message stringlengths 2 15.4k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 45 107k | cluster float64 21 21 | __index_level_0__ int64 90 214k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that |s| as the length of string s. A strict prefix s[1... l] (1≤ l< |s|) of a string s = s_1s_2... s_{|s|} is string s_1s_2... s_l. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string s containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in s independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
Input
The first line contains a single integer |s| (1≤ |s|≤ 3 ⋅ 10^5), the length of the string.
The second line contains a string s, containing only "(", ")" and "?".
Output
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
Examples
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
Note
It can be proved that there is no solution for the second sample, so print ":(".
Submitted Solution:
```
n=int(input())
S=list(input())
def solve():
if n%2:
return ':('
L=[0]*n
maxcnt=n//2-S.count('(')
cnt=0
for i in range(n):
if S[i]=='(':
L[i]=1
elif S[i]==')':
L[i]=-1
else:
if cnt<maxcnt:
L[i]=1; S[i]='('
cnt+=1
else:
L[i]=-1; S[i]=')'
judge=0
for i in range(n-1):
judge+=L[i]
if judge<=0:
return ':('
judge+=L[-1]
if judge!=0:
return ':('
return ''.join(S)
print(solve())
``` | instruction | 0 | 107,088 | 21 | 214,176 |
Yes | output | 1 | 107,088 | 21 | 214,177 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that |s| as the length of string s. A strict prefix s[1... l] (1≤ l< |s|) of a string s = s_1s_2... s_{|s|} is string s_1s_2... s_l. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string s containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in s independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
Input
The first line contains a single integer |s| (1≤ |s|≤ 3 ⋅ 10^5), the length of the string.
The second line contains a string s, containing only "(", ")" and "?".
Output
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
Examples
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
Note
It can be proved that there is no solution for the second sample, so print ":(".
Submitted Solution:
```
from math import ceil
import sys
input=sys.stdin.readline
from collections import defaultdict as dd
n=int(input())
s=input().split()[0]
lol=0
res=[0]*n
if(n%2 or s[0]==')' or s[-1]=='('):
print(':(')
else:
st=[]
stt=[]
res[0]='('
res[-1]=')'
for i in range(1,n-1):
if(s[i]=='('):
st.append(i)
res[i]='('
elif(s[i]=='?'):
stt.append(i)
else:
if(len(st)!=0):
st.pop()
res[i]=')'
else:
if(stt):
ind=stt.pop()
res[ind]='('
res[i]=')'
else:
lol=1
break
#print(lol)
x=len(stt)
#print(len(st),x)
if(len(st)>x):
lol=1
elif((x-len(st))%2):
lol=1
else:
cou=0
dif=len(st)
for i in range(dif):
a=st.pop()
b=stt.pop()
if(a>b):
lol=1
break
else:
res[b]=')'
while stt:
if(cou%2==0):
res[stt.pop()]=')'
else:
res[stt.pop()]='('
cou+=1
#print(lol)
if(lol):
print(':(')
else:
print(*res,sep="")
``` | instruction | 0 | 107,089 | 21 | 214,178 |
Yes | output | 1 | 107,089 | 21 | 214,179 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that |s| as the length of string s. A strict prefix s[1... l] (1≤ l< |s|) of a string s = s_1s_2... s_{|s|} is string s_1s_2... s_l. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string s containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in s independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
Input
The first line contains a single integer |s| (1≤ |s|≤ 3 ⋅ 10^5), the length of the string.
The second line contains a string s, containing only "(", ")" and "?".
Output
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
Examples
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
Note
It can be proved that there is no solution for the second sample, so print ":(".
