Yes`; otherwise, print `No`.
Examples
Input
2
)
(()
Output
Yes
Input
2
)(
()
Output
No
Input
4
((()))
((((((
))))))
()()()
Output
Yes
Input
3
(((
)
)
Output
No | instruction | 0 | 1,446 | 21 | 2,892 |
"Correct Solution:
```
N = int(input())
Up = []
Down = []
for _ in range(N):
S = input()
L = [0]
mi = 0
now = 0
for __ in range(len(S)):
if S[__] == '(':
now += 1
else:
now -= 1
mi = min(mi, now)
if now > 0:
Up.append((mi, now))
else:
Down.append((mi - now, mi, now))
Up.sort(reverse=True)
Down.sort()
now = 0
for i, j in Up:
if now + i < 0:
print('No')
exit()
else:
now += j
for _, i, j in Down:
if now + i < 0:
print('No')
exit()
else:
now += j
if now == 0:
print('Yes')
else:
print('No')
``` | output | 1 | 1,446 | 21 | 2,893 |
Provide a correct Python 3 solution for this coding contest problem.
A bracket sequence is a string that is one of the following:
1. An empty string;
2. The concatenation of `(`, A, and `)` in this order, for some bracket sequence A ;
3. The concatenation of A and B in this order, for some non-empty bracket sequences A and B /
Given are N strings S_i. Can a bracket sequence be formed by concatenating all the N strings in some order?
Constraints
* 1 \leq N \leq 10^6
* The total length of the strings S_i is at most 10^6.
* S_i is a non-empty string consisting of `(` and `)`.
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If a bracket sequence can be formed by concatenating all the N strings in some order, print `Yes`; otherwise, print `No`.
Examples
Input
2
)
(()
Output
Yes
Input
2
)(
()
Output
No
Input
4
((()))
((((((
))))))
()()()
Output
Yes
Input
3
(((
)
)
Output
No | instruction | 0 | 1,447 | 21 | 2,894 |
"Correct Solution:
```
N=int(input())
S=[]
L=0
R=0
for i in range(N):
s=input()
n=len(s)
data=[s[j] for j in range(n)]
l=0
r=0
yabai=float("inf")
for j in range(n):
l+=(s[j]=="(")
r+=(s[j]==")")
yabai=min(l-r,yabai)
S.append([yabai,l-r])
L+=l
R+=r
if L!=R:
print("No")
else:
first=[]
last=[]
for i in range(N):
yabai,gap=S[i]
if gap>0:
first.append(S[i])
else:
last.append([gap-yabai,yabai])
first.sort(reverse=True)
last.sort(reverse=True)
G=0
for i in range(len(first)):
yabai,gap=first[i]
if 0>G+yabai:
print("No")
exit()
else:
G+=gap
for j in range(len(last)):
gapminus,yabai=last[j]
if 0>G+yabai:
print("No")
break
else:
G+=gapminus+yabai
else:
print("Yes")
``` | output | 1 | 1,447 | 21 | 2,895 |
Provide a correct Python 3 solution for this coding contest problem.
A bracket sequence is a string that is one of the following:
1. An empty string;
2. The concatenation of `(`, A, and `)` in this order, for some bracket sequence A ;
3. The concatenation of A and B in this order, for some non-empty bracket sequences A and B /
Given are N strings S_i. Can a bracket sequence be formed by concatenating all the N strings in some order?
Constraints
* 1 \leq N \leq 10^6
* The total length of the strings S_i is at most 10^6.
* S_i is a non-empty string consisting of `(` and `)`.
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If a bracket sequence can be formed by concatenating all the N strings in some order, print `Yes`; otherwise, print `No`.
