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.
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 heapq
N=int(input())
S = [input() for _ in range(N)]
p1 = 0
p2 = []
p3 = []
for i,s in enumerate(S):
depth = 0
max_depth = 0
min_depth = 0
for chr in s:
if chr=='(':
depth += 1
if depth > max_depth:
max_depth = depth
else:
depth -= 1
if depth < min_depth:
min_depth = depth
if depth >= 0 and min_depth >= 0:
p1 += depth
elif depth > 0 and min_depth < 0:
p2.append((min_depth,depth))
else:
p3.append((min_depth,depth))
pp2 = sorted(p2,reverse=True)
pp3 = sorted(p3,reverse=True)
# print(p1)
# print(p2)
# print(p3)
depth = p1
for p in pp2:
if depth+p[0] < 0:
print('No')
exit()
depth += p[1]
for p in pp3:
if depth+p[0] < 0:
print('No')
exit()
depth += p[1]
if depth != 0:
print('No')
else:
print('Yes')
``` | instruction | 0 | 1,458 | 21 | 2,916 |
No | output | 1 | 1,458 | 21 | 2,917 |
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
n=int(input())
s=[input() for i in range(n)]
t=[2*(i.count("("))-len(i) for i in s]
if sum(t)!=0:
print("No")
sys.exit()
st=list(map(list,zip(s,t)))
for i in range(n):
u=st[i][0]
l=len(u)
now,mi=0,0
for j in range(l):
if u[j]=="(":
now+=1
else:
now-=1
mi=min(mi,now)
st[i].append(mi)
st.sort(reverse=True,key=lambda z:z[1])
st.sort(reverse=True,key=lambda z:z[2])
now2=0
for i in range(n):
if now2+st[i][2]<0:
print("No")
break
now2+=st[i][1]
else:
print("Yes")
``` | instruction | 0 | 1,459 | 21 | 2,918 |
No | output | 1 | 1,459 | 21 | 2,919 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bracket sequence is a string, containing only characters "(", ")", "[" and "]".
A correct 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 correct (the resulting expressions are: "(1)+[1]", "([1+1]+1)"), and "](" and "[" are not. The empty string is a correct bracket sequence by definition.
A substring s[l... r] (1 ≤ l ≤ r ≤ |s|) of string s = s1s2... s|s| (where |s| is the length of string s) is the string slsl + 1... sr. The empty string is a substring of any string by definition.
You are given a bracket sequence, not necessarily correct. Find its substring which is a correct bracket sequence and contains as many opening square brackets «[» as possible.
Input
The first and the only line contains the bracket sequence as a string, consisting only of characters "(", ")", "[" and "]". It is guaranteed that the string is non-empty and its length doesn't exceed 105 characters.
Output
In the first line print a single integer — the number of brackets «[» in the required bracket sequence. In the second line print the optimal sequence. If there are more than one optimal solutions print any of them.
Examples
Input
([])
Output
1
([])
Input
(((
Output
0 | instruction | 0 | 3,648 | 21 | 7,296 |
Tags: data structures, expression parsing, implementation
Correct Solution:
```
s, st, v, vi, vj, vc = input(), [], [], 0, 0, 0
for i, ch in enumerate(s):
if ch in '[(':
st.append(i)
else:
if st and s[st[-1]] + ch in ('()', '[]'):
b = (st[-1], i + 1)
if v and v[-1][1] == i:
v[-1] = b
else:
v.append(b)
if len(v) >= 2 and v[-2][1] == v[-1][0]:
v[-2:] = [(v[-2][0], v[-1][1])]
st.pop()
else:
st = []
for b in v:
c = s.count('[', b[0], b[1])
if c > vc:
vi, vj, vc = b[0], b[1], c
print(vc)
print(s[vi:vj])
``` | output | 1 | 3,648 | 21 | 7,297 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bracket sequence is a string, containing only characters "(", ")", "[" and "]".
A correct 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 correct (the resulting expressions are: "(1)+[1]", "([1+1]+1)"), and "](" and "[" are not. The empty string is a correct bracket sequence by definition.
A substring s[l... r] (1 ≤ l ≤ r ≤ |s|) of string s = s1s2... s|s| (where |s| is the length of string s) is the string slsl + 1... sr. The empty string is a substring of any string by definition.
You are given a bracket sequence, not necessarily correct. Find its substring which is a correct bracket sequence and contains as many opening square brackets «[» as possible.
Input
The first and the only line contains the bracket sequence as a string, consisting only of characters "(", ")", "[" and "]". It is guaranteed that the string is non-empty and its length doesn't exceed 105 characters.
Output
In the first line print a single integer — the number of brackets «[» in the required bracket sequence. In the second line print the optimal sequence. If there are more than one optimal solutions print any of them.
Examples
Input
([])
Output
1
([])
Input
(((
Output
0 | instruction | 0 | 3,649 | 21 | 7,298 |
Tags: data structures, expression parsing, implementation
Correct Solution:
```
string = list(input())
d, p = [], set(range(len(string)))
for j, q in enumerate(string):
if q in '([': d.append((j,q))
elif d:
i, x = d.pop()
if x+q in '(][)': d = []
else: p -= {i, j}
n, s = 0, ''
for i in p: string[i] = ' '
for k in "".join(string).split():
if k.count("[") > n:
n = k.count("[")
s = k
print(n, s)
``` | output | 1 | 3,649 | 21 | 7,299 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bracket sequence is a string, containing only characters "(", ")", "[" and "]".
A correct 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 correct (the resulting expressions are: "(1)+[1]", "([1+1]+1)"), and "](" and "[" are not. The empty string is a correct bracket sequence by definition.
A substring s[l... r] (1 ≤ l ≤ r ≤ |s|) of string s = s1s2... s|s| (where |s| is the length of string s) is the string slsl + 1... sr. The empty string is a substring of any string by definition.
You are given a bracket sequence, not necessarily correct. Find its substring which is a correct bracket sequence and contains as many opening square brackets «[» as possible.
Input
The first and the only line contains the bracket sequence as a string, consisting only of characters "(", ")", "[" and "]". It is guaranteed that the string is non-empty and its length doesn't exceed 105 characters.
Output
In the first line print a single integer — the number of brackets «[» in the required bracket sequence. In the second line print the optimal sequence. If there are more than one optimal solutions print any of them.
Examples
Input
([])
Output
1
([])
Input
(((
Output
0 | instruction | 0 | 3,650 | 21 | 7,300 |
Tags: data structures, expression parsing, implementation
Correct Solution:
```
import sys
import math
import string
import operator
import functools
import fractions
import collections
sys.setrecursionlimit(10**7)
dX= [-1, 1, 0, 0,-1, 1,-1, 1]
dY= [ 0, 0,-1, 1, 1,-1,-1, 1]
RI=lambda: list(map(int,input().split()))
RS=lambda: input().rstrip().split()
#################################################
s=RS()[0]
st=[]
closeInd=[0]*len(s)
def rev(c):
return "(["[c==']']
for i in range(len(s)):
if len(st) and s[i] in ")]" and st[-1][0]==rev(s[i]):
temp=st[-1]
st.pop()
closeInd[temp[1]]=i
else:
st.append((s[i],i))
maxSq=0
maxX,maxY=0,0
i=0
while i<len(s):
sq=0
start=i
while i<len(s) and closeInd[i]:
j=i
while i<=closeInd[j]:
if s[i]=='[':
sq+=1
i+=1
else:
i+=1
if sq>maxSq:
maxSq=sq
maxX=start
maxY=i-1
print(maxSq, s[maxX:maxY], sep='\n')
``` | output | 1 | 3,650 | 21 | 7,301 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bracket sequence is a string, containing only characters "(", ")", "[" and "]".
A correct 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 correct (the resulting expressions are: "(1)+[1]", "([1+1]+1)"), and "](" and "[" are not. The empty string is a correct bracket sequence by definition.
A substring s[l... r] (1 ≤ l ≤ r ≤ |s|) of string s = s1s2... s|s| (where |s| is the length of string s) is the string slsl + 1... sr. The empty string is a substring of any string by definition.
You are given a bracket sequence, not necessarily correct. Find its substring which is a correct bracket sequence and contains as many opening square brackets «[» as possible.
Input
The first and the only line contains the bracket sequence as a string, consisting only of characters "(", ")", "[" and "]". It is guaranteed that the string is non-empty and its length doesn't exceed 105 characters.
Output
In the first line print a single integer — the number of brackets «[» in the required bracket sequence. In the second line print the optimal sequence. If there are more than one optimal solutions print any of them.
Examples
Input
([])
Output
1
([])
Input
(((
Output
0 | instruction | 0 | 3,651 | 21 | 7,302 |
Tags: data structures, expression parsing, implementation
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
INF = 10 ** 18
MOD = 10**9+7
Ri = lambda : [int(x) for x in sys.stdin.readline().split()]
ri = lambda : sys.stdin.readline().strip()
s = ri()
sta = []
l,r = -1,-2
lis = []
for i in range(len(s)):
if s[i] == '(':
sta.append(['(', i])
elif s[i] == '[':
sta.append(['[', i])
elif s[i] == ')':
if len(sta) == 0 or sta[-1][0] != '(':
sta = []
else:
temp = sta.pop()
lis.append([temp[1], i])
elif s[i] == ']':
if len(sta) == 0 or sta[-1][0] != '[':
sta = []
else:
temp = sta.pop()
lis.append([temp[1], i])
lis.sort(key = lambda x : (x[0], -x[1]))
cur = []
totcnt =0
cnt = 0
# print(lis)
for i in range(len(lis)):
if len(cur) == 0:
cur = lis[i][:]
if s[lis[i][0]] == '[':
cnt+=1
elif cur[0] <= lis[i][0] and lis[i][1] <= cur[1]:
if s[lis[i][0]] == '[':
cnt+=1
else:
if lis[i][0] == cur[1]+1:
cur[1] = lis[i][1]
if s[lis[i][0]] == '[':
cnt+=1
else:
if cnt > totcnt:
l,r = cur[0],cur[1]
totcnt = cnt
cur = lis[i][:]
cnt = 0
if s[lis[i][0]] == '[':
cnt+=1
if cnt > totcnt:
l,r = cur[0],cur[1]
if l == -1:
print(0)
else:
cnt =0
for i in range(l, r+1):
if s[i] == '[':
cnt+=1
print(cnt)
print(s[l:r+1])
``` | output | 1 | 3,651 | 21 | 7,303 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bracket sequence is a string, containing only characters "(", ")", "[" and "]".
A correct 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 correct (the resulting expressions are: "(1)+[1]", "([1+1]+1)"), and "](" and "[" are not. The empty string is a correct bracket sequence by definition.
A substring s[l... r] (1 ≤ l ≤ r ≤ |s|) of string s = s1s2... s|s| (where |s| is the length of string s) is the string slsl + 1... sr. The empty string is a substring of any string by definition.
You are given a bracket sequence, not necessarily correct. Find its substring which is a correct bracket sequence and contains as many opening square brackets «[» as possible.
Input
The first and the only line contains the bracket sequence as a string, consisting only of characters "(", ")", "[" and "]". It is guaranteed that the string is non-empty and its length doesn't exceed 105 characters.
Output
In the first line print a single integer — the number of brackets «[» in the required bracket sequence. In the second line print the optimal sequence. If there are more than one optimal solutions print any of them.
Examples
Input
([])
Output
1
([])
Input
(((
Output
0 | instruction | 0 | 3,652 | 21 | 7,304 |
Tags: data structures, expression parsing, implementation
Correct Solution:
```
s = input()
st, v, vi, vj, vc = [], [], 0, 0, 0
for i, c in enumerate(s):
if c in '[(':
st.append(i)
continue
if st and s[st[-1]] + c in ('()', '[]'):
b = (st[-1], i+1)
if v and v[-1][1] == i: v[-1] = b
else: v.append(b)
if len(v) >= 2 and v[-2][1] == v[-1][0]:
v[-2:] = [(v[-2][0], v[-1][1])]
st.pop()
else:
st = []
for b in v:
c = s.count('[', b[0], b[1])
if c > vc:
vi, vj, vc = b[0], b[1], c
print(vc)
print(s[vi:vj])
``` | output | 1 | 3,652 | 21 | 7,305 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bracket sequence is a string, containing only characters "(", ")", "[" and "]".
A correct 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 correct (the resulting expressions are: "(1)+[1]", "([1+1]+1)"), and "](" and "[" are not. The empty string is a correct bracket sequence by definition.
A substring s[l... r] (1 ≤ l ≤ r ≤ |s|) of string s = s1s2... s|s| (where |s| is the length of string s) is the string slsl + 1... sr. The empty string is a substring of any string by definition.
You are given a bracket sequence, not necessarily correct. Find its substring which is a correct bracket sequence and contains as many opening square brackets «[» as possible.
Input
The first and the only line contains the bracket sequence as a string, consisting only of characters "(", ")", "[" and "]". It is guaranteed that the string is non-empty and its length doesn't exceed 105 characters.
Output
In the first line print a single integer — the number of brackets «[» in the required bracket sequence. In the second line print the optimal sequence. If there are more than one optimal solutions print any of them.