Submitted Solution:
```
import sys
num = int(input())
raw = input()
inputs = [item for item in raw]
result = [item for item in raw]
if(num % 2 == 0):
found = True
a = 0
b = 0
c = 0
for i in range(num):
if(inputs[i] == '('):
a += 1
elif(inputs[i] == ')'):
b += 1
else:
c += 1
if((abs(a - b) + c)) % 2 != 0 or (a - b + c) < 0 or (b - a + c) < 0:
print(":(")
else:
if(a > b):
n = a - b
for i in range(num-1, 0, -1):
if(n == 0):
break
if(inputs[i] == '?'):
inputs[i] = ')'
n -= 1
elif(a < b):
n = b - a
for i in range(0, num):
if(n == 0):
break
if(inputs[i] == '?'):
inputs[i] = '('
n -= 1
n = (c - abs(a - b)) / 2
for i in range(0, num):
if(n == 0):
break
if(inputs[i] == '?'):
inputs[i] = '('
n -= 1
n = (c - abs(a - b)) / 2
for i in range(num-1, 0, -1):
if(n == 0):
break
if(inputs[i] == '?'):
inputs[i] = ')'
n -= 1
sum = (inputs[0] == '(') - (inputs[0] == ')')
temp = 0
for i in range(1, num - 1):
sum = sum + (inputs[i] == '(') - (inputs[i] == ')')
if(sum <= 0):
print(":(")
sys.exit(0)
print(''.join(inputs))
else:
print(":(")
``` | instruction | 0 | 107,090 | 21 | 214,180 |
Yes | output | 1 | 107,090 | 21 | 214,181 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that |s| as the length of string s. A strict prefix s[1... l] (1≤ l< |s|) of a string s = s_1s_2... s_{|s|} is string s_1s_2... s_l. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string s containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in s independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
Input
The first line contains a single integer |s| (1≤ |s|≤ 3 ⋅ 10^5), the length of the string.
The second line contains a string s, containing only "(", ")" and "?".
Output
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
Examples
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
Note
It can be proved that there is no solution for the second sample, so print ":(".
Submitted Solution:
```
n = int(input())
string = input()
if n % 2 != 0:
print(":(")
else:
ms = ""
sum = 0
o = 0
c = 0
h = n // 2
for i in string:
if i == "(":
o += 1
elif i == ")":
c += 1
# print(o, c, h)
for i in range(n):
if sum <= 0 and i != 0:
sum = 0
break
if sum == 0 and i == 0:
if string[i] == "?":
ms += "("
o += 1
sum += 1
elif string[i] == "(":
sum += 1
ms += "("
else:
sum -= 1
ms += ")"
if sum > 0 and i != 0:
if string[i] == "?":
if h - o > 0:
ms += "("
o += 1
sum += 1
elif h - c > 0:
ms += ")"
c += 1
sum -= 1
if sum == 0 and i < n-1:
print(":(")
else:
print(ms)
``` | instruction | 0 | 107,091 | 21 | 214,182 |
No | output | 1 | 107,091 | 21 | 214,183 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that |s| as the length of string s. A strict prefix s[1... l] (1≤ l< |s|) of a string s = s_1s_2... s_{|s|} is string s_1s_2... s_l. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string s containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in s independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
Input
The first line contains a single integer |s| (1≤ |s|≤ 3 ⋅ 10^5), the length of the string.
The second line contains a string s, containing only "(", ")" and "?".
Output
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
Examples
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
Note
It can be proved that there is no solution for the second sample, so print ":(".
Submitted Solution:
```
import sys
import math
from collections import defaultdict,deque
import heapq
n=int(sys.stdin.readline())
s = sys.stdin.readline()[:-1]
cnt=0
if s[0]==')':
print(":(")
sys.exit()
cnt+=1
ans=[0 for _ in range(n)]
ans[0]='('
#print(cnt,'cnt')
if n%2:
print(':(')
ans[0]='('
ans[1]='('
cnt=0
cnto=0
cntc=0
for i in range(n):
if s[i]=='(':
cnt+=1
rem=n//2-(cnt)
if s[0]=='?':
rem-=1
if s[1]=='?':
rem-=1
#print(rem,'rem')
for i in range(2,n-1):
if s[i]=='?':
if cnto<rem:
ans[i]='('
cnto+=1
else:
ans[i]=')'
else:
if s[i]=='(':
cnto+=1
ans[i]=s[i]
if s[-1]!='?':
if s[-1]=='(':
print(':(')
sys.exit()
ans[-1]=')'
res=0
#print(ans,'ans')
for i in range(n):
if res<0:
print(":(")
sys.exit()
if ans[i]=='(':
res+=1
else:
res-=1
if res>0:
print(':(')
sys.exit()
print(''.join(x for x in ans))
``` | instruction | 0 | 107,092 | 21 | 214,184 |
No | output | 1 | 107,092 | 21 | 214,185 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that |s| as the length of string s. A strict prefix s[1... l] (1≤ l< |s|) of a string s = s_1s_2... s_{|s|} is string s_1s_2... s_l. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string s containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in s independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
Input
The first line contains a single integer |s| (1≤ |s|≤ 3 ⋅ 10^5), the length of the string.