Examples
Input
2
)
(()
Output
Yes
Input
2
)(
()
Output
No
Input
4
((()))
((((((
))))))
()()()
Output
Yes
Input
3
(((
)
)
Output
No | instruction | 0 | 1,448 | 21 | 2,896 |
"Correct Solution:
```
n = int(input())
left = []
mid = []
midminus = []
right = []
L = []
R = []
for i in range(n):
s = input()
l = 0
r = 0
for x in s:
if x == '(':
l += 1
else:
if l > 0:
l -= 1
else:
r += 1
if l > 0 and r == 0:
left.append((l, r))
elif l > 0 and r > 0:
if l > r:
mid.append((r, l)) # a,b-a
else:
midminus.append((r,l))
elif l == 0 and r > 0:
right.append((l, r))
L.append(l)
R.append(r)
if sum(L) != sum(R):
print('No')
exit()
A = 0
B = 0
for x in left:
A += x[0]
for x in right:
B += x[1]
mid = sorted(mid, key=lambda x: (x[0],-x[1]))
midminus = sorted(midminus,key= lambda x:x[0]-x[1])
mid += midminus
l = A
r = 0
for a, b in mid:
if l < a:
print('No')
exit()
l -= a
l += b
print('Yes')
``` | output | 1 | 1,448 | 21 | 2,897 |
Provide a correct Python 3 solution for this coding contest problem.
A bracket sequence is a string that is one of the following:
1. An empty string;
2. The concatenation of `(`, A, and `)` in this order, for some bracket sequence A ;
3. The concatenation of A and B in this order, for some non-empty bracket sequences A and B /
Given are N strings S_i. Can a bracket sequence be formed by concatenating all the N strings in some order?
Constraints
* 1 \leq N \leq 10^6
* The total length of the strings S_i is at most 10^6.
* S_i is a non-empty string consisting of `(` and `)`.
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If a bracket sequence can be formed by concatenating all the N strings in some order, print `Yes`; otherwise, print `No`.
Examples
Input
2
)
(()
Output
Yes
Input
2
)(
()
Output
No
Input
4
((()))
((((((
))))))
()()()
Output
Yes
Input
3
(((
)
)
Output
No | instruction | 0 | 1,449 | 21 | 2,898 |
"Correct Solution:
```
N = int(input())
#INF = float('inf')
def check(A):
th = 0 #高さの合計
for i in range(len(A)):
b = A[i][0]
h = A[i][1]
if th + b < 0:
return False
else:
th += h
return True
L = []
R = []
goukei = 0
for i in range(N):
temp = str(input())
b= 0 #最下点
h = 0 #ゴールの位置
for j in range(len(temp)):
if temp[j] == "(":
h += 1
else:
h -= 1
b = min(b,h)
if h > 0:
L.append([b,h])
else:
R.append([b-h,-h])
goukei += h
L.sort(reverse=True)
R.sort(reverse=True)
#print(L,R)
if check(L) and check(R) and goukei == 0:
print("Yes")
else:
print("No")
``` | output | 1 | 1,449 | 21 | 2,899 |
Provide a correct Python 3 solution for this coding contest problem.
A bracket sequence is a string that is one of the following:
1. An empty string;
2. The concatenation of `(`, A, and `)` in this order, for some bracket sequence A ;
3. The concatenation of A and B in this order, for some non-empty bracket sequences A and B /
Given are N strings S_i. Can a bracket sequence be formed by concatenating all the N strings in some order?
Constraints
* 1 \leq N \leq 10^6
* The total length of the strings S_i is at most 10^6.
* S_i is a non-empty string consisting of `(` and `)`.
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If a bracket sequence can be formed by concatenating all the N strings in some order, print `Yes`; otherwise, print `No`.
Examples
Input
2
)
(()
Output
Yes
Input
2
)(
()
Output
No
Input
4
((()))
((((((
))))))
()()()
Output
Yes
Input
3
(((
)
)
Output
No | instruction | 0 | 1,450 | 21 | 2,900 |
"Correct Solution:
```
import sys
input = sys.stdin.readline
N = int(input())
D, E = [], []
t,l = 0, 0
res = 0
for _ in range(N):
S = input().rstrip()
x,y = 0,0
for s in S:
if s=="(": x+=1
else: x=max(x-1,0)
for s in reversed(S):
if s==")": y+=1
else: y=max(y-1,0)
D.append((x,y))
D.sort(key=lambda x:x[1])
t = 0
for x,y in D:
if x-y>=0:
if t>=y: t+=x-y
else: print("No"); exit()
D.sort(key=lambda x:x[0])
s = 0
for x,y in D:
if y-x>=0:
if s>=x: s+=y-x
else: print("No"); exit()
if t!=s: print("No")
else: print("Yes")
``` | output | 1 | 1,450 | 21 | 2,901 |
Provide a correct Python 3 solution for this coding contest problem.