Examples
Input
([])
Output
1
([])
Input
(((
Output
0 | instruction | 0 | 3,653 | 21 | 7,306 |
Tags: data structures, expression parsing, implementation
Correct Solution:
```
s = input()
a = []
for i in range(len(s)):
if len(a) != 0 and ((s[a[-1]] == '(' and s[i] == ')') or (s[a[-1]] == '[' and s[i] == ']')):
a.pop()
else:
a.append(i)
if len(a) == 0:
print(s.count('['))
print(s)
else:
s1 = s[0: a[0]]
le = s[0: a[0]].count('[')
for i in range(len(a) - 1):
le1 = s[a[i] + 1: a[i + 1]].count('[')
if le1 > le:
s1 = s[a[i] + 1: a[i + 1]]
le = le1
le1 = s[a[-1] + 1:].count('[')
if le1 > le:
s1 = s[a[-1] + 1:]
le = le1
print(le)
print(s1)
``` | output | 1 | 3,653 | 21 | 7,307 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bracket sequence is a string, containing only characters "(", ")", "[" and "]".
A correct 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 correct (the resulting expressions are: "(1)+[1]", "([1+1]+1)"), and "](" and "[" are not. The empty string is a correct bracket sequence by definition.
A substring s[l... r] (1 ≤ l ≤ r ≤ |s|) of string s = s1s2... s|s| (where |s| is the length of string s) is the string slsl + 1... sr. The empty string is a substring of any string by definition.
You are given a bracket sequence, not necessarily correct. Find its substring which is a correct bracket sequence and contains as many opening square brackets «[» as possible.
Input
The first and the only line contains the bracket sequence as a string, consisting only of characters "(", ")", "[" and "]". It is guaranteed that the string is non-empty and its length doesn't exceed 105 characters.
Output
In the first line print a single integer — the number of brackets «[» in the required bracket sequence. In the second line print the optimal sequence. If there are more than one optimal solutions print any of them.
Examples
Input
([])
Output
1
([])
Input
(((
Output
0 | instruction | 0 | 3,654 | 21 | 7,308 |
Tags: data structures, expression parsing, implementation
Correct Solution:
```
s=list(input())
st=[]
a=[0]*len(s)
for i in range(len(s)):
if s[i]==']':
if len(st)>0 and st[-1][0]=='[':
a[st[-1][1]]=i
st.pop()
else:
st=[]
elif s[i]==')':
if len(st)>0 and st[-1][0]=='(':
a[st[-1][1]]=i
st.pop()
else:
st=[]
else:
st.append([s[i],i])
mans=0
ml=0
mr=0
def kkk(s,l,r):
ans=0
for i in range(l,r+1):
if s[i]=='[':
ans+=1
return ans
i=0
while i<len(s):
while i<len(s) and a[i]==0: i+=1
if i<len(s):
l=i
r=a[i]
while r+1<len(s) and a[r+1]>0: r=a[r+1]
t=kkk(s,l,r)
if t>mans:
mans=t
ml=l
mr=r
i=r+1
print(mans)
if mans>0:
for i in range(ml,mr+1):
print(s[i],end='')
``` | output | 1 | 3,654 | 21 | 7,309 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bracket sequence is a string, containing only characters "(", ")", "[" and "]".
A correct 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 correct (the resulting expressions are: "(1)+[1]", "([1+1]+1)"), and "](" and "[" are not. The empty string is a correct bracket sequence by definition.
A substring s[l... r] (1 ≤ l ≤ r ≤ |s|) of string s = s1s2... s|s| (where |s| is the length of string s) is the string slsl + 1... sr. The empty string is a substring of any string by definition.
You are given a bracket sequence, not necessarily correct. Find its substring which is a correct bracket sequence and contains as many opening square brackets «[» as possible.
Input
The first and the only line contains the bracket sequence as a string, consisting only of characters "(", ")", "[" and "]". It is guaranteed that the string is non-empty and its length doesn't exceed 105 characters.
Output
In the first line print a single integer — the number of brackets «[» in the required bracket sequence. In the second line print the optimal sequence. If there are more than one optimal solutions print any of them.
Examples
Input
([])
Output
1
([])
Input
(((
Output
0 | instruction | 0 | 3,655 | 21 | 7,310 |
Tags: data structures, expression parsing, implementation
Correct Solution:
```
import sys
from math import gcd,sqrt,ceil,log2
from collections import defaultdict,Counter,deque
from bisect import bisect_left,bisect_right
import math
import heapq
from itertools import permutations
# input=sys.stdin.readline
# def print(x):
# sys.stdout.write(str(x)+"\n")
# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# import sys
# import io, os
# input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
def get_sum(bit,i):
s = 0
i+=1
while i>0:
s+=bit[i]
i-=i&(-i)
return s
def update(bit,n,i,v):
i+=1
while i<=n:
bit[i]+=v
i+=i&(-i)
def modInverse(b,m):
g = math.gcd(b, m)
if (g != 1):
return -1
else:
return pow(b, m - 2, m)
def primeFactors(n):
sa = set()
sa.add(n)
while n % 2 == 0:
sa.add(2)
n = n // 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
sa.add(i)
n = n // i
# sa.add(n)
return sa
def seive(n):
pri = [True]*(n+1)
p = 2
while p*p<=n:
if pri[p] == True:
for i in range(p*p,n+1,p):
pri[i] = False
p+=1
return pri
def check_prim(n):
if n<0:
return False
for i in range(2,int(sqrt(n))+1):
if n%i == 0:
return False
return True
def getZarr(string, z):
n = len(string)
# [L,R] make a window which matches
# with prefix of s
l, r, k = 0, 0, 0
for i in range(1, n):
# if i>R nothing matches so we will calculate.
# Z[i] using naive way.
if i > r:
l, r = i, i
# R-L = 0 in starting, so it will start
# checking from 0'th index. For example,
# for "ababab" and i = 1, the value of R
# remains 0 and Z[i] becomes 0. For string
# "aaaaaa" and i = 1, Z[i] and R become 5
while r < n and string[r - l] == string[r]:
r += 1
z[i] = r - l
r -= 1
else:
# k = i-L so k corresponds to number which
# matches in [L,R] interval.
k = i - l
# if Z[k] is less than remaining interval
# then Z[i] will be equal to Z[k].
# For example, str = "ababab", i = 3, R = 5
# and L = 2
if z[k] < r - i + 1:
z[i] = z[k]
# For example str = "aaaaaa" and i = 2,
# R is 5, L is 0
else:
# else start from R and check manually
l = i
while r < n and string[r - l] == string[r]:
r += 1
z[i] = r - l
r -= 1
def search(text, pattern):
# Create concatenated string "P$T"
concat = pattern + "$" + text
l = len(concat)
z = [0] * l
getZarr(concat, z)
ha = []
for i in range(l):
if z[i] == len(pattern):
ha.append(i - len(pattern) - 1)
return ha
# n,k = map(int,input().split())
# l = list(map(int,input().split()))
#
# n = int(input())
# l = list(map(int,input().split()))
#
# hash = defaultdict(list)
# la = []
#
# for i in range(n):
# la.append([l[i],i+1])
#
# la.sort(key = lambda x: (x[0],-x[1]))
# ans = []
# r = n
# flag = 0
# lo = []
# ha = [i for i in range(n,0,-1)]
# yo = []
# for a,b in la:
#
# if a == 1:
# ans.append([r,b])
# # hash[(1,1)].append([b,r])
# lo.append((r,b))
# ha.pop(0)
# yo.append([r,b])
# r-=1
#
# elif a == 2:
# # print(yo,lo)
# # print(hash[1,1])
# if lo == []:
# flag = 1
# break
# c,d = lo.pop(0)
# yo.pop(0)
# if b>=d:
# flag = 1
# break
# ans.append([c,b])
# yo.append([c,b])
#
#
#
# elif a == 3:
#
# if yo == []:
# flag = 1
# break
# c,d = yo.pop(0)
# if b>=d:
# flag = 1
# break
# if ha == []:
# flag = 1
# break
#
# ka = ha.pop(0)
#
# ans.append([ka,b])
# ans.append([ka,d])
# yo.append([ka,b])
#
# if flag:
# print(-1)
# else:
# print(len(ans))
# for a,b in ans:
# print(a,b)
def mergeIntervals(arr):
# Sorting based on the increasing order
# of the start intervals
arr.sort(key = lambda x: x[0])
# array to hold the merged intervals
m = []
s = -10000
max = -100000
for i in range(len(arr)):
a = arr[i]
if a[0] > max:
if i != 0:
m.append([s,max])
max = a[1]
s = a[0]
else:
if a[1] >= max:
max = a[1]
#'max' value gives the last point of
# that particular interval
# 's' gives the starting point of that interval
# 'm' array contains the list of all merged intervals
if max != -100000 and [s, max] not in m:
m.append([s, max])
return m
s = input()
n = len(s)
stack = []
i = 0
ans = []
pre = [0]
for i in s:
if i == '[':
pre.append(pre[-1]+1)
else:
pre.append(pre[-1])
i = 0
while i<n:
if s[i] == '(' or s[i] == '[':
stack.append(i)
elif stack!=[] and s[i] == ')' and s[stack[-1]] == '(':
z = stack.pop()
ans.append((z,i))
elif stack!=[] and s[i] == ')' and s[stack[-1]] == '[':
stack = []
elif stack!=[] and s[i] == ']' and s[stack[-1]] == '[':
z = stack.pop()
ans.append((z,i))
elif stack!=[] and s[i] == ']' and s[stack[-1]] == '(':
stack = []
i+=1
ans.sort()
x,y = -1,-1
maxi = 0
lo = []
i = 1
# print(ans)
ans = mergeIntervals(ans)
if ans == []:
print(0)
print()
exit()
a,b = ans[i-1]
lo.append([a,b])
# print(ans)
while i<=len(ans):
a,b = ans[i-1]
while i<len(ans) and ans[i][0]-ans[i-1][1] == 1:
i+=1
lo.append([a,ans[i-1][1]])
i+=1
ans = lo
# print(lo)
for i in range(len(ans)):
a,b = ans[i]
a+=1
b+=1
z = pre[b] - pre[a-1]
if z>maxi:
maxi = z
a-=1
b-=1
x,y = a,b
if ans == []:
print(0)
print()
else:
print(maxi)
print(s[x:y+1])
``` | output | 1 | 3,655 | 21 | 7,311 |
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, containing only characters "(", ")", "[" and "]".
A correct 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 correct (the resulting expressions are: "(1)+[1]", "([1+1]+1)"), and "](" and "[" are not. The empty string is a correct bracket sequence by definition.
A substring s[l... r] (1 ≤ l ≤ r ≤ |s|) of string s = s1s2... s|s| (where |s| is the length of string s) is the string slsl + 1... sr. The empty string is a substring of any string by definition.
You are given a bracket sequence, not necessarily correct. Find its substring which is a correct bracket sequence and contains as many opening square brackets «[» as possible.
Input
The first and the only line contains the bracket sequence as a string, consisting only of characters "(", ")", "[" and "]". It is guaranteed that the string is non-empty and its length doesn't exceed 105 characters.
Output
In the first line print a single integer — the number of brackets «[» in the required bracket sequence. In the second line print the optimal sequence. If there are more than one optimal solutions print any of them.
Examples
Input
([])
Output
1
([])
Input
(((
Output
0
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
INF = 10 ** 18
MOD = 10**9+7
Ri = lambda : [int(x) for x in sys.stdin.readline().split()]
ri = lambda : sys.stdin.readline().strip()
s = ri()
sta1 = [];sta2 = []
least = -1
l,r = -1,-2
for i in range(len(s)):
if s[i] == '(':
sta1.append(['(', i])
elif s[i] == '[':
sta2.append(['[', i])
if least == -1:
least = i
elif s[i] == ')':
if len(sta1) == 0:
sta1 = []
# if flag :
sta2 = []
least = -1
else:
sta1.pop()
elif s[i] == ']':
if len(sta2) == 0 or sta2[-1][0] != '[':
sta2 = []
sta1 = []
else:
if len(sta2) == 1:
temp = sta2.pop()
tl,tr = least, i
if l == -1:
l,r = tl,tr
if tr - tl +1 > r - l +1:
l,r= tl,tr
else:
temp = sta2.pop()
tl,tr = temp[1], i
if l == -1:
l,r = tl,tr
if tr - tl +1 > r - l +1:
l,r= tl,tr
if l == -1:
print(0)
else:
cnt =0
for i in range(l, r+1):
if s[i] == '[':
cnt+=1
print(cnt)
print(s[l:r+1])
``` | instruction | 0 | 3,656 | 21 | 7,312 |
No | output | 1 | 3,656 | 21 | 7,313 |
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, containing only characters "(", ")", "[" and "]".
A correct 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 correct (the resulting expressions are: "(1)+[1]", "([1+1]+1)"), and "](" and "[" are not. The empty string is a correct bracket sequence by definition.
A substring s[l... r] (1 ≤ l ≤ r ≤ |s|) of string s = s1s2... s|s| (where |s| is the length of string s) is the string slsl + 1... sr. The empty string is a substring of any string by definition.
You are given a bracket sequence, not necessarily correct. Find its substring which is a correct bracket sequence and contains as many opening square brackets «[» as possible.
Input
The first and the only line contains the bracket sequence as a string, consisting only of characters "(", ")", "[" and "]". It is guaranteed that the string is non-empty and its length doesn't exceed 105 characters.
Output
In the first line print a single integer — the number of brackets «[» in the required bracket sequence. In the second line print the optimal sequence. If there are more than one optimal solutions print any of them.