The second line contains a string s, containing only "(", ")" and "?".
Output
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
Examples
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
Note
It can be proved that there is no solution for the second sample, so print ":(".
Submitted Solution:
```
import sys
import math
from collections import defaultdict,deque
import heapq
n=int(sys.stdin.readline())
s = sys.stdin.readline()[:-1]
cnt=0
if s[0]==')':
print(":(")
sys.exit()
cnt+=1
ans=[0 for _ in range(n)]
ans[0]='('
#print(cnt,'cnt')
if n%2:
print(':(')
ans[0]='('
ans[1]='('
cnt=0
cnto=0
cntc=0
for i in range(n):
if s[i]=='(':
cnt+=1
rem=n//2-(cnt)
if s[0]=='?':
rem-=1
if s[1]=='?':
rem-=1
print(rem,'rem')
for i in range(2,n-1):
if s[i]=='?':
if cnto<rem:
ans[i]='('
cnto+=1
else:
ans[i]=')'
else:
if s[i]=='(':
cnto+=1
ans[i]=s[i]
if s[-1]!='?':
if s[-1]=='(':
print(':(')
sys.exit()
ans[-1]=')'
res=0
#print(ans,'ans')
for i in range(n):
if res<0:
print(":(")
sys.exit()
if ans[i]=='(':
res+=1
else:
res-=1
if res>0:
print(':(')
sys.exit()
print(''.join(x for x in ans))
``` | instruction | 0 | 107,093 | 21 | 214,186 |
No | output | 1 | 107,093 | 21 | 214,187 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that |s| as the length of string s. A strict prefix s[1... l] (1≤ l< |s|) of a string s = s_1s_2... s_{|s|} is string s_1s_2... s_l. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string s containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in s independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
Input
The first line contains a single integer |s| (1≤ |s|≤ 3 ⋅ 10^5), the length of the string.
The second line contains a string s, containing only "(", ")" and "?".
Output
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
Examples
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
Note
It can be proved that there is no solution for the second sample, so print ":(".
Submitted Solution:
```
n = int(input())
s = str(input()[:n])
a = [0]*n
count = 0 # count open parenthesis
pos = 1
if n%2 !=0:
print(":(")
else:
for i in range(1, n):
if s[i] == '(':
a[i] = 1
elif s[i] == ')':
a[i] = -1
a[0] = 1
#print(a)
for pos in range(1, n//2):
if a[pos] == 1:
count += 1
elif a[pos] == 0:
if count > 1:
a[pos] = -1
count -= 1
else:
a[pos] = 1
count += 1
for pos in range(n//2, n-1):
if a[pos] == 0 and count > 0:
a[pos] = -1
count -= 1
else:
a[pos] = 1
count += 1
#print(a)
if s[n-1] == '(':
print(':(')
quit()
else:
a[n - 1] = -1
for i in range(n):
if sum(a) == 0:
if a[i] == 1:
print('(', end='')
elif a[i] == -1:
print(')', end='')
else:
print(':(')
break
#print(a)
``` | instruction | 0 | 107,094 | 21 | 214,188 |
No | output | 1 | 107,094 | 21 | 214,189 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()())) | instruction | 0 | 107,127 | 21 | 214,254 |
Tags: dp, strings, two pointers
Correct Solution:
```
S, T = input(), input()
s1 = [1 if c=='(' else -1 for c in S]
s2 = [1 if c=='(' else -1 for c in T]
len1, len2 = len(s1), len(s2)
inf = 10**9
dp = [[[inf]*(len1+len2+1) for _ in range(len2+1)] for _ in range(len1+1)]
dp[0][0][0] = 0
for i in range(len1+1):
for j in range(len2+1):