A bracket sequence is a string that is one of the following:
1. An empty string;
2. The concatenation of `(`, A, and `)` in this order, for some bracket sequence A ;
3. The concatenation of A and B in this order, for some non-empty bracket sequences A and B /
Given are N strings S_i. Can a bracket sequence be formed by concatenating all the N strings in some order?
Constraints
* 1 \leq N \leq 10^6
* The total length of the strings S_i is at most 10^6.
* S_i is a non-empty string consisting of `(` and `)`.
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If a bracket sequence can be formed by concatenating all the N strings in some order, print `Yes`; otherwise, print `No`.
Examples
Input
2
)
(()
Output
Yes
Input
2
)(
()
Output
No
Input
4
((()))
((((((
))))))
()()()
Output
Yes
Input
3
(((
)
)
Output
No | instruction | 0 | 1,451 | 21 | 2,902 |
"Correct Solution:
```
n = int(input())
ls, rs = [], []
def check(sl):
h = 0
for t in sl:
b = h + t[0]
if b < 0:
return False
h += t[1]
return True
total = 0
for i in range(n):
b, h = 0, 0
S = input()
for s in S:
if s == '(':
h += 1
else:
h -= 1
b = min(b, h)
if h > 0:
ls.append((b, h))
else:
rs.append((b - h, -h))
total += h
ls.sort(key=lambda x: x[0], reverse=True)
rs.sort(key=lambda x: x[0], reverse=True)
print('Yes' if check(ls) and check(rs) and total == 0 else 'No')
``` | output | 1 | 1,451 | 21 | 2,903 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bracket sequence is a string that is one of the following:
1. An empty string;
2. The concatenation of `(`, A, and `)` in this order, for some bracket sequence A ;
3. The concatenation of A and B in this order, for some non-empty bracket sequences A and B /
Given are N strings S_i. Can a bracket sequence be formed by concatenating all the N strings in some order?
Constraints
* 1 \leq N \leq 10^6
* The total length of the strings S_i is at most 10^6.
* S_i is a non-empty string consisting of `(` and `)`.
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If a bracket sequence can be formed by concatenating all the N strings in some order, print `Yes`; otherwise, print `No`.
Examples
Input
2
)
(()
Output
Yes
Input
2
)(
()
Output
No
Input
4
((()))
((((((
))))))
()()()
Output
Yes
Input
3
(((
)
)
Output
No
Submitted Solution:
```
from sys import exit
from itertools import accumulate,chain
n,*s=open(0).read().split()
t=[2*(i.count("("))-len(i) for i in s]
if sum(t)!=0:
print("No");exit()
#pypyではinitialだめ
st=[[t_,min(accumulate(s_,lambda a,b: a+(1 if b=="(" else -1),initial=0))] for s_,t_ in zip(s,t)]
now=0
v=list(filter(lambda x:x[0]>=0,st))
w=list(filter(lambda x:x[0]<0,st))
v.sort(reverse=True,key=lambda z:z[1])
w.sort(key=lambda z:z[0]-z[1],reverse=True)
for sub in chain(v,w):
if now+sub[1]<0:
print("No");exit()
now+=sub[0]
print("Yes")
``` | instruction | 0 | 1,452 | 21 | 2,904 |
Yes | output | 1 | 1,452 | 21 | 2,905 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bracket sequence is a string that is one of the following:
1. An empty string;
2. The concatenation of `(`, A, and `)` in this order, for some bracket sequence A ;
3. The concatenation of A and B in this order, for some non-empty bracket sequences A and B /
Given are N strings S_i. Can a bracket sequence be formed by concatenating all the N strings in some order?
Constraints
* 1 \leq N \leq 10^6
* The total length of the strings S_i is at most 10^6.