Examples
Input
([])
Output
1
([])
Input
(((
Output
0
Submitted Solution:
```
s=list(input())
st=[]
a=[0]*len(s)
for i in range(len(s)):
if s[i]==']':
if len(st)>0 and st[-1][0]=='[':
a[st[-1][1]]=i
else:
st=[]
elif s[i]==')':
if len(st)>0 and st[-1][0]=='(':
a[st[-1][1]]=i
else:
st=[]
else:
st.append([s[i],i])
mans=0
ml=0
mr=0
def kkk(s,l,r):
ans=0
for i in range(l,r+1):
if s[i]=='[':
ans+=1
return ans
i=0
while i<len(s):
while i<len(s) and a[i]==0: i+=1
if i<len(s):
l=i
r=a[i]
while r+1<len(s) and a[r+1]>0: r=a[r+1]
t=kkk(s,l,r)
if t>mans:
mans=t
ml=l
mr=r
i=r+1
print(mans)
if mans>0:
for i in range(ml,mr+1):
print(s[i],end='')
``` | instruction | 0 | 3,657 | 21 | 7,314 |
No | output | 1 | 3,657 | 21 | 7,315 |
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, containing only characters "(", ")", "[" and "]".
A correct 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 correct (the resulting expressions are: "(1)+[1]", "([1+1]+1)"), and "](" and "[" are not. The empty string is a correct bracket sequence by definition.
A substring s[l... r] (1 ≤ l ≤ r ≤ |s|) of string s = s1s2... s|s| (where |s| is the length of string s) is the string slsl + 1... sr. The empty string is a substring of any string by definition.
You are given a bracket sequence, not necessarily correct. Find its substring which is a correct bracket sequence and contains as many opening square brackets «[» as possible.
Input
The first and the only line contains the bracket sequence as a string, consisting only of characters "(", ")", "[" and "]". It is guaranteed that the string is non-empty and its length doesn't exceed 105 characters.
Output
In the first line print a single integer — the number of brackets «[» in the required bracket sequence. In the second line print the optimal sequence. If there are more than one optimal solutions print any of them.
Examples
Input
([])
Output
1
([])
Input
(((
Output
0
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
INF = 10 ** 18
MOD = 10**9+7
Ri = lambda : [int(x) for x in sys.stdin.readline().split()]
ri = lambda : sys.stdin.readline().strip()
s = ri()
sta = []
l,r = -1,-2
for i in range(len(s)):
if s[i] == '(':
sta.append(['(', i])
elif s[i] == '[':
sta.append(['[', i])
elif s[i] == ')':
if len(sta) == 0 or sta[-1][0] != '(':
sta = []
else:
sta.pop()
elif s[i] == ']':
if len(sta) == 0 or sta[-1][0] != '[':
sta = []
else:
temp = sta.pop()
tl,tr = temp[1], i
if l == -1:
l,r = tl,tr
if tr - tl +1 > r - l +1:
l,r= tl,tr
if l == -1:
print(0)
else:
cnt =0
for i in range(l, r+1):
if s[i] == '[':
cnt+=1
print(cnt)
print(s[l:r+1])
``` | instruction | 0 | 3,658 | 21 | 7,316 |
No | output | 1 | 3,658 | 21 | 7,317 |
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, containing only characters "(", ")", "[" and "]".
A correct 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 correct (the resulting expressions are: "(1)+[1]", "([1+1]+1)"), and "](" and "[" are not. The empty string is a correct bracket sequence by definition.
A substring s[l... r] (1 ≤ l ≤ r ≤ |s|) of string s = s1s2... s|s| (where |s| is the length of string s) is the string slsl + 1... sr. The empty string is a substring of any string by definition.
You are given a bracket sequence, not necessarily correct. Find its substring which is a correct bracket sequence and contains as many opening square brackets «[» as possible.
Input
The first and the only line contains the bracket sequence as a string, consisting only of characters "(", ")", "[" and "]". It is guaranteed that the string is non-empty and its length doesn't exceed 105 characters.
Output
In the first line print a single integer — the number of brackets «[» in the required bracket sequence. In the second line print the optimal sequence. If there are more than one optimal solutions print any of them.
Examples
Input
([])
Output
1
([])
Input
(((
Output
0
Submitted Solution:
```
s=list(input())
st=[]
a=[0]*len(s)
ans=0
for i in range(len(s)):
if s[i]==']':
if len(st)==0:
pass
else:
ans+=1
a[i]=1
a[st[-1]]=1
st.pop()
elif s[i]=='[':
st.append(i)
print(ans)
for i in range(len(s)):
if a[i]==1:
print(s[i],end='')
``` | instruction | 0 | 3,659 | 21 | 7,318 |
No | output | 1 | 3,659 | 21 | 7,319 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence of square brackets is regular if by inserting symbols "+" and "1" into it, you can get a regular mathematical expression from it. For example, sequences "[[]][]", "[]" and "[[][[]]]" — are regular, at the same time "][", "[[]" and "[[]]][" — are irregular.
Draw the given sequence using a minimalistic pseudographics in the strip of the lowest possible height — use symbols '+', '-' and '|'. For example, the sequence "[[][]][]" should be represented as:
+- -++- -+
|+- -++- -+|| |
|| || ||| |
|+- -++- -+|| |
+- -++- -+
Each bracket should be represented with the hepl of one or more symbols '|' (the vertical part) and symbols '+' and '-' as on the example which is given above.
Brackets should be drawn without spaces one by one, only dividing pairs of consecutive pairwise brackets with a single-space bar (so that the two brackets do not visually merge into one symbol). The image should have the minimum possible height.
The enclosed bracket is always smaller than the surrounding bracket, but each bracket separately strives to maximize the height of the image. So the pair of final brackets in the example above occupies the entire height of the image.
Study carefully the examples below, they adequately explain the condition of the problem. Pay attention that in this problem the answer (the image) is unique.
Input
The first line contains an even integer n (2 ≤ n ≤ 100) — the length of the sequence of brackets.
The second line contains the sequence of brackets — these are n symbols "[" and "]". It is guaranteed that the given sequence of brackets is regular.
Output
Print the drawn bracket sequence in the format which is given in the condition. Don't print extra (unnecessary) spaces.
Examples
Input
8
[[][]][]
Output
+- -++- -+
|+- -++- -+|| |
|| || ||| |
|+- -++- -+|| |
+- -++- -+
Input
6
[[[]]]
Output
+- -+
|+- -+|
||+- -+||
||| |||
||+- -+||
|+- -+|
+- -+
Input
6
[[][]]
Output
+- -+
|+- -++- -+|
|| || ||
|+- -++- -+|
+- -+
Input
2
[]
Output
+- -+
| |
+- -+
Input
4
[][]
Output
+- -++- -+
| || |
+- -++- -+ | instruction | 0 | 4,683 | 21 | 9,366 |
Tags: *special, implementation
Correct Solution:
```
""" Created by Henrikh Kantuni and Shahen Kosyan on 3/12/17 """
from itertools import groupby
n = int(input())
seq = list(input())
answer = []
stack = []
depth = -1
for i in range(len(seq)):
if seq[i] == '[':
stack.append('[')
depth += 2
answer.append(depth)
else:
stack.pop()
answer.append(depth)
depth -= 2
if len(stack) == 0:
answer.append('*')
# break down answer into nested lists using * separator
answer = [list(group) for k, group in groupby(answer, lambda x: x != "*") if k]
max_depth = 0
for i in range(len(answer)):
if max_depth < max(answer[i]):
max_depth = max(answer[i])
for i in range(len(answer)):
answer[i] = [(max_depth + 1 - a) for a in answer[i]]
# flatten
answer = [s for sub in answer for s in sub]
out = ""
stack = []
for i in range(len(answer)):
line = ""
if seq[i] == "[" and len(stack) == 0:
line += " " * ((max_depth - answer[i]) // 2)
line += "+"
line += "|" * answer[i]
line += "+"
line += " " * ((max_depth - answer[i]) // 2)
if seq[i + 1] == "]":
line += " " * ((max_depth - answer[i]) // 2)
line += "-"
line += " " * answer[i]
line += "-"
line += " " * ((max_depth - answer[i]) // 2)
stack.append(answer[i])
elif seq[i] == "[" and len(stack) > 0:
if seq[i - 1] == "[":
line += " " * ((max_depth - answer[i] - 2) // 2)
line += "-"
else:
line += " " * ((max_depth - answer[i]) // 2)
line += "+"
line += "|" * answer[i]
line += "+"
if seq[i - 1] == "[":
line += "-"
line += " " * ((max_depth - answer[i] - 2) // 2)
else:
line += " " * ((max_depth - answer[i]) // 2)
if seq[i + 1] == "]":
line += " " * ((max_depth - answer[i]) // 2)
line += "-"
line += " " * answer[i]
line += "-"
line += " " * ((max_depth - answer[i]) // 2)
stack.append(answer[i])
elif seq[i] == "]":
if seq[i - 1] == "[":
line += " " * ((max_depth - answer[i]) // 2)
line += "-"
line += " " * answer[i]
line += "-"
line += " " * ((max_depth - answer[i]) // 2)
if i < len(answer) - 1 and seq[i + 1] == "]":
line += " " * ((max_depth - answer[i] - 2) // 2)
line += "-"
line += "+"
line += "|" * answer[i]
line += "+"
line += "-"
line += " " * ((max_depth - answer[i] - 2) // 2)
else:
line += " " * ((max_depth - answer[i]) // 2)
line += "+"
line += "|" * answer[i]
line += "+"
line += " " * ((max_depth - answer[i]) // 2)
stack.pop()
out += line
# divide into chunks of size = max_depth + 2
out = [out[i:i + max_depth + 2] for i in range(0, len(out), max_depth + 2)]
# add a new line between - - and - -
i = 0
while i < len(out):
if out[i].replace(" ", "") == "--" and out[i + 1].replace(" ", "") == "--":
out.insert(i + 1, " " * len(out[i]))
i += 1
vertical = [list(r) for r in out]
# horizontal printing
final = ""
for i in range(len(vertical[0])):
for j in range(len(vertical)):
final += vertical[j][i]
final += "\n"
# remove the trailing space
final = final[:-1]
print(final)
``` | output | 1 | 4,683 | 21 | 9,367 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence of square brackets is regular if by inserting symbols "+" and "1" into it, you can get a regular mathematical expression from it. For example, sequences "[[]][]", "[]" and "[[][[]]]" — are regular, at the same time "][", "[[]" and "[[]]][" — are irregular.
Draw the given sequence using a minimalistic pseudographics in the strip of the lowest possible height — use symbols '+', '-' and '|'. For example, the sequence "[[][]][]" should be represented as:
+- -++- -+
|+- -++- -+|| |
|| || ||| |
|+- -++- -+|| |
+- -++- -+
Each bracket should be represented with the hepl of one or more symbols '|' (the vertical part) and symbols '+' and '-' as on the example which is given above.
Brackets should be drawn without spaces one by one, only dividing pairs of consecutive pairwise brackets with a single-space bar (so that the two brackets do not visually merge into one symbol). The image should have the minimum possible height.
The enclosed bracket is always smaller than the surrounding bracket, but each bracket separately strives to maximize the height of the image. So the pair of final brackets in the example above occupies the entire height of the image.
Study carefully the examples below, they adequately explain the condition of the problem. Pay attention that in this problem the answer (the image) is unique.
Input
The first line contains an even integer n (2 ≤ n ≤ 100) — the length of the sequence of brackets.
The second line contains the sequence of brackets — these are n symbols "[" and "]". It is guaranteed that the given sequence of brackets is regular.
Output
Print the drawn bracket sequence in the format which is given in the condition. Don't print extra (unnecessary) spaces.
Examples
Input
8
[[][]][]
Output
+- -++- -+
|+- -++- -+|| |
|| || ||| |
|+- -++- -+|| |
+- -++- -+
Input
6
[[[]]]
Output
+- -+
|+- -+|
||+- -+||
||| |||
||+- -+||
|+- -+|
+- -+
Input
6
[[][]]
Output
+- -+
|+- -++- -+|
|| || ||
|+- -++- -+|
+- -+
Input
2
[]
Output
+- -+
| |
+- -+
Input
4
[][]
Output
+- -++- -+
| || |
+- -++- -+ | instruction | 0 | 4,684 | 21 | 9,368 |
Tags: *special, implementation
Correct Solution:
```
n = int(input())
s = input()
def parse(s, i=0):
ss = i
start = i
h = 0
while i < len(s) and s[i] == '[':
i, nh = parse(s, i+1)
h = max(h, nh+2)
#print(s[start:i], h)
start = i
#print("parse {}:{} {}".format(start, i+1, h))
if i == ss:
h = 1
return i+1, h
_, h = parse(s)
l = 4 * n
mid = (h-1) // 2
buf = [[' ' for i in range(l)] for i in range(h)]
def draw(s, h, i=0, p=0):
if i >= len(s):
return
if s[i] == '[':
r = (h-1)//2
for j in range(r):
buf[mid+j][p] = '|'
buf[mid-j][p] = '|'
buf[mid+r][p] = '+'
buf[mid-r][p] = '+'
buf[mid+r][p+1] = '-'
buf[mid-r][p+1] = '-'
draw(s, h-2, i+1, p+1)
else:
h += 2
if s[i-1] == '[':
p += 3
r = (h-1)//2
for j in range(r):
buf[mid+j][p] = '|'
buf[mid-j][p] = '|'
buf[mid+r][p] = '+'
buf[mid-r][p] = '+'
buf[mid+r][p-1] = '-'
buf[mid-r][p-1] = '-'
draw(s, h, i+1, p+1)
draw(s, h)
for i in range(h):
l = -1
for j in range(len(buf[i])):
if buf[i][j] != ' ':
l = j+1
buf[i] = buf[i][:l]
for i in buf:
print(''.join(i))
``` | output | 1 | 4,684 | 21 | 9,369 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence of square brackets is regular if by inserting symbols "+" and "1" into it, you can get a regular mathematical expression from it. For example, sequences "[[]][]", "[]" and "[[][[]]]" — are regular, at the same time "][", "[[]" and "[[]]][" — are irregular.