for k in range(len1+len2):
if i < len1:
k1 = k + s1[i]
if k1 < 0:
dp[i+1][j][k] = min(dp[i+1][j][k], dp[i][j][k]+2)
else:
dp[i+1][j][k1] = min(dp[i+1][j][k1], dp[i][j][k]+1)
if j < len2:
k2 = k + s2[j]
if k2 < 0:
dp[i][j+1][k] = min(dp[i][j+1][k], dp[i][j][k]+2)
else:
dp[i][j+1][k2] = min(dp[i][j+1][k2], dp[i][j][k]+1)
if i < len1 and j < len2:
if s1[i] == s2[j]:
if k1 < 0:
dp[i+1][j+1][k] = min(dp[i+1][j+1][k], dp[i][j][k]+2)
else:
dp[i+1][j+1][k1] = min(dp[i+1][j+1][k1], dp[i][j][k] + 1)
k = 0
length = inf
for i in range(len1+len2+1):
if dp[len1][len2][i] + i < length:
length = dp[len1][len2][i] + 1
k = i
i, j = len1, len2
ans = [')']*k
while i > 0 or j > 0:
if i > 0 and j > 0 and s1[i-1] == s2[j-1]:
if dp[i-1][j-1][k-s1[i-1]] + 1 == dp[i][j][k]:
ans.append(S[i-1])
k -= s1[i-1]
i -= 1
j -= 1
continue
elif k == 0 and s1[i-1] == -1 and dp[i-1][j-1][k] + 2 == dp[i][j][k]:
ans.extend((S[i-1], '('))
i -= 1
j -= 1
continue
if i > 0:
if dp[i-1][j][k-s1[i-1]] + 1 == dp[i][j][k]:
ans.append(S[i-1])
k -= s1[i-1]
i -= 1
continue
elif k == 0 and s1[i-1] == -1 and dp[i-1][j][k] + 2 == dp[i][j][k]:
ans.extend((S[i-1], '('))
i -= 1
continue
if j > 0:
if dp[i][j-1][k-s2[j-1]] + 1 == dp[i][j][k]:
ans.append(T[j-1])
k -= s2[j-1]
j -= 1
continue
elif k == 0 and s2[j-1] == -1 and dp[i][j-1][k] + 2 == dp[i][j][k]:
ans.extend((T[j-1], '('))
j -= 1
continue
raise Exception(i, j, k, S[i-1], T[j-1])
print(*ans[::-1], sep='')
``` | output | 1 | 107,127 | 21 | 214,255 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()())) | instruction | 0 | 107,128 | 21 | 214,256 |
Tags: dp, strings, two pointers
Correct Solution:
```
s1 = input();
s2 = input();
inf = 10**8;
n = len(s1);
m = len(s2);
d = [[[inf for x in range(n+m+10)] for y in range(m+10)] for z in range(n+10)];
tata = [[[0 for x in range(n+m+10)] for y in range(m+10)] for z in range(n+10)];
d[0][0][0]=0;
for i in range(n+1):
for j in range(m+1):
i1 = i + (i<n and s1[i] == '(');
j1 = j + (j<m and s2[j] == '(');
for k in range(n+m+1):
if d[i][j][k]==inf:
continue;
if d[i][j][k]+1<d[i1][j1][k+1]:
d[i1][j1][k+1]=d[i][j][k]+1;
tata[i1][j1][k+1]=1;
i1 = i + (i<n and s1[i] == ')');
j1 = j + (j<m and s2[j] == ')');
for k in reversed(range(1,n+m+1)):
if d[i][j][k]==inf:
continue;
if d[i][j][k]+1<d[i1][j1][k-1]:
d[i1][j1][k-1]=d[i][j][k]+1;
tata[i1][j1][k-1]=-1;
lmin=inf;
pz=0;
for i in range(n+m+1):
if d[n][m][i]+i<lmin:
lmin=d[n][m][i]+i;
pz=i;
ans = "";
for i in range(pz):
ans+=')';
i = n;
j = m;
k = pz;
while(i>0 or j>0 or k>0):
val = d[i][j][k]-1;
if tata[i][j][k]==1:
c='(';
k-=1;
else:
c=')';
k+=1;
ans+=c;
i1 = (i>0 and s1[i-1]==c);
j1 = (j>0 and s2[j-1]==c);
if d[i][j][k]==val:
continue;
elif i1>0 and d[i-1][j][k]==val:
i-=1;
elif j1>0 and d[i][j-1][k]==val:
j-=1;
else:
i-=1;
j-=1;
ans = ans[::-1];
print(ans);
``` | output | 1 | 107,128 | 21 | 214,257 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()())) | instruction | 0 | 107,129 | 21 | 214,258 |
Tags: dp, strings, two pointers
Correct Solution:
```
def scs(str1, str2):
"""Shortest Common Supersequence"""
INF = 10 ** 9
dp = [[[INF] * 210 for _ in range(210)] for _ in range(210)]
dp[0][0][0] = 0