* S_i is a non-empty string consisting of `(` and `)`.
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If a bracket sequence can be formed by concatenating all the N strings in some order, print `Yes`; otherwise, print `No`.
Examples
Input
2
)
(()
Output
Yes
Input
2
)(
()
Output
No
Input
4
((()))
((((((
))))))
()()()
Output
Yes
Input
3
(((
)
)
Output
No
Submitted Solution:
```
N=int(input())
if N==0:
print("Yes")
exit()
P=[]
M=[]
for i in range(N):
s=input()
f=0
m=0
cnt=0
for j in range(len(s)):
if s[j]=="(":
cnt+=1
else:
cnt-=1
if cnt<m:
m=cnt
if cnt>=0:
P.append([m,cnt])
else:
M.append([m-cnt,-cnt])
P.sort(reverse=True)
M.sort(reverse=True)
#print(P)
#print(M)
SUM=0
for i,j in P:
SUM+=j
for i,j in M:
SUM-=j
if SUM!=0:
print("No")
exit()
SUMP=0
for i,j in P:
if SUMP>=(-i):
SUMP+=j
else:
print("No")
exit()
SUMM=0
for i,j in M:
if SUMM>=(-i):
SUMM+=j
else:
print("No")
exit()
print("Yes")
``` | instruction | 0 | 1,453 | 21 | 2,906 |
Yes | output | 1 | 1,453 | 21 | 2,907 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bracket sequence is a string that is one of the following:
1. An empty string;
2. The concatenation of `(`, A, and `)` in this order, for some bracket sequence A ;
3. The concatenation of A and B in this order, for some non-empty bracket sequences A and B /
Given are N strings S_i. Can a bracket sequence be formed by concatenating all the N strings in some order?
Constraints
* 1 \leq N \leq 10^6
* The total length of the strings S_i is at most 10^6.
* S_i is a non-empty string consisting of `(` and `)`.
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If a bracket sequence can be formed by concatenating all the N strings in some order, print `Yes`; otherwise, print `No`.
Examples
Input
2
)
(()
Output
Yes
Input
2
)(
()
Output
No
Input
4
((()))
((((((
))))))
()()()
Output
Yes
Input
3
(((
)
)
Output
No
Submitted Solution:
```
n = int(input())
s = [input() for _ in range(n)]
def bracket(x):
# f: final sum of brackets '(':+1, ')': -1
# m: min value of f
f = m = 0
for i in range(len(x)):
if x[i] == '(':
f += 1
else:
f -= 1
m = min(m, f)
# m <= 0
return f, m
def func(l):
# l = [(f, m)]
l.sort(key=lambda x: -x[1])
v = 0
for fi, mi in l:
if v + mi >= 0:
v += fi
else:
return -1
return v
l1 = []
l2 = []
for i in range(n):
fi, mi = bracket(s[i])
if fi >= 0:
l1.append((fi, mi))
else:
l2.append((-fi, mi - fi))
v1 = func(l1)
v2 = func(l2)
if v1 == -1 or v2 == -1:
ans = 'No'
else:
ans = 'Yes' if v1 == v2 else 'No'
print(ans)
``` | instruction | 0 | 1,454 | 21 | 2,908 |
Yes | output | 1 | 1,454 | 21 | 2,909 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bracket sequence is a string that is one of the following:
1. An empty string;
2. The concatenation of `(`, A, and `)` in this order, for some bracket sequence A ;
3. The concatenation of A and B in this order, for some non-empty bracket sequences A and B /
Given are N strings S_i. Can a bracket sequence be formed by concatenating all the N strings in some order?
Constraints
* 1 \leq N \leq 10^6
* The total length of the strings S_i is at most 10^6.
* S_i is a non-empty string consisting of `(` and `)`.
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If a bracket sequence can be formed by concatenating all the N strings in some order, print `Yes`; otherwise, print `No`.