Draw the given sequence using a minimalistic pseudographics in the strip of the lowest possible height — use symbols '+', '-' and '|'. For example, the sequence "[[][]][]" should be represented as:
+- -++- -+
|+- -++- -+|| |
|| || ||| |
|+- -++- -+|| |
+- -++- -+
Each bracket should be represented with the hepl of one or more symbols '|' (the vertical part) and symbols '+' and '-' as on the example which is given above.
Brackets should be drawn without spaces one by one, only dividing pairs of consecutive pairwise brackets with a single-space bar (so that the two brackets do not visually merge into one symbol). The image should have the minimum possible height.
The enclosed bracket is always smaller than the surrounding bracket, but each bracket separately strives to maximize the height of the image. So the pair of final brackets in the example above occupies the entire height of the image.
Study carefully the examples below, they adequately explain the condition of the problem. Pay attention that in this problem the answer (the image) is unique.
Input
The first line contains an even integer n (2 ≤ n ≤ 100) — the length of the sequence of brackets.
The second line contains the sequence of brackets — these are n symbols "[" and "]". It is guaranteed that the given sequence of brackets is regular.
Output
Print the drawn bracket sequence in the format which is given in the condition. Don't print extra (unnecessary) spaces.
Examples
Input
8
[[][]][]
Output
+- -++- -+
|+- -++- -+|| |
|| || ||| |
|+- -++- -+|| |
+- -++- -+
Input
6
[[[]]]
Output
+- -+
|+- -+|
||+- -+||
||| |||
||+- -+||
|+- -+|
+- -+
Input
6
[[][]]
Output
+- -+
|+- -++- -+|
|| || ||
|+- -++- -+|
+- -+
Input
2
[]
Output
+- -+
| |
+- -+
Input
4
[][]
Output
+- -++- -+
| || |
+- -++- -+ | instruction | 0 | 4,685 | 21 | 9,370 |
Tags: *special, implementation
Correct Solution:
```
from itertools import groupby
n = int(input())
seq = list(input())
answer = []
stack = []
depth = -1
for i in range(len(seq)):
if seq[i] == '[':
stack.append('[')
depth += 2
answer.append(depth)
else:
stack.pop()
answer.append(depth)
depth -= 2
if len(stack) == 0:
answer.append('*')
# break down answer into nested lists using * separator
answer = [list(group) for k, group in groupby(answer, lambda x: x != "*") if k]
max_depth = 0
for i in range(len(answer)):
if max_depth < max(answer[i]):
max_depth = max(answer[i])
for i in range(len(answer)):
answer[i] = [(max_depth + 1 - a) for a in answer[i]]
# flatten
answer = [s for sub in answer for s in sub]
out = ""
stack = []
for i in range(len(answer)):
line = ""
if seq[i] == "[" and len(stack) == 0:
line += " " * ((max_depth - answer[i]) // 2)
line += "+"
line += "|" * answer[i]
line += "+"
line += " " * ((max_depth - answer[i]) // 2)
if seq[i + 1] == "]":
line += " " * ((max_depth - answer[i]) // 2)
line += "-"
line += " " * answer[i]
line += "-"
line += " " * ((max_depth - answer[i]) // 2)
stack.append(answer[i])
elif seq[i] == "[" and len(stack) > 0:
if seq[i - 1] == "[":
line += " " * ((max_depth - answer[i] - 2) // 2)
line += "-"
else:
line += " " * ((max_depth - answer[i]) // 2)
line += "+"
line += "|" * answer[i]
line += "+"
if seq[i - 1] == "[":
line += "-"
line += " " * ((max_depth - answer[i] - 2) // 2)
else:
line += " " * ((max_depth - answer[i]) // 2)
if seq[i + 1] == "]":
line += " " * ((max_depth - answer[i]) // 2)
line += "-"
line += " " * answer[i]
line += "-"
line += " " * ((max_depth - answer[i]) // 2)
stack.append(answer[i])
elif seq[i] == "]":
if seq[i - 1] == "[":
line += " " * ((max_depth - answer[i]) // 2)
line += "-"
line += " " * answer[i]
line += "-"
line += " " * ((max_depth - answer[i]) // 2)
if i < len(answer) - 1 and seq[i + 1] == "]":
line += " " * ((max_depth - answer[i] - 2) // 2)
line += "-"
line += "+"
line += "|" * answer[i]
line += "+"
line += "-"
line += " " * ((max_depth - answer[i] - 2) // 2)
else:
line += " " * ((max_depth - answer[i]) // 2)
line += "+"
line += "|" * answer[i]
line += "+"
line += " " * ((max_depth - answer[i]) // 2)
stack.pop()
out += line
# divide into chunks of size = max_depth + 2
out = [out[i:i + max_depth + 2] for i in range(0, len(out), max_depth + 2)]
# add a new line between - - and - -
i = 0
while i < len(out):
if out[i].replace(" ", "") == "--" and out[i + 1].replace(" ", "") == "--":
out.insert(i + 1, " " * len(out[i]))
i += 1
vertical = [list(r) for r in out]
# horizontal printing
final = ""
for i in range(len(vertical[0])):
for j in range(len(vertical)):
final += vertical[j][i]
final += "\n"
# remove the trailing space
final = final[:-1]
print(final)
``` | output | 1 | 4,685 | 21 | 9,371 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence of square brackets is regular if by inserting symbols "+" and "1" into it, you can get a regular mathematical expression from it. For example, sequences "[[]][]", "[]" and "[[][[]]]" — are regular, at the same time "][", "[[]" and "[[]]][" — are irregular.
Draw the given sequence using a minimalistic pseudographics in the strip of the lowest possible height — use symbols '+', '-' and '|'. For example, the sequence "[[][]][]" should be represented as:
+- -++- -+
|+- -++- -+|| |
|| || ||| |
|+- -++- -+|| |
+- -++- -+
Each bracket should be represented with the hepl of one or more symbols '|' (the vertical part) and symbols '+' and '-' as on the example which is given above.
Brackets should be drawn without spaces one by one, only dividing pairs of consecutive pairwise brackets with a single-space bar (so that the two brackets do not visually merge into one symbol). The image should have the minimum possible height.
The enclosed bracket is always smaller than the surrounding bracket, but each bracket separately strives to maximize the height of the image. So the pair of final brackets in the example above occupies the entire height of the image.
Study carefully the examples below, they adequately explain the condition of the problem. Pay attention that in this problem the answer (the image) is unique.
Input
The first line contains an even integer n (2 ≤ n ≤ 100) — the length of the sequence of brackets.
The second line contains the sequence of brackets — these are n symbols "[" and "]". It is guaranteed that the given sequence of brackets is regular.
Output
Print the drawn bracket sequence in the format which is given in the condition. Don't print extra (unnecessary) spaces.
Examples
Input
8
[[][]][]
Output
+- -++- -+
|+- -++- -+|| |
|| || ||| |
|+- -++- -+|| |
+- -++- -+
Input
6
[[[]]]
Output
+- -+
|+- -+|
||+- -+||
||| |||
||+- -+||
|+- -+|
+- -+
Input
6
[[][]]
Output
+- -+
|+- -++- -+|
|| || ||
|+- -++- -+|
+- -+
Input
2
[]
Output
+- -+
| |
+- -+
Input
4
[][]
Output
+- -++- -+
| || |
+- -++- -+ | instruction | 0 | 4,686 | 21 | 9,372 |
Tags: *special, implementation
Correct Solution:
```
input()
a = input()
field = [[" " for j in range(500)] for i in range(500)]
mxb = -1
b = 0
for i in range(len(a)):
if a[i] == "[":
b += 1
else:
b -= 1
mxb = max(mxb, b)
m = (mxb * 2 + 1) // 2
def opn(curpos, curb):
up = mxb - curb + 1
for i in range(up):
field[m + i][curpos] = "|"
for i in range(up):
field[m - i][curpos] = "|"
field[m + up][curpos] = "+"
field[m - up][curpos] = "+"
field[m + up][curpos + 1] = "-"
field[m - up][curpos + 1] = "-"
def clos(curpos,curb):
up = mxb - curb + 1
for i in range(up):
field[m + i][curpos] = "|"
for i in range(up):
field[m - i][curpos] = "|"
field[m + up][curpos] = "+"
field[m - up][curpos] = "+"
field[m + up][curpos - 1] = "-"
field[m - up][curpos - 1] = "-"
curb = 0
pos = 0
for i in range(len(a)):
if a[i] == "[":
curb += 1
opn(pos,curb)
pos += 1
else:
if a[i - 1] == "[":
pos += 3
clos(pos,curb)
curb -= 1
pos += 1
ans = []
for i in range(len(field)):
lst = -1
for j in range(len(field[0])):
if field[i][j] != " ":
lst = j
ans.append(lst + 1)
for i in range(len(field)):
for j in range(ans[i]):
print(field[i][j],end="")
if ans[i]:
print()
``` | output | 1 | 4,686 | 21 | 9,373 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence of square brackets is regular if by inserting symbols "+" and "1" into it, you can get a regular mathematical expression from it. For example, sequences "[[]][]", "[]" and "[[][[]]]" — are regular, at the same time "][", "[[]" and "[[]]][" — are irregular.
Draw the given sequence using a minimalistic pseudographics in the strip of the lowest possible height — use symbols '+', '-' and '|'. For example, the sequence "[[][]][]" should be represented as:
+- -++- -+
|+- -++- -+|| |
|| || ||| |
|+- -++- -+|| |
+- -++- -+
Each bracket should be represented with the hepl of one or more symbols '|' (the vertical part) and symbols '+' and '-' as on the example which is given above.
Brackets should be drawn without spaces one by one, only dividing pairs of consecutive pairwise brackets with a single-space bar (so that the two brackets do not visually merge into one symbol). The image should have the minimum possible height.
The enclosed bracket is always smaller than the surrounding bracket, but each bracket separately strives to maximize the height of the image. So the pair of final brackets in the example above occupies the entire height of the image.
Study carefully the examples below, they adequately explain the condition of the problem. Pay attention that in this problem the answer (the image) is unique.
Input
The first line contains an even integer n (2 ≤ n ≤ 100) — the length of the sequence of brackets.
The second line contains the sequence of brackets — these are n symbols "[" and "]". It is guaranteed that the given sequence of brackets is regular.
Output
Print the drawn bracket sequence in the format which is given in the condition. Don't print extra (unnecessary) spaces.
Examples
Input
8
[[][]][]
Output
+- -++- -+
|+- -++- -+|| |
|| || ||| |
|+- -++- -+|| |
+- -++- -+
Input
6
[[[]]]
Output
+- -+
|+- -+|
||+- -+||
||| |||
||+- -+||
|+- -+|
+- -+
Input
6
[[][]]
Output
+- -+
|+- -++- -+|
|| || ||
|+- -++- -+|
+- -+
Input
2
[]
Output
+- -+
| |
+- -+
Input
4
[][]
Output
+- -++- -+
| || |
+- -++- -+ | instruction | 0 | 4,687 | 21 | 9,374 |
Tags: *special, implementation
Correct Solution:
```
n = int(input())
s = input()
bal = 0
m = 0
v = [0] * n
for i in range(n):
if s[i] == '[':
bal += 1
v[i] = bal
m = max(m, bal)
else:
v[i] = -bal
bal -= 1
for j in range(1, 2 * m + 2):
for i in range(0, n):
if abs(v[i]) == j or abs(v[i]) == 2 * (m + 1) - j:
if v[i] > 0:
print('+-', end = '')
else:
print('-+', end = '')
elif abs(v[i]) < j and abs(v[i]) < 2 * (m + 1) - j:
print('|', end = '')
if i + 1 < n and v[i + 1] < 0 and v[i] == -v[i + 1]:
print(' ', end = '')
else:
print(' ', end = '')
if i > 0 and abs(v[i - 1]) >= abs(v[i]) and i + 1 < n and abs(v[i + 1]) >= abs(v[i]):
print(' ', end = '')
if i + 1 < n and v[i + 1] < 0 and v[i] == -v[i + 1]:
print(' ', end = '')
print()
``` | output | 1 | 4,687 | 21 | 9,375 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence of square brackets is regular if by inserting symbols "+" and "1" into it, you can get a regular mathematical expression from it. For example, sequences "[[]][]", "[]" and "[[][[]]]" — are regular, at the same time "][", "[[]" and "[[]]][" — are irregular.
Draw the given sequence using a minimalistic pseudographics in the strip of the lowest possible height — use symbols '+', '-' and '|'. For example, the sequence "[[][]][]" should be represented as:
+- -++- -+
|+- -++- -+|| |
|| || ||| |
|+- -++- -+|| |
+- -++- -+
Each bracket should be represented with the hepl of one or more symbols '|' (the vertical part) and symbols '+' and '-' as on the example which is given above.
Brackets should be drawn without spaces one by one, only dividing pairs of consecutive pairwise brackets with a single-space bar (so that the two brackets do not visually merge into one symbol). The image should have the minimum possible height.