prv = [[[None] * 210 for _ in range(210)] for _ in range(210)]
len_str1 = len(str1)
len_str2 = len(str2)
str1 += "#"
str2 += "#"
for i in range(len_str1 + 1):
for j in range(len_str2 + 1):
for k in range(201):
# add "("
k2 = k + 1
i2 = i + (str1[i] == "(")
j2 = j + (str2[j] == "(")
if dp[i2][j2][k2] > dp[i][j][k] + 1:
dp[i2][j2][k2] = dp[i][j][k] + 1
prv[i2][j2][k2] = (i, j, k)
for k in range(1, 201):
# add ")"
k2 = k - 1
i2 = i + (str1[i] == ")")
j2 = j + (str2[j] == ")")
if dp[i2][j2][k2] > dp[i][j][k] + 1:
dp[i2][j2][k2] = dp[i][j][k] + 1
prv[i2][j2][k2] = (i, j, k)
ans = INF
cnt = 0
for c, val in enumerate(dp[len_str1][len_str2]):
if c + val < ans + cnt:
ans = val
cnt = c
res = []
i, j, k = len_str1, len_str2, cnt
while i > 0 or j > 0 or k > 0:
prv_i, prv_j, prv_k = prv[i][j][k]
if prv_k < k:
res.append("(")
else:
res.append(")")
i, j, k = prv_i, prv_j, prv_k
return "(" * (ans - len(res) - cnt) + "".join(res[::-1]) + ")" * cnt
s = input()
t = input()
print(scs(s, t))
``` | output | 1 | 107,129 | 21 | 214,259 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()())) | instruction | 0 | 107,130 | 21 | 214,260 |
Tags: dp, strings, two pointers
Correct Solution:
```
import sys
import time
def play(s1, s2):
#start_time = time.time()
n = len(s1)
m = len(s2)
# баланс скобок не может быть > 200 (всегда можно построить правильную последовательность, не обяхателбно минимальной длины с балансом
# < 200)
maxBalance = 200
#print(time.time() - start_time)
#start_time = time.time()
dp = []
p = []
for i in range (0, n + 1):
dp.append([])
p.append([])
for j in range (0, m + 1):
dp[i].append([])
p[i].append([])
for a in range (0, maxBalance + 1):
dp[i][j].append(sys.maxsize)
p[i][j].append(None)
#dp = [[[{'len': sys.maxsize, 'i': 0, 'j': 0, 'a': 0, 'c': ''} for a in range(maxBalance + 1)] for i in range(m + 1)] for j in range(n + 1)]
#print(time.time() - start_time)
# dp[i][j][a] - это минимальная длина итоговой последовательности после прочтения
# i символов строки s1,
# j символов строки s2,
# a - баланс скобок (количество открытых, но не закрытых скобок)
#start_time = time.time()
dp[0][0][0] = 0
#print(time.time() - start_time)
#start_time = time.time()
for i in range(0, n + 1):
for j in range(0, m + 1):
for a in range(0, maxBalance + 1):
if dp[i][j][a] == sys.maxsize:
continue
# сейчас мы находимся в следующем положении:
# 1. В позиции i строки s1 (длина i + 1)
# 2. В позицию j строки s2 (длина j + 1)
# 3. имеем баланс a
# Возможны всегда 2 варианта действия:
# Добавляем в выходную строку '(' или ')'
# Добавляем '(' => баланс увеличится на 1. Если в s1 открывающаяся скобка, то можем дигаться по ней далее
# но если же там была закрывающаяся скобка, то стоим где сейчас. Поэтому:
next_i = i + (i < n and s1[i] == '(')
next_j = j + (j < m and s2[j] == '(')
if a < maxBalance and dp[next_i][next_j][a + 1] > dp[i][j][a] + 1:
dp[next_i][next_j][a + 1] = dp[i][j][a] + 1
p[next_i][next_j][a + 1] = (i, j, a, '(')
# Второй вариант - добавление ')'. Рассуждения аналогичны. При добавлении закрывающей скобки баланс всегда уменьшается на 1
next_i = i + (i < n and s1[i] == ')')
next_j = j + (j < m and s2[j] == ')')