Examples
Input
2
)
(()
Output
Yes
Input
2
)(
()
Output
No
Input
4
((()))
((((((
))))))
()()()
Output
Yes
Input
3
(((
)
)
Output
No
Submitted Solution:
```
import sys
input = lambda: sys.stdin.readline().rstrip()
N = int(input())
def I():
s = input()
mi = 0
su = 0
t = 0
for a in s:
if a == "(":
t += 1
else:
t -= 1
mi = min(mi, t)
return (mi, t)
X = [I() for _ in range(N)]
if sum([x[1] for x in X]):
print("No")
exit()
X = sorted(X, key = lambda x: -10**10 if x[0] == 0 else -10**9 - x[0] if x[1] > 0 else x[0] - x[1])
T = 0
for mi, t in X:
if T + mi < 0:
print("No")
exit()
T += t
print("Yes")
``` | instruction | 0 | 1,455 | 21 | 2,910 |
Yes | output | 1 | 1,455 | 21 | 2,911 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bracket sequence is a string that is one of the following:
1. An empty string;
2. The concatenation of `(`, A, and `)` in this order, for some bracket sequence A ;
3. The concatenation of A and B in this order, for some non-empty bracket sequences A and B /
Given are N strings S_i. Can a bracket sequence be formed by concatenating all the N strings in some order?
Constraints
* 1 \leq N \leq 10^6
* The total length of the strings S_i is at most 10^6.
* S_i is a non-empty string consisting of `(` and `)`.
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If a bracket sequence can be formed by concatenating all the N strings in some order, print `Yes`; otherwise, print `No`.
Examples
Input
2
)
(()
Output
Yes
Input
2
)(
()
Output
No
Input
4
((()))
((((((
))))))
()()()
Output
Yes
Input
3
(((
)
)
Output
No
Submitted Solution:
```
# import sys
# input = sys.stdin.readline
def solve():
N = int(input())
brackets_gen = (input() for _ in range(N))
grads_positive = list()
grads_negative = list()
for brackets in brackets_gen:
elevation, bottom = 0, 0
for bk in brackets:
elevation += 1 if bk == '(' else -1
bottom = min(bottom, elevation)
if elevation >= 0:
grads_positive.append((bottom, elevation))
elif elevation < 0:
grads_negative.append((bottom - elevation, -elevation))
grads_positive.sort(reverse=True)
grads_negative.sort()
def is_good(grads):
elevation, bottom = 0, 0
for grad in grads:
bottom = elevation + grad[0]
if bottom < 0:
return False
elevation += grad[1]
if elevation == 0:
return True
else:
return False
if is_good(grads_positive) and is_good(grads_negative):
return True
else:
return False
def main():
ok = solve()
if ok:
print('Yes')
else:
print('No')
if __name__ == "__main__":
main()
``` | instruction | 0 | 1,456 | 21 | 2,912 |
No | output | 1 | 1,456 | 21 | 2,913 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bracket sequence is a string that is one of the following:
1. An empty string;
2. The concatenation of `(`, A, and `)` in this order, for some bracket sequence A ;
3. The concatenation of A and B in this order, for some non-empty bracket sequences A and B /
Given are N strings S_i. Can a bracket sequence be formed by concatenating all the N strings in some order?
Constraints
* 1 \leq N \leq 10^6
* The total length of the strings S_i is at most 10^6.
* S_i is a non-empty string consisting of `(` and `)`.
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If a bracket sequence can be formed by concatenating all the N strings in some order, print `Yes`; otherwise, print `No`.
Examples
Input
2
)
(()
Output
Yes
Input
2
)(
()
Output
No
Input
4
((()))
((((((
))))))
()()()
Output
Yes
Input
3
(((
)
)
Output
No
Submitted Solution:
```
def main():
N = int(input())
S = []
for i in range(N):
_ = input()
S.append([ _.count('(') - _.count(')'), _ ])
if sum([S[_][0] for _ in range(N)]) != 0:
print('No')
return
S.sort(reverse=True)
cnt = 0
for s in S:
for c in s[1]:
cnt += c.count('(') - c.count(')')
if cnt < 0:
print('No')
return
print('Yes')
if __name__ == "__main__":
main()
``` | instruction | 0 | 1,457 | 21 | 2,914 |
No | output | 1 | 1,457 | 21 | 2,915 |
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