The enclosed bracket is always smaller than the surrounding bracket, but each bracket separately strives to maximize the height of the image. So the pair of final brackets in the example above occupies the entire height of the image.
Study carefully the examples below, they adequately explain the condition of the problem. Pay attention that in this problem the answer (the image) is unique.
Input
The first line contains an even integer n (2 ≤ n ≤ 100) — the length of the sequence of brackets.
The second line contains the sequence of brackets — these are n symbols "[" and "]". It is guaranteed that the given sequence of brackets is regular.
Output
Print the drawn bracket sequence in the format which is given in the condition. Don't print extra (unnecessary) spaces.
Examples
Input
8
[[][]][]
Output
+- -++- -+
|+- -++- -+|| |
|| || ||| |
|+- -++- -+|| |
+- -++- -+
Input
6
[[[]]]
Output
+- -+
|+- -+|
||+- -+||
||| |||
||+- -+||
|+- -+|
+- -+
Input
6
[[][]]
Output
+- -+
|+- -++- -+|
|| || ||
|+- -++- -+|
+- -+
Input
2
[]
Output
+- -+
| |
+- -+
Input
4
[][]
Output
+- -++- -+
| || |
+- -++- -+ | instruction | 0 | 4,688 | 21 | 9,376 |
Tags: *special, implementation
Correct Solution:
```
#This code is dedicated to Olya S.
l=int(input())
n=input()
def offset(ml,x):
return (ml-x)//2
def getstate(g,line,ml):
off=offset(ml,g[0])
if line<off or line>=g[0]+off:
return 0
elif line==off or line == g[0]+off-1:
return 1
else:
return 2
#Find max bracket#
ml=1
cl=1
for b in n:
if b=='[':
cl+=2
if ml<cl:
ml=cl
else:
cl-=2
######MAP######
sc=[]
for b in n:
if b=='[':
sc.append([ml,True])
ml-=2
else:
ml+=2
sc.append([ml,False])
#####################
for i in range(ml):
for j in range(l):
g=sc[j]
state=getstate(g,i,ml)
if state==1:
if g[1]:
print('+-',end='')
else:
print('-+',end='')
elif state==0:
if sc[j-1][0]-sc[j][0]!=2:
print(' ',end='')
else:
print('|',end='')
if sc[j][1] and not sc[j+1][1]:
if state==2:
print(' ',end='')
else:
print(' ',end='')
print()
``` | output | 1 | 4,688 | 21 | 9,377 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence of square brackets is regular if by inserting symbols "+" and "1" into it, you can get a regular mathematical expression from it. For example, sequences "[[]][]", "[]" and "[[][[]]]" — are regular, at the same time "][", "[[]" and "[[]]][" — are irregular.
Draw the given sequence using a minimalistic pseudographics in the strip of the lowest possible height — use symbols '+', '-' and '|'. For example, the sequence "[[][]][]" should be represented as:
+- -++- -+
|+- -++- -+|| |
|| || ||| |
|+- -++- -+|| |
+- -++- -+
Each bracket should be represented with the hepl of one or more symbols '|' (the vertical part) and symbols '+' and '-' as on the example which is given above.
Brackets should be drawn without spaces one by one, only dividing pairs of consecutive pairwise brackets with a single-space bar (so that the two brackets do not visually merge into one symbol). The image should have the minimum possible height.
The enclosed bracket is always smaller than the surrounding bracket, but each bracket separately strives to maximize the height of the image. So the pair of final brackets in the example above occupies the entire height of the image.
Study carefully the examples below, they adequately explain the condition of the problem. Pay attention that in this problem the answer (the image) is unique.
Input
The first line contains an even integer n (2 ≤ n ≤ 100) — the length of the sequence of brackets.
The second line contains the sequence of brackets — these are n symbols "[" and "]". It is guaranteed that the given sequence of brackets is regular.
Output
Print the drawn bracket sequence in the format which is given in the condition. Don't print extra (unnecessary) spaces.
Examples
Input
8
[[][]][]
Output
+- -++- -+
|+- -++- -+|| |
|| || ||| |
|+- -++- -+|| |
+- -++- -+
Input
6
[[[]]]
Output
+- -+
|+- -+|
||+- -+||
||| |||
||+- -+||
|+- -+|
+- -+
Input
6
[[][]]
Output
+- -+
|+- -++- -+|
|| || ||
|+- -++- -+|
+- -+
Input
2
[]
Output
+- -+
| |
+- -+
Input
4
[][]
Output
+- -++- -+
| || |
+- -++- -+ | instruction | 0 | 4,689 | 21 | 9,378 |
Tags: *special, implementation
Correct Solution:
```
n = int(input())
x = input()
d = [0] * n
cd = 0
xp = []
for i in range(n):
if x[i] == '[':
d[i] = cd
cd = cd + 1
else:
cd = cd - 1
d[i] = cd
for i in range(n-1):
xp.append((x[i], d[i]))
if x[i] == '[' and x[i+1] == ']':
xp.extend([(' ', d[i]), (' ', d[i]), (' ', d[i])])
xp.append((x[n-1], d[n-1]))
md = max(d)
h = md * 2 + 3
res = []
for i in range(h):
l = [' ' for j in xp]
res.append(l)
for i in range(len(xp)):
for j in range(h):
if xp[i][0] == '[' and j > xp[i][1] and j < h - xp[i][1] - 1:
res[j][i] = '|'
elif xp[i][0] == ']' and j > xp[i][1] and j < h - xp[i][1] - 1:
res[j][i] = '|'
elif xp[i][0] == '[' and (j == xp[i][1] or j == h - xp[i][1] - 1):
res[j][i] = '+'
res[j][i+1] = '-'
elif xp[i][0] == ']' and (j == xp[i][1] or j == h - xp[i][1] - 1):
res[j][i] = '+'
res[j][i-1] = '-'
for i in range(h):
print(''.join(res[i]))
``` | output | 1 | 4,689 | 21 | 9,379 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence of square brackets is regular if by inserting symbols "+" and "1" into it, you can get a regular mathematical expression from it. For example, sequences "[[]][]", "[]" and "[[][[]]]" — are regular, at the same time "][", "[[]" and "[[]]][" — are irregular.
Draw the given sequence using a minimalistic pseudographics in the strip of the lowest possible height — use symbols '+', '-' and '|'. For example, the sequence "[[][]][]" should be represented as:
+- -++- -+
|+- -++- -+|| |
|| || ||| |
|+- -++- -+|| |
+- -++- -+
Each bracket should be represented with the hepl of one or more symbols '|' (the vertical part) and symbols '+' and '-' as on the example which is given above.
Brackets should be drawn without spaces one by one, only dividing pairs of consecutive pairwise brackets with a single-space bar (so that the two brackets do not visually merge into one symbol). The image should have the minimum possible height.
The enclosed bracket is always smaller than the surrounding bracket, but each bracket separately strives to maximize the height of the image. So the pair of final brackets in the example above occupies the entire height of the image.
Study carefully the examples below, they adequately explain the condition of the problem. Pay attention that in this problem the answer (the image) is unique.
Input
The first line contains an even integer n (2 ≤ n ≤ 100) — the length of the sequence of brackets.
The second line contains the sequence of brackets — these are n symbols "[" and "]". It is guaranteed that the given sequence of brackets is regular.
Output
Print the drawn bracket sequence in the format which is given in the condition. Don't print extra (unnecessary) spaces.
Examples
Input
8
[[][]][]
Output
+- -++- -+
|+- -++- -+|| |
|| || ||| |
|+- -++- -+|| |
+- -++- -+
Input
6
[[[]]]
Output
+- -+
|+- -+|
||+- -+||
||| |||
||+- -+||
|+- -+|
+- -+
Input
6
[[][]]
Output
+- -+
|+- -++- -+|
|| || ||
|+- -++- -+|
+- -+
Input
2
[]
Output
+- -+
| |
+- -+
Input
4
[][]
Output
+- -++- -+
| || |
+- -++- -+ | instruction | 0 | 4,690 | 21 | 9,380 |
Tags: *special, implementation
Correct Solution:
```
import base64
code = "aW5wdXQgKCkjbGluZToxCmJyID1pbnB1dCAoKSNsaW5lOjIKZCA9W10jbGluZTozCmNvbSA9MCAjbGluZTo0CmhtYXggPTAgI2xpbmU6NQpmb3IgaSBpbiByYW5nZSAobGVuIChiciApKTojbGluZTo2CglpZiBiciBbaSBdPT0nWyc6I2xpbmU6NwoJCWQgLmFwcGVuZCAoeydvcGVuJzpUcnVlICwnY29tJzpjb20gfSkjbGluZToxMQoJCWNvbSArPTIgI2xpbmU6MTIKCWVsc2UgOiNsaW5lOjEzCgkJY29tIC09MiAjbGluZToxNAoJCWQgLmFwcGVuZCAoeydvcGVuJzpGYWxzZSAsJ2NvbSc6Y29tIH0pI2xpbmU6MTgKCWlmIGNvbSA+aG1heCA6I2xpbmU6MjAKCQlobWF4ID1jb20gI2xpbmU6MjEKaG1heCAtPTEgI2xpbmU6MjMKYSA9W1tdZm9yIE8wTzAwME8wTzAwTzAwMDBPIGluIHJhbmdlIChobWF4ICsyICldI2xpbmU6MjUKeSA9MCAjbGluZToyNgp4ID0wICNsaW5lOjI3CmFwID1UcnVlICNsaW5lOjI4CmZvciBqIGluIHJhbmdlIChsZW4gKGQgKSk6I2xpbmU6MjkKCWlmIGFwIDojbGluZTozMQoJCWZvciBrZWsgaW4gcmFuZ2UgKGxlbiAoYSApKTojbGluZTozMgoJCQlhIFtrZWsgXS5hcHBlbmQgKCcgJykjbGluZTozMwoJZWxzZSA6I2xpbmU6MzQKCQlhcCA9VHJ1ZSAjbGluZTozNQoJaSA9ZCBbaiBdI2xpbmU6MzcKCXkgPWkgWydjb20nXS8vMiAjbGluZTozOAoJeTAgPXkgI2xpbmU6MzkKCWEgW3kgXVt4IF09JysnI2xpbmU6NDAKCWZvciBfIGluIHJhbmdlIChobWF4IC1pIFsnY29tJ10pOiNsaW5lOjQxCgkJeSArPTEgI2xpbmU6NDIKCQlhIFt5IF1beCBdPSd8JyNsaW5lOjQzCgl5ICs9MSAjbGluZTo0NAoJYSBbeSBdW3ggXT0nKycjbGluZTo0NQoJaWYgaSBbJ29wZW4nXTojbGluZTo0NwoJCWZvciBrZWsgaW4gcmFuZ2UgKGxlbiAoYSApKTojbGluZTo0OAoJCQlhIFtrZWsgXS5hcHBlbmQgKCcgJykjbGluZTo0OQoJCWFwID1GYWxzZSAjbGluZTo1MAoJCWEgW3kwIF1beCArMSBdPSctJyNsaW5lOjUxCgkJYSBbeSBdW3ggKzEgXT0nLScjbGluZTo1MgoJZWxzZSA6I2xpbmU6NTMKCQlhIFt5MCBdW3ggLTEgXT0nLScjbGluZTo1NAoJCWEgW3kgXVt4IC0xIF09Jy0nI2xpbmU6NTUKCXRyeSA6I2xpbmU6NTYKCQlpZiBpIFsnb3BlbiddYW5kIG5vdCBkIFtqICsxIF1bJ29wZW4nXTojbGluZTo1NwoJCQl4ICs9MyAjbGluZTo1OAoJCQlmb3Iga2VrIGluIHJhbmdlIChsZW4gKGEgKSk6I2xpbmU6NTkKCQkJCWZvciBfIGluIHJhbmdlICgzICk6I2xpbmU6NjAKCQkJCQlhIFtrZWsgXS5hcHBlbmQgKCcgJykjbGluZTo2MQoJZXhjZXB0IDojbGluZTo2MgoJCXBhc3MgI2xpbmU6NjMKCXggKz0xICNsaW5lOjY1CmZvciBpIGluIGEgOiNsaW5lOjY3CglwcmludCAoKmkgLHNlcCA9JycpCg=="
eval(compile(base64.b64decode(code),'<string>','exec'))
``` | output | 1 | 4,690 | 21 | 9,381 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A sequence of square brackets is regular if by inserting symbols "+" and "1" into it, you can get a regular mathematical expression from it. For example, sequences "[[]][]", "[]" and "[[][[]]]" — are regular, at the same time "][", "[[]" and "[[]]][" — are irregular.
Draw the given sequence using a minimalistic pseudographics in the strip of the lowest possible height — use symbols '+', '-' and '|'. For example, the sequence "[[][]][]" should be represented as:
+- -++- -+
|+- -++- -+|| |
|| || ||| |
|+- -++- -+|| |
+- -++- -+
Each bracket should be represented with the hepl of one or more symbols '|' (the vertical part) and symbols '+' and '-' as on the example which is given above.
Brackets should be drawn without spaces one by one, only dividing pairs of consecutive pairwise brackets with a single-space bar (so that the two brackets do not visually merge into one symbol). The image should have the minimum possible height.
The enclosed bracket is always smaller than the surrounding bracket, but each bracket separately strives to maximize the height of the image. So the pair of final brackets in the example above occupies the entire height of the image.
Study carefully the examples below, they adequately explain the condition of the problem. Pay attention that in this problem the answer (the image) is unique.
Input
The first line contains an even integer n (2 ≤ n ≤ 100) — the length of the sequence of brackets.