if a > 0 and dp[next_i][next_j][a - 1] > dp[i][j][a] + 1:
dp[next_i][next_j][a - 1] = dp[i][j][a] + 1
p[next_i][next_j][a - 1] = (i, j, a, ')')
#print(time.time() - start_time)
aa = 0
nn = n
mm = m
#start_time = time.time()
for a in range(0, maxBalance + 1):
if dp[n][m][a] + a < dp[n][m][aa] + aa:
aa = a
#print(time.time() - start_time)
#start_time = time.time()
ret = ')' * aa
while nn > 0 or mm > 0 or aa > 0:
d = p[nn][mm][aa]
nn = d[0]
mm = d[1]
aa = d[2]
ret += d[3]
#print(time.time() - start_time)
return ret[::-1]
def main():
#start_time = time.time()
#print(play(input(), input()))
print(play(input(), input()))
#print(time.time() - start_time)
main()
#print(play('(())(()','()))()'))
#print(play('((',')'))
``` | output | 1 | 107,130 | 21 | 214,261 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()())) | instruction | 0 | 107,131 | 21 | 214,262 |
Tags: dp, strings, two pointers
Correct Solution:
```
def limitbal(a, b):
min_level = 0
level = 0
for aa in a:
if aa == '(':
level += 1
else:
level -= 1
if level < min_level:
min_level = level
for aa in b:
if aa == '(':
level += 1
else:
level -= 1
if level < min_level:
min_level = level
return -min_level + len(a) + len(b) + (level - min_level)
def go():
a = input()
b = input()
lena = len(a)
lenb = len(b)
a+='X'
b+='X'
bal_lim = limitbal(a, b)//2 + 3
tab = [[[None] * (bal_lim+2) for _ in range(lenb + 1)] for _ in range(lena+1)]
par = [[[None] * (bal_lim+2) for _ in range(lenb + 1)] for _ in range(lena+1)]
tab[0][0][0] = 0
que = [(0, 0, 0)]
index = 0
while tab[lena][lenb][0] is None:
i, j, bal = que[index]
ai = a[i]
bj = b[j]
if bal < bal_lim and (bal==0 or not (ai==bj==')')):
# Add (
ii = i
jj = j
if ai == '(':
ii = i + 1
if bj == '(':
jj = j + 1
if tab[ii][jj][bal + 1] is None:
tab[ii][jj][bal + 1] = tab[i][j][bal] + 1
par[ii][jj][bal + 1] = i, j, bal, '('
que.append((ii, jj, bal + 1))
if bal > 0 and not (ai==bj=='('):
ii = i
jj = j
# Add )
if ai == ')':
ii = i + 1
if bj == ')':
jj = j + 1
if tab[ii][jj][bal - 1] is None:
tab[ii][jj][bal - 1] = tab[i][j][bal] + 1
par[ii][jj][bal - 1] = i, j, bal, ')'
que.append((ii, jj, bal - 1))
index += 1
i = lena
j = lenb
bal = 0
answer = []
while (i, j, bal) != (0, 0, 0):
# print (i,j,bal)
i, j, bal, symb = par[i][j][bal]
answer.append(symb)
print(''.join(reversed(answer)))
go()
``` | output | 1 | 107,131 | 21 | 214,263 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()())) | instruction | 0 | 107,132 | 21 | 214,264 |
Tags: dp, strings, two pointers
Correct Solution:
```
import sys,math,itertools
from collections import Counter,deque,defaultdict
from bisect import bisect_left,bisect_right
from heapq import heappop,heappush,heapify, nlargest
from copy import deepcopy
mod = 10**9+7
INF = float('inf')
def inp(): return int(sys.stdin.readline())
def inpl(): return list(map(int, sys.stdin.readline().split()))
def inpl_1(): return list(map(lambda x:int(x)-1, sys.stdin.readline().split()))
def inps(): return sys.stdin.readline()
def inpsl(x): tmp = sys.stdin.readline(); return list(tmp[:x])
def err(x): print(x); exit()
s = list(input())
n = len(s)
t = list(input())
m = len(t)
d = {'(':1 , ')':-1}
s = [1 if x=='(' else -1 for x in s]