The second line contains the sequence of brackets — these are n symbols "[" and "]". It is guaranteed that the given sequence of brackets is regular.
Output
Print the drawn bracket sequence in the format which is given in the condition. Don't print extra (unnecessary) spaces.
Examples
Input
8
[[][]][]
Output
+- -++- -+
|+- -++- -+|| |
|| || ||| |
|+- -++- -+|| |
+- -++- -+
Input
6
[[[]]]
Output
+- -+
|+- -+|
||+- -+||
||| |||
||+- -+||
|+- -+|
+- -+
Input
6
[[][]]
Output
+- -+
|+- -++- -+|
|| || ||
|+- -++- -+|
+- -+
Input
2
[]
Output
+- -+
| |
+- -+
Input
4
[][]
Output
+- -++- -+
| || |
+- -++- -+
Submitted Solution:
```
n = int(input())
s = input()
t = set()
a = [0]*len(s)
q = 0
m = 0
ans = []
for i in range(len(s)):
if(s[i]=='['):
q+=1
a[i]=q
m=max(m,q)
else:
a[i]=q
q-=1
m = max(m,q)
if(s[i-1]=='['):
t.add(i-1)
for i in range(m+1):
e=''
for j in range(len(s)):
if(a[j]-1==i):
if(s[j]=='['):
e+='+-'
if(j in t):
e+=' '
else:
e+='-+'
elif(a[j]-1<i):
if(s[j]=='['):
e+='|'
if(j in t):
e+=' '
elif(s[j]==']'):
if(j-1 in t):
e+=' '
e+='|'
else:
if(s[j]=='['):
if(j>0 and s[j-1]==']' and a[j-1]==a[j]):
e+=' '
e+=' '
if(j in t):
e+=' '
else:
if(j!=len(s)-1 and s[j+1]!=']'):
e+=' '
e+=' '
ans+=[e]
for i in range(len(ans)):
print(ans[i])
for i in range(len(ans)-2,-1,-1):
print(ans[i])
``` | instruction | 0 | 4,691 | 21 | 9,382 |
Yes | output | 1 | 4,691 | 21 | 9,383 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A sequence of square brackets is regular if by inserting symbols "+" and "1" into it, you can get a regular mathematical expression from it. For example, sequences "[[]][]", "[]" and "[[][[]]]" — are regular, at the same time "][", "[[]" and "[[]]][" — are irregular.
Draw the given sequence using a minimalistic pseudographics in the strip of the lowest possible height — use symbols '+', '-' and '|'. For example, the sequence "[[][]][]" should be represented as:
+- -++- -+
|+- -++- -+|| |
|| || ||| |
|+- -++- -+|| |
+- -++- -+
Each bracket should be represented with the hepl of one or more symbols '|' (the vertical part) and symbols '+' and '-' as on the example which is given above.
Brackets should be drawn without spaces one by one, only dividing pairs of consecutive pairwise brackets with a single-space bar (so that the two brackets do not visually merge into one symbol). The image should have the minimum possible height.
The enclosed bracket is always smaller than the surrounding bracket, but each bracket separately strives to maximize the height of the image. So the pair of final brackets in the example above occupies the entire height of the image.
Study carefully the examples below, they adequately explain the condition of the problem. Pay attention that in this problem the answer (the image) is unique.
Input
The first line contains an even integer n (2 ≤ n ≤ 100) — the length of the sequence of brackets.
The second line contains the sequence of brackets — these are n symbols "[" and "]". It is guaranteed that the given sequence of brackets is regular.
Output
Print the drawn bracket sequence in the format which is given in the condition. Don't print extra (unnecessary) spaces.
Examples
Input
8
[[][]][]
Output
+- -++- -+
|+- -++- -+|| |
|| || ||| |
|+- -++- -+|| |
+- -++- -+
Input
6
[[[]]]
Output
+- -+
|+- -+|
||+- -+||
||| |||
||+- -+||
|+- -+|
+- -+
Input
6
[[][]]
Output
+- -+
|+- -++- -+|
|| || ||
|+- -++- -+|
+- -+
Input
2
[]
Output
+- -+
| |
+- -+
Input
4
[][]
Output
+- -++- -+
| || |
+- -++- -+
Submitted Solution:
```
d = dict()
d['['] = 1
d[']'] = -1
n = int(input())
s = input()
bal = 0
best = -1
for elem in s:
bal += d[elem]
best = max(best, bal)
size = 2 * best + 1
ans = [[' '] * (3 * n) for i in range(size)]
last_type = 'close'
last_size = size
curr_state = 0
for elem in s:
if last_type == 'close' and elem == '[':
curr_state += 1
up = best - last_size // 2
down = best + last_size // 2
ans[up][curr_state] = '+'
ans[down][curr_state] = '+'
for i in range(up + 1, down):
ans[i][curr_state] = '|'
ans[up][curr_state + 1] = '-'
ans[down][curr_state + 1] = '-'
last_type = 'open'
elif last_type == 'close' and elem == ']':
curr_state += 1
up = best - last_size // 2 - 1
down = best + last_size // 2 + 1
ans[up][curr_state] = '+'
ans[down][curr_state] = '+'
for i in range(up + 1, down):
ans[i][curr_state] = '|'
ans[up][curr_state - 1] = '-'
ans[down][curr_state - 1] = '-'
last_size += 2
elif last_type == 'open' and elem == '[':
curr_state += 1
up = best - last_size // 2 + 1
down = best + last_size // 2 - 1
ans[up][curr_state] = '+'
ans[down][curr_state] = '+'
for i in range(up + 1, down):
ans[i][curr_state] = '|'
ans[up][curr_state + 1] = '-'
ans[down][curr_state + 1] = '-'
last_size -= 2
else:
curr_state += 4
up = best - last_size // 2
down = best + last_size // 2
ans[up][curr_state] = '+'
ans[down][curr_state] = '+'
for i in range(up + 1, down):
ans[i][curr_state] = '|'
ans[up][curr_state - 1] = '-'
ans[down][curr_state - 1] = '-'
last_type = 'close'
for i in range(size):
for j in range(1, curr_state + 1):
print(ans[i][j], end = '')
print()
``` | instruction | 0 | 4,692 | 21 | 9,384 |
Yes | output | 1 | 4,692 | 21 | 9,385 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A sequence of square brackets is regular if by inserting symbols "+" and "1" into it, you can get a regular mathematical expression from it. For example, sequences "[[]][]", "[]" and "[[][[]]]" — are regular, at the same time "][", "[[]" and "[[]]][" — are irregular.
Draw the given sequence using a minimalistic pseudographics in the strip of the lowest possible height — use symbols '+', '-' and '|'. For example, the sequence "[[][]][]" should be represented as:
+- -++- -+
|+- -++- -+|| |
|| || ||| |
|+- -++- -+|| |
+- -++- -+
Each bracket should be represented with the hepl of one or more symbols '|' (the vertical part) and symbols '+' and '-' as on the example which is given above.
Brackets should be drawn without spaces one by one, only dividing pairs of consecutive pairwise brackets with a single-space bar (so that the two brackets do not visually merge into one symbol). The image should have the minimum possible height.
The enclosed bracket is always smaller than the surrounding bracket, but each bracket separately strives to maximize the height of the image. So the pair of final brackets in the example above occupies the entire height of the image.
Study carefully the examples below, they adequately explain the condition of the problem. Pay attention that in this problem the answer (the image) is unique.
Input
The first line contains an even integer n (2 ≤ n ≤ 100) — the length of the sequence of brackets.
The second line contains the sequence of brackets — these are n symbols "[" and "]". It is guaranteed that the given sequence of brackets is regular.
Output
Print the drawn bracket sequence in the format which is given in the condition. Don't print extra (unnecessary) spaces.
Examples
Input
8
[[][]][]
Output
+- -++- -+
|+- -++- -+|| |
|| || ||| |
|+- -++- -+|| |
+- -++- -+
Input
6
[[[]]]
Output
+- -+
|+- -+|
||+- -+||
||| |||
||+- -+||
|+- -+|
+- -+
Input
6
[[][]]
Output
+- -+
|+- -++- -+|
|| || ||
|+- -++- -+|
+- -+
Input
2
[]
Output
+- -+
| |
+- -+
Input
4
[][]
Output
+- -++- -+
| || |
+- -++- -+
Submitted Solution:
```
n = input()
s = input()
b = 0
maxb = 0
for c in s:
if c == '[':
b += 1
else:
b -= 1
if b > maxb:
maxb = b
res = [""] * (maxb * 2 + 1)
b = maxb
pred = ""
for k in range(len(s)):
c = s[k]
if c == '[':
if k != len(s) - 1:
if s[k+1] == ']':
sep = '| '
else:
sep = '|'
else:
sep = '|'
i = maxb - b
for j in range(i):
res[j] += ' '
res[i] += '+-'
for j in range(i + 1, len(res) - i - 1):
res[j] += sep
res[len(res) - i - 1] += '+-'
for j in range(len(res) - i, len(res)):
res[j] += ' '
pred = '['
b -= 1
elif c == ']':
if k != len(s) - 1:
if s[k+1] == '[':
space = ' '
else:
space = ' '
else:
space = ' '
b += 1
if pred == '[':
sep = ' |'
for j in range(len(res)):
res[j] += ' '
else:
sep = '|'
i = maxb - b
for j in range(i):
res[j] += space
res[i] += '-+'
for j in range(i + 1, len(res) - i - 1):
res[j] += sep
res[len(res) - i - 1] += '-+'
for j in range(len(res) - i, len(res)):
res[j] += space
pred = ']'
for i in res:
print(i)
``` | instruction | 0 | 4,693 | 21 | 9,386 |
Yes | output | 1 | 4,693 | 21 | 9,387 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A sequence of square brackets is regular if by inserting symbols "+" and "1" into it, you can get a regular mathematical expression from it. For example, sequences "[[]][]", "[]" and "[[][[]]]" — are regular, at the same time "][", "[[]" and "[[]]][" — are irregular.
Draw the given sequence using a minimalistic pseudographics in the strip of the lowest possible height — use symbols '+', '-' and '|'. For example, the sequence "[[][]][]" should be represented as:
+- -++- -+
|+- -++- -+|| |
|| || ||| |
|+- -++- -+|| |
+- -++- -+
Each bracket should be represented with the hepl of one or more symbols '|' (the vertical part) and symbols '+' and '-' as on the example which is given above.
Brackets should be drawn without spaces one by one, only dividing pairs of consecutive pairwise brackets with a single-space bar (so that the two brackets do not visually merge into one symbol). The image should have the minimum possible height.
The enclosed bracket is always smaller than the surrounding bracket, but each bracket separately strives to maximize the height of the image. So the pair of final brackets in the example above occupies the entire height of the image.
Study carefully the examples below, they adequately explain the condition of the problem. Pay attention that in this problem the answer (the image) is unique.
Input
The first line contains an even integer n (2 ≤ n ≤ 100) — the length of the sequence of brackets.
The second line contains the sequence of brackets — these are n symbols "[" and "]". It is guaranteed that the given sequence of brackets is regular.
Output
Print the drawn bracket sequence in the format which is given in the condition. Don't print extra (unnecessary) spaces.
Examples
Input
8
[[][]][]
Output
+- -++- -+
|+- -++- -+|| |
|| || ||| |
|+- -++- -+|| |
+- -++- -+
Input
6
[[[]]]
Output
+- -+
|+- -+|
||+- -+||
||| |||
||+- -+||
|+- -+|
+- -+
Input
6
[[][]]
Output
+- -+
|+- -++- -+|
|| || ||
|+- -++- -+|
+- -+
Input
2
[]
Output
+- -+
| |
+- -+
Input
4
[][]
Output
+- -++- -+
| || |
+- -++- -+
Submitted Solution:
```
class Screen:
def __init__(self, n_rows):
self.rows = [[] for _ in range(n_rows)]
self.height = n_rows
def draw(self, x, y, c):
row = self.rows[y]
while x > len(row) - 1:
row.append(' ')
row[x] = c
def draw_open(self, x, height):
middle = self.height // 2
self.draw(x, middle, '|')
for dy in range(1, height + 1):
self.draw(x, middle + dy, '|')
self.draw(x, middle - dy, '|')
self.draw(x, middle + height + 1, '+')
self.draw(x + 1, middle + height + 1, '-')
self.draw(x, middle - height - 1, '+')
self.draw(x + 1, middle - height - 1, '-')
def draw_close(self, x, height):
middle = self.height // 2
self.draw(x, middle, '|')
for dy in range(1, height + 1):
self.draw(x, middle + dy, '|')
self.draw(x, middle - dy, '|')
self.draw(x, middle + height + 1, '+')
self.draw(x - 1, middle + height + 1, '-')
self.draw(x, middle - height - 1, '+')
self.draw(x - 1, middle - height - 1, '-')
def strings(self):
return [''.join(row) for row in self.rows]
def to_heights(seq):
depths = []
cur_depth = 0
for par in seq:
if par == '[':
depths.append(cur_depth)
cur_depth += 1
else:
cur_depth -= 1
depths.append(cur_depth)
max_depth = max(depths)
heights = [max_depth - depth for depth in depths]
return heights
def to_strings(seq, heights):
n_rows = 2 * (max(heights) + 1) + 1
screen = Screen(n_rows)
cur_x = 0
prev_par = None
for par, height in zip(seq, heights):
if par == '[':
screen.draw_open(cur_x, height)
cur_x += 1
if par == ']':
if prev_par == '[':
cur_x += 3
screen.draw_close(cur_x, height)
cur_x += 1
prev_par = par
return screen.strings()
n = int(input())
seq = input()
heights = to_heights(seq)
strings = to_strings(seq, heights)
for string in strings:
print(string)
``` | instruction | 0 | 4,694 | 21 | 9,388 |
Yes | output | 1 | 4,694 | 21 | 9,389 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A sequence of square brackets is regular if by inserting symbols "+" and "1" into it, you can get a regular mathematical expression from it. For example, sequences "[[]][]", "[]" and "[[][[]]]" — are regular, at the same time "][", "[[]" and "[[]]][" — are irregular.
Draw the given sequence using a minimalistic pseudographics in the strip of the lowest possible height — use symbols '+', '-' and '|'. For example, the sequence "[[][]][]" should be represented as:
+- -++- -+
|+- -++- -+|| |
|| || ||| |
|+- -++- -+|| |
+- -++- -+
Each bracket should be represented with the hepl of one or more symbols '|' (the vertical part) and symbols '+' and '-' as on the example which is given above.
Brackets should be drawn without spaces one by one, only dividing pairs of consecutive pairwise brackets with a single-space bar (so that the two brackets do not visually merge into one symbol). The image should have the minimum possible height.
The enclosed bracket is always smaller than the surrounding bracket, but each bracket separately strives to maximize the height of the image. So the pair of final brackets in the example above occupies the entire height of the image.
Study carefully the examples below, they adequately explain the condition of the problem. Pay attention that in this problem the answer (the image) is unique.
Input
The first line contains an even integer n (2 ≤ n ≤ 100) — the length of the sequence of brackets.
The second line contains the sequence of brackets — these are n symbols "[" and "]". It is guaranteed that the given sequence of brackets is regular.
Output
Print the drawn bracket sequence in the format which is given in the condition. Don't print extra (unnecessary) spaces.
Examples
Input
8
[[][]][]
Output
+- -++- -+
|+- -++- -+|| |
|| || ||| |
|+- -++- -+|| |
+- -++- -+
Input
6
[[[]]]
Output
+- -+
|+- -+|
||+- -+||
||| |||
||+- -+||
|+- -+|
+- -+
Input
6
[[][]]
Output
+- -+
|+- -++- -+|
|| || ||
|+- -++- -+|
+- -+
Input
2
[]
Output
+- -+
| |
+- -+
Input
4
[][]
Output
+- -++- -+
| || |
+- -++- -+
Submitted Solution:
```
s = input()
t = set()
a = [0]*len(s)
q = 0
m = 0
ans = []
for i in range(len(s)):
if(s[i]=='['):
q+=1
a[i]=q
m=max(m,q)
else:
a[i]=q
q-=1
m = max(m,q)
if(s[i-1]=='['):
t.add(i-1)
for i in range(m+1):
e=''
for j in range(len(s)):
if(a[j]-1==i):
if(s[j]=='['):
e+='+-'
if(j in t):
e+=' '
else:
e+='-+'
elif(a[j]-1<i):
if(s[j]=='['):
e+='|'
if(j in t):
e+=' '
elif(s[j]==']'):
if(j-1 in t):
e+=' '
e+='|'
else:
if(s[j]=='['):
if(j>0 and s[j-1]==']' and a[j-1]==a[j]):
e+=' '
e+=' '
if(j in t):
e+=' '
else:
if(j!=len(s)-1 and s[j+1]!=']'):
e+=' '
e+=' '
ans+=[e]
for i in range(len(ans)):
print(ans[i])
for i in range(len(ans)-2,-1,-1):
print(ans[i])
``` | instruction | 0 | 4,695 | 21 | 9,390 |
No | output | 1 | 4,695 | 21 | 9,391 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A sequence of square brackets is regular if by inserting symbols "+" and "1" into it, you can get a regular mathematical expression from it. For example, sequences "[[]][]", "[]" and "[[][[]]]" — are regular, at the same time "][", "[[]" and "[[]]][" — are irregular.
Draw the given sequence using a minimalistic pseudographics in the strip of the lowest possible height — use symbols '+', '-' and '|'. For example, the sequence "[[][]][]" should be represented as:
+- -++- -+
|+- -++- -+|| |
|| || ||| |
|+- -++- -+|| |
+- -++- -+
Each bracket should be represented with the hepl of one or more symbols '|' (the vertical part) and symbols '+' and '-' as on the example which is given above.
Brackets should be drawn without spaces one by one, only dividing pairs of consecutive pairwise brackets with a single-space bar (so that the two brackets do not visually merge into one symbol). The image should have the minimum possible height.
The enclosed bracket is always smaller than the surrounding bracket, but each bracket separately strives to maximize the height of the image. So the pair of final brackets in the example above occupies the entire height of the image.
Study carefully the examples below, they adequately explain the condition of the problem. Pay attention that in this problem the answer (the image) is unique.
Input
The first line contains an even integer n (2 ≤ n ≤ 100) — the length of the sequence of brackets.
The second line contains the sequence of brackets — these are n symbols "[" and "]". It is guaranteed that the given sequence of brackets is regular.
Output
Print the drawn bracket sequence in the format which is given in the condition. Don't print extra (unnecessary) spaces.
Examples
Input
8
[[][]][]
Output
+- -++- -+
|+- -++- -+|| |
|| || ||| |
|+- -++- -+|| |
+- -++- -+
Input
6
[[[]]]
Output
+- -+
|+- -+|
||+- -+||
||| |||
||+- -+||
|+- -+|
+- -+
Input
6
[[][]]
Output
+- -+
|+- -++- -+|
|| || ||
|+- -++- -+|
+- -+
Input
2
[]
Output
+- -+
| |
+- -+
Input
4
[][]
Output
+- -++- -+
| || |
+- -++- -+
Submitted Solution:
```
import fileinput
s = fileinput.input().readline()[:-1]
ln = len(s)
l = [0] * ln
c = 0
for i, x in enumerate(s):
if x == '[':
c += 1
l[i] = c
if x == ']':
c -= 1
enc = max(l)
h_max = enc * 2 + 1
for h_ in range(h_max):
h = min(h_, 2*enc - h_)
for i in range(ln):
if h+1 == l[i]:
if s[i] == '[':
print('+', end='')
elif s[i] == ']':
print('+', end='')
elif h+2 == l[i] and (s[i] == s[i-1] == '[' or s[i] == s[i+1] == ']'):
print('-', end='')
elif h >= l[i]:
print('|', end='')
else:
print(' ', end='')
if s[i] == '[' and s[i+1] ==']':
if h+1 == l[i]:
print('- -', end='')
else:
print(' ', end='')
print()
``` | instruction | 0 | 4,696 | 21 | 9,392 |
No | output | 1 | 4,696 | 21 | 9,393 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A sequence of square brackets is regular if by inserting symbols "+" and "1" into it, you can get a regular mathematical expression from it. For example, sequences "[[]][]", "[]" and "[[][[]]]" — are regular, at the same time "][", "[[]" and "[[]]][" — are irregular.
Draw the given sequence using a minimalistic pseudographics in the strip of the lowest possible height — use symbols '+', '-' and '|'. For example, the sequence "[[][]][]" should be represented as:
+- -++- -+
|+- -++- -+|| |
|| || ||| |
|+- -++- -+|| |
+- -++- -+
Each bracket should be represented with the hepl of one or more symbols '|' (the vertical part) and symbols '+' and '-' as on the example which is given above.
Brackets should be drawn without spaces one by one, only dividing pairs of consecutive pairwise brackets with a single-space bar (so that the two brackets do not visually merge into one symbol). The image should have the minimum possible height.
The enclosed bracket is always smaller than the surrounding bracket, but each bracket separately strives to maximize the height of the image. So the pair of final brackets in the example above occupies the entire height of the image.
Study carefully the examples below, they adequately explain the condition of the problem. Pay attention that in this problem the answer (the image) is unique.
Input
The first line contains an even integer n (2 ≤ n ≤ 100) — the length of the sequence of brackets.
The second line contains the sequence of brackets — these are n symbols "[" and "]". It is guaranteed that the given sequence of brackets is regular.
Output
Print the drawn bracket sequence in the format which is given in the condition. Don't print extra (unnecessary) spaces.
Examples
Input
8
[[][]][]
Output
+- -++- -+
|+- -++- -+|| |
|| || ||| |
|+- -++- -+|| |
+- -++- -+
Input
6
[[[]]]
Output
+- -+
|+- -+|
||+- -+||
||| |||
||+- -+||
|+- -+|
+- -+
Input
6
[[][]]
Output
+- -+
|+- -++- -+|
|| || ||
|+- -++- -+|
+- -+
Input
2
[]
Output
+- -+
| |
+- -+
Input
4
[][]
Output
+- -++- -+
| || |
+- -++- -+
Submitted Solution:
```
def depths(S):
current_max = 1
n = len(S)
for i in range(n):
if S[i] == '[':
yield current_max, 'L'
current_max += 2
elif S[i] == ']':
yield current_max - 2, 'R'
current_max -= 2
n = input()
s = input()
full_rev_l = list(depths(s))
rev_l = [i[0] for i in full_rev_l]
set_rev_l = list(set(rev_l))
l = [(set_rev_l[set_rev_l.index(val) * -1 - 1], full_rev_l[ind][1]) for ind, val in enumerate(rev_l)]
m = max(set_rev_l) + 2
paint = [[' ' for x in range(len(l)*3)] for y in range(m)]
i = 0
pre = ''
for br, side in l:
start = int((m - br) / 2 - 1)
end = int(m - (m - br) / 2)
if side == 'L':
# paint head
paint[start][i] = '+'
paint[start][i + 1] = '-'
# paint body
for j in range(start + 1, end):
paint[j][i] = '|'
# paint foot
paint[end][i] = '+'
paint[end][i + 1] = '-'
# i ++
else:
if pre == 'L': i += 3
# paint head
paint[start][i] = '+'
paint[start][i - 1] = '-'
# paint body
for j in range(start + 1, end):
paint[j][i] = '|'
# paint foot
paint[end][i] = '+'
paint[end][i - 1] = '-'
i += 1
pre = side
for s in paint:
print(''.join(s).strip())
``` | instruction | 0 | 4,697 | 21 | 9,394 |
No | output | 1 | 4,697 | 21 | 9,395 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A sequence of square brackets is regular if by inserting symbols "+" and "1" into it, you can get a regular mathematical expression from it. For example, sequences "[[]][]", "[]" and "[[][[]]]" — are regular, at the same time "][", "[[]" and "[[]]][" — are irregular.
Draw the given sequence using a minimalistic pseudographics in the strip of the lowest possible height — use symbols '+', '-' and '|'. For example, the sequence "[[][]][]" should be represented as:
+- -++- -+
|+- -++- -+|| |
|| || ||| |
|+- -++- -+|| |
+- -++- -+
Each bracket should be represented with the hepl of one or more symbols '|' (the vertical part) and symbols '+' and '-' as on the example which is given above.
Brackets should be drawn without spaces one by one, only dividing pairs of consecutive pairwise brackets with a single-space bar (so that the two brackets do not visually merge into one symbol). The image should have the minimum possible height.
The enclosed bracket is always smaller than the surrounding bracket, but each bracket separately strives to maximize the height of the image. So the pair of final brackets in the example above occupies the entire height of the image.
Study carefully the examples below, they adequately explain the condition of the problem. Pay attention that in this problem the answer (the image) is unique.
Input
The first line contains an even integer n (2 ≤ n ≤ 100) — the length of the sequence of brackets.
The second line contains the sequence of brackets — these are n symbols "[" and "]". It is guaranteed that the given sequence of brackets is regular.
Output
Print the drawn bracket sequence in the format which is given in the condition. Don't print extra (unnecessary) spaces.
Examples
Input
8
[[][]][]
Output
+- -++- -+
|+- -++- -+|| |
|| || ||| |
|+- -++- -+|| |
+- -++- -+
Input
6
[[[]]]
Output
+- -+
|+- -+|
||+- -+||
||| |||
||+- -+||
|+- -+|
+- -+
Input
6
[[][]]
Output
+- -+
|+- -++- -+|
|| || ||
|+- -++- -+|
+- -+
Input
2
[]
Output
+- -+
| |
+- -+
Input
4
[][]
Output
+- -++- -+
| || |
+- -++- -+
Submitted Solution:
```
def draw(ind, type, size):
global ans, mid
ans[mid][ind] = '|'
for i in range(size // 2):
ans[mid + i][ind] = '|'
ans[mid - i][ind] = '|'
ans[mid + size // 2][ind] = '+'
ans[mid - size // 2][ind] = '+'
if type == "left":
ans[mid + size // 2][ind - 1] = '-'
ans[mid - size // 2][ind - 1] = '-'
else:
ans[mid + size // 2][ind + 1] = '-'
ans[mid - size // 2][ind + 1] = '-'
n = int(input())
s = input()
mx = 0
cs = 0
for i in range(n):
if s[i] == "[":
cs += 1
else:
cs -= 1
mx = max(mx, cs)
mx -= 1
a = [mx]
for i in range(1, n):
if (s[i] == "[" and s[i - 1] == "]") or (s[i - 1] == "[" and s[i] == "]"):
a.append(a[-1])
elif s[i] == "[":
a.append(a[-1] - 1)
else:
a.append(a[-1] + 1)
ans = [[' ' for i in range(n * 3)] for j in range(3 + mx * 2)]
mid = mx + 1
draw(0, "right", mx * 2 + 3)
li = 0
for i in range(1, n):
if s[i] == "[" and s[i - 1] == "]":
li += 1
draw(li, "right", a[i] * 2 + 3)
elif s[i - 1] == "[" and s[i] == "]":
li += 4
draw(li, "left", a[i] * 2 + 3)
elif s[i] == "[":
li += 1
draw(li, "right", a[i] * 2 + 3)
else:
li += 1
draw(li, "left", a[i] * 2 + 3)
for i in ans:
for j in i:
print(j, end="")
print()
``` | instruction | 0 | 4,698 | 21 | 9,396 |
No | output | 1 | 4,698 | 21 | 9,397 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given string s consists of opening and closing brackets of four kinds <>, {}, [], (). There are two types of brackets: opening and closing. You can replace any bracket by another of the same type. For example, you can replace < by the bracket {, but you can't replace it by ) or >.