t = [1 if x=='(' else -1 for x in t]
s.append(0); t.append(0)
K = 210
dp = [[[INF]*K for _ in range(m+2)] for i in range(n+2)]
pa = [[[-1 for k in range(K)] for j in range(m+1)] for i in range(n+1)]
dp[0][0][0] = 0
for i in range(n+1):
for j in range(m+1):
for k in range(201):
nk = k+1
ni = i+(s[i]==1)
nj = j+(t[j]==1)
if dp[ni][nj][nk] > dp[i][j][k]+1:
dp[ni][nj][nk] = dp[i][j][k]+1
pa[ni][nj][nk] = (i,j,k)
for k in range(1,201):
nk = k-1
ni = i+(s[i]==-1)
nj = j+(t[j]==-1)
if dp[ni][nj][nk] > dp[i][j][k]+1:
dp[ni][nj][nk] = dp[i][j][k]+1
pa[ni][nj][nk] = (i,j,k)
rem = -1
mi = INF
for k,num in enumerate(dp[n][m]):
if k+num < mi:
rem = k; mi = k+num
res = []
i,j,k = n,m,rem
while i>0 or j>0 or k>0:
pi,pj,pk = pa[i][j][k]
if k > pk: res.append('(')
else: res.append(')')
i,j,k = pi,pj,pk
res = res[::-1]
for i in range(rem): res.append(')')
print(''.join(res))
``` | output | 1 | 107,132 | 21 | 214,265 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
Submitted Solution:
```
s=input()
s1=input()
o=max(s.count("("),s1.count("("))
p=max(s.count(")"),s1.count(")"))
if(o>=p):
print("("*o + ")"*o)
else:
print("(" * p + ")" * p)
``` | instruction | 0 | 107,133 | 21 | 214,266 |
No | output | 1 | 107,133 | 21 | 214,267 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
Submitted Solution:
```
import sys,math,itertools
from collections import Counter,deque,defaultdict
from bisect import bisect_left,bisect_right
from heapq import heappop,heappush,heapify, nlargest
from copy import deepcopy
mod = 10**9+7
INF = float('inf')
def inp(): return int(sys.stdin.readline())
def inpl(): return list(map(int, sys.stdin.readline().split()))
def inpl_1(): return list(map(lambda x:int(x)-1, sys.stdin.readline().split()))
def inps(): return sys.stdin.readline()
def inpsl(x): tmp = sys.stdin.readline(); return list(tmp[:x])
def err(x): print(x); exit()
s = list(input())
n = len(s)
t = list(input())
m = len(t)
d = {'(':1 , ')':-1}
s = [d[x] for x in s]
t = [d[x] for x in t]
s += [0]; t += [0]
K = 210
dp = [[[INF]*K for _ in range(m+2)] for i in range(n+2)]
pa = [[[(-1,-1,-1) for k in range(K)] for j in range(m+1)] for i in range(n+1)]
dp[0][0][0] = 0
for i in range(n+1):
for j in range(m+1):
for k in range(5):
nk = k+1
ni = i+(s[i]==1)
nj = j+(t[j]==1)
if dp[ni][nj][nk] > dp[i][j][k]+1:
dp[ni][nj][nk] = dp[i][j][k]+1
pa[ni][nj][nk] = (i,j,k)
for k in range(1,5):
nk = k-1
ni = i+(s[i]==-1)
nj = j+(t[j]==-1)
if dp[ni][nj][nk] > dp[i][j][k]+1:
dp[ni][nj][nk] = dp[i][j][k]+1
pa[ni][nj][nk] = (i,j,k)
rem = -1
mi = INF
for k,num in enumerate(dp[n][m]):
if k+num < mi:
rem = k; mi = k+num
res = []
i,j,k = n,m,rem
while i>0 or j>0 or k>0:
pi,pj,pk = pa[i][j][k]
if k > pk: res.append('(')
else: res.append(')')
i,j,k = pi,pj,pk
print(''.join(res[::-1]) + ')'*rem)
``` | instruction | 0 | 107,134 | 21 | 214,268 |
No | output | 1 | 107,134 | 21 | 214,269 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
Submitted Solution:
```
s = input()
t = input()
d = {'(':1, ')':-1}
if s[0] != '(': s = '(' + s
if t[0] != '(': t = '(' + t
for i in range(min(len(s),len(t))):
if s[i] != t[i]:
if t[i] == '(':
tmp = s
s = t
t = tmp
break
b = []
c = []
for i in range(len(s)+1):
newc = []
newb = []
for j in range(len(t)+1):
newc.append(0)
newb.append('')
c.append(newc)
b.append(newb)
for i in range(len(s)+1):
c[i][0] = i
b[i][0] = s[:i]
for i in range(len(t)+1):
c[0][i] = i
b[0][i] = t[:i]
for i in range(1,1+len(s)):
for j in range(1,1+len(t)):
if s[i-1]==t[j-1]:
c[i][j] = c[i-1][j-1]
b[i][j] = b[i-1][j-1]+s[i-1]
else:
if c[i-1][j]+1<c[i][j-1]+1:
c[i][j]=c[i-1][j]+1
b[i][j]=b[i-1][j]+s[i-1]
else:
c[i][j]=c[i][j-1]+1
b[i][j]=b[i][j-1]+t[j-1]
subtotal = b[-1][-1]
# tt = 0
# for i in range(len(subtotal)):
# tt += d[subtotal[i]]
# for x in range(tt):
# subtotal = subtotal + ')'
# for x in range(-tt):
# subtotal = '(' + subtotal
tt = 0
add = 0
for i in range(len(subtotal)):
tt += d[subtotal[i]]
if tt<0:
add += 1
tt += 1
for i in range(add):
subtotal = '(' + subtotal
print(subtotal)
# print(c[len(s)][len(t)])
# s1.append(d[s[i]])
# total1 += d[s[i]]
# for i in range(len(t)):
# s2.append(d[t[i]])
# total2 += d[t[i]]
# if total > 0:
# for i in range(total): s1.append(-1)
# elif total < 0:
# for i in range(-total): s1.insert(0, 1)
``` | instruction | 0 | 107,135 | 21 | 214,270 |
No | output | 1 | 107,135 | 21 | 214,271 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).
Recall what is the regular bracket sequence:
* () is the regular bracket sequence;
* if S is the regular bracket sequence, then (S) is a regular bracket sequence;
* if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence.
Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.
Input
The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.
The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.
Output
Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.
Examples
Input
(())(()
()))()
Output
(())()()
Input
)
((
Output
(())
Input
)
)))
Output
((()))
Input
())
(()(()(()(
Output
(()()()(()()))
Submitted Solution:
```
s = input()
t = input()
d = {'(':1, ')':-1}
if '(' in s and '(' in t:
if s.index('(') > t.index('('):
tmp = s
s = t
t = tmp
elif '(' not in s:
tmp = s
s = t
t = tmp
b = []
c = []
for i in range(len(s)+1):
newc = []
newb = []
for j in range(len(t)+1):
newc.append(0)
newb.append('')
c.append(newc)
b.append(newb)
for i in range(len(s)+1):
c[i][0] = i
b[i][0] = s[:i]
for i in range(len(t)+1):
c[0][i] = i
b[0][i] = t[:i]
for i in range(1,1+len(s)):
for j in range(1,1+len(t)):
if s[i-1]==t[j-1]:
c[i][j] = c[i-1][j-1]
b[i][j] = b[i-1][j-1]+s[i-1]
else:
if c[i-1][j]+1<c[i][j-1]+1:
c[i][j]=c[i-1][j]+1
b[i][j]=b[i-1][j]+s[i-1]
else:
c[i][j]=c[i][j-1]+1
b[i][j]=b[i][j-1]+t[j-1]
subtotal = b[-1][-1]
print(subtotal)
tt = 0
for i in range(len(subtotal)):
tt += d[subtotal[i]]
for x in range(tt):
subtotal = subtotal + ')'
for y in range(-tt):
subtotal = '(' + subtotal
print(subtotal)
# print(c[len(s)][len(t)])
# s1.append(d[s[i]])
# total1 += d[s[i]]
# for i in range(len(t)):
# s2.append(d[t[i]])
# total2 += d[t[i]]
# if total > 0:
# for i in range(total): s1.append(-1)
# elif total < 0:
# for i in range(-total): s1.insert(0, 1)
``` | instruction | 0 | 107,136 | 21 | 214,272 |
No | output | 1 | 107,136 | 21 | 214,273 |
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