The following definition of a regular bracket sequence is well-known, so you can be familiar with it.
Let's define a regular bracket sequence (RBS). Empty string is RBS. Let s1 and s2 be a RBS then the strings <s1>s2, {s1}s2, [s1]s2, (s1)s2 are also RBS.
For example the string "[[(){}]<>]" is RBS, but the strings "[)()" and "][()()" are not.
Determine the least number of replaces to make the string s RBS.
Input
The only line contains a non empty string s, consisting of only opening and closing brackets of four kinds. The length of s does not exceed 106.
Output
If it's impossible to get RBS from s print Impossible.
Otherwise print the least number of replaces needed to get RBS from s.
Examples
Input
[<}){}
Output
2
Input
{()}[]
Output
0
Input
]]
Output
Impossible | instruction | 0 | 6,221 | 21 | 12,442 |
Tags: data structures, expression parsing, math
Correct Solution:
```
bracket = input()
stack = []
bracketType = {
'[': (0, ']'),
'(': (0, ')'),
'<': (0, '>'),
'{': (0, '}'),
']': (1,),
')': (1,),
'>': (1,),
'}': (1,)
}
res = 0
for b in bracket:
if bracketType.get(b) == None:
break
elif bracketType.get(b)[0] == 0:
stack.append(b)
elif bracketType.get(b)[0] == 1:
if len(stack) == 0:
res = 'Impossible'
break
else:
if b != bracketType[stack.pop()][1]:
res += 1
if len(stack):
print('Impossible')
else:
print(res)
``` | output | 1 | 6,221 | 21 | 12,443 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given string s consists of opening and closing brackets of four kinds <>, {}, [], (). There are two types of brackets: opening and closing. You can replace any bracket by another of the same type. For example, you can replace < by the bracket {, but you can't replace it by ) or >.
The following definition of a regular bracket sequence is well-known, so you can be familiar with it.
Let's define a regular bracket sequence (RBS). Empty string is RBS. Let s1 and s2 be a RBS then the strings <s1>s2, {s1}s2, [s1]s2, (s1)s2 are also RBS.
For example the string "[[(){}]<>]" is RBS, but the strings "[)()" and "][()()" are not.
Determine the least number of replaces to make the string s RBS.
Input
The only line contains a non empty string s, consisting of only opening and closing brackets of four kinds. The length of s does not exceed 106.
Output
If it's impossible to get RBS from s print Impossible.
Otherwise print the least number of replaces needed to get RBS from s.
Examples
Input
[<}){}
Output
2
Input
{()}[]
Output
0
Input
]]
Output
Impossible | instruction | 0 | 6,222 | 21 | 12,444 |
Tags: data structures, expression parsing, math
Correct Solution:
```
s = input().strip()
braces = {'<' : '>',
'>' : '<',
'{' : '}',
'}' : '{',
'[' : ']',
']' : '[',
'(' : ')',
')' : '('}
stack = []
answer = 0
for char in s:
if char in "(<{[":
stack.append(char)
elif char in ")>}]":
if not stack:
print('Impossible')
exit()
lastBrace = stack.pop()
if char != braces[lastBrace]:
answer += 1
if not stack:
print(answer)
else:
print("Impossible")
``` | output | 1 | 6,222 | 21 | 12,445 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given string s consists of opening and closing brackets of four kinds <>, {}, [], (). There are two types of brackets: opening and closing. You can replace any bracket by another of the same type. For example, you can replace < by the bracket {, but you can't replace it by ) or >.
The following definition of a regular bracket sequence is well-known, so you can be familiar with it.
Let's define a regular bracket sequence (RBS). Empty string is RBS. Let s1 and s2 be a RBS then the strings <s1>s2, {s1}s2, [s1]s2, (s1)s2 are also RBS.
For example the string "[[(){}]<>]" is RBS, but the strings "[)()" and "][()()" are not.
Determine the least number of replaces to make the string s RBS.
Input
The only line contains a non empty string s, consisting of only opening and closing brackets of four kinds. The length of s does not exceed 106.
Output
If it's impossible to get RBS from s print Impossible.
Otherwise print the least number of replaces needed to get RBS from s.
Examples
Input
[<}){}
Output
2
Input
{()}[]
Output
0
Input
]]
Output
Impossible | instruction | 0 | 6,223 | 21 | 12,446 |
Tags: data structures, expression parsing, math
Correct Solution:
```
#! /usr/bin/env python
# -*- coding: utf-8 -*-
# vim:fenc=utf-8
#
# Copyright © 2016 missingdays <missingdays@missingdays>
#
# Distributed under terms of the MIT license.
"""
"""
opening = {
"[": "]",
"<": ">",
"{": "}",
"(": ")",
}
closing = {
"]": "[",
">": "<",
"}": "{",
")": "(",
}
s = input()
stack = []
answ = 0
for c in s:
if c in opening:
stack.append(c)
else:
if len(stack) == 0:
print("Impossible")
exit()
op = stack.pop()
if c != opening[op]:
answ += 1
if len(stack) != 0:
print("Impossible")
exit()
print(answ)
``` | output | 1 | 6,223 | 21 | 12,447 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given string s consists of opening and closing brackets of four kinds <>, {}, [], (). There are two types of brackets: opening and closing. You can replace any bracket by another of the same type. For example, you can replace < by the bracket {, but you can't replace it by ) or >.
The following definition of a regular bracket sequence is well-known, so you can be familiar with it.
Let's define a regular bracket sequence (RBS). Empty string is RBS. Let s1 and s2 be a RBS then the strings <s1>s2, {s1}s2, [s1]s2, (s1)s2 are also RBS.
For example the string "[[(){}]<>]" is RBS, but the strings "[)()" and "][()()" are not.
Determine the least number of replaces to make the string s RBS.
Input
The only line contains a non empty string s, consisting of only opening and closing brackets of four kinds. The length of s does not exceed 106.
Output
If it's impossible to get RBS from s print Impossible.
Otherwise print the least number of replaces needed to get RBS from s.
Examples
Input
[<}){}
Output
2
Input
{()}[]
Output
0
Input
]]
Output
Impossible | instruction | 0 | 6,224 | 21 | 12,448 |
Tags: data structures, expression parsing, math
Correct Solution:
```
import sys
s = input()
stack = []
piar = {'{' : '}', '(' : ')', '<' : '>', '[':']'}
ans = 0
for ch in s:
if ch in piar.keys():
stack.append(ch)
else:
if len(stack) == 0:
print("Impossible")
sys.exit()
if piar[stack.pop()] != ch:
ans+=1
if len(stack) != 0:
print("Impossible")
sys.exit()
print(ans)
``` | output | 1 | 6,224 | 21 | 12,449 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given string s consists of opening and closing brackets of four kinds <>, {}, [], (). There are two types of brackets: opening and closing. You can replace any bracket by another of the same type. For example, you can replace < by the bracket {, but you can't replace it by ) or >.
The following definition of a regular bracket sequence is well-known, so you can be familiar with it.
Let's define a regular bracket sequence (RBS). Empty string is RBS. Let s1 and s2 be a RBS then the strings <s1>s2, {s1}s2, [s1]s2, (s1)s2 are also RBS.
For example the string "[[(){}]<>]" is RBS, but the strings "[)()" and "][()()" are not.
Determine the least number of replaces to make the string s RBS.
Input
The only line contains a non empty string s, consisting of only opening and closing brackets of four kinds. The length of s does not exceed 106.
Output
If it's impossible to get RBS from s print Impossible.
Otherwise print the least number of replaces needed to get RBS from s.
Examples
Input
[<}){}
Output
2
Input
{()}[]
Output
0
Input
]]
Output
Impossible | instruction | 0 | 6,225 | 21 | 12,450 |
Tags: data structures, expression parsing, math
Correct Solution:
```
from collections import *
from itertools import *
from random import *
from bisect import *
from string import *
from queue import *
from heapq import *
from math import *
from re import *
from sys import *
def fast(): return stdin.readline().strip()
def zzz(): return [int(i) for i in fast().split()]
z, zz = input, lambda: list(map(int, z().split()))
szz, graph, mod, szzz = lambda: sorted(
zz()), {}, 10**9 + 7, lambda: sorted(zzz())
def lcd(xnum1, xnum2): return (xnum1 * xnum2 // gcd(xnum1, xnum2))
def output(answer): stdout.write(str(answer))
dx = [-1, 1, 0, 0, 1, -1, 1, -1]
dy = [0, 0, 1, -1, 1, -1, -1, 1]
###########################---Test-Case---#################################
"""
If you Know me , Then you probably don't know me !
"""
###########################---START-CODING---##############################
cnt = 0
arr = fast()
opn = {'(': ')', '<': '>', '[': ']', '{': '}'}
openBract = 0
que = deque()
for i in arr:
if i in opn:
openBract += 1
que.append(opn[i])
else:
try:
x = que.pop()
if i != x:
cnt += 1
que.appendleft(x)
except:
print("Impossible")
exit()
print(cnt if openBract == len(arr) - openBract else "Impossible")
``` | output | 1 | 6,225 | 21 | 12,451 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given string s consists of opening and closing brackets of four kinds <>, {}, [], (). There are two types of brackets: opening and closing. You can replace any bracket by another of the same type. For example, you can replace < by the bracket {, but you can't replace it by ) or >.
The following definition of a regular bracket sequence is well-known, so you can be familiar with it.
Let's define a regular bracket sequence (RBS). Empty string is RBS. Let s1 and s2 be a RBS then the strings <s1>s2, {s1}s2, [s1]s2, (s1)s2 are also RBS.
For example the string "[[(){}]<>]" is RBS, but the strings "[)()" and "][()()" are not.
Determine the least number of replaces to make the string s RBS.
Input
The only line contains a non empty string s, consisting of only opening and closing brackets of four kinds. The length of s does not exceed 106.
Output
If it's impossible to get RBS from s print Impossible.
Otherwise print the least number of replaces needed to get RBS from s.
Examples
Input
[<}){}
Output
2
Input
{()}[]
Output
0
Input
]]
Output
Impossible | instruction | 0 | 6,226 | 21 | 12,452 |
Tags: data structures, expression parsing, math
Correct Solution:
```
#!/usr/bin/env python3
import sys
s = input()
OPENING = ('<', '{', '[', '(')
CLOSING = ('>', '}', ']', ')')
result = 0
stack = []
for c in s:
if c in OPENING:
stack.append(c)
else:
if stack:
last_br = stack.pop()
if c != CLOSING[OPENING.index(last_br)]:
result += 1
else:
print("Impossible")
sys.exit(0)
print("Impossible" if stack else result)
``` | output | 1 | 6,226 | 21 | 12,453 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given string s consists of opening and closing brackets of four kinds <>, {}, [], (). There are two types of brackets: opening and closing. You can replace any bracket by another of the same type. For example, you can replace < by the bracket {, but you can't replace it by ) or >.
The following definition of a regular bracket sequence is well-known, so you can be familiar with it.
Let's define a regular bracket sequence (RBS). Empty string is RBS. Let s1 and s2 be a RBS then the strings <s1>s2, {s1}s2, [s1]s2, (s1)s2 are also RBS.
For example the string "[[(){}]<>]" is RBS, but the strings "[)()" and "][()()" are not.
Determine the least number of replaces to make the string s RBS.
Input
The only line contains a non empty string s, consisting of only opening and closing brackets of four kinds. The length of s does not exceed 106.
Output
If it's impossible to get RBS from s print Impossible.
Otherwise print the least number of replaces needed to get RBS from s.
Examples
Input
[<}){}
Output
2
Input
{()}[]
Output
0
Input
]]
Output
Impossible | instruction | 0 | 6,227 | 21 | 12,454 |
Tags: data structures, expression parsing, math
Correct Solution:
```
import sys
s=input()
stack = []
c=0
brackets = {')':'(',']':'[','}':'{','>':'<'}
for char in s:
if char in brackets.values():
stack.append(char)
elif char in brackets.keys():
if stack==[]:
print('Impossible')
sys.exit()
if brackets[char] != stack.pop():
c=c+1
if len(stack)!=0:
print('Impossible')
sys.exit(0)
print(c)
``` | output | 1 | 6,227 | 21 | 12,455 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given string s consists of opening and closing brackets of four kinds <>, {}, [], (). There are two types of brackets: opening and closing. You can replace any bracket by another of the same type. For example, you can replace < by the bracket {, but you can't replace it by ) or >.
The following definition of a regular bracket sequence is well-known, so you can be familiar with it.
Let's define a regular bracket sequence (RBS). Empty string is RBS. Let s1 and s2 be a RBS then the strings <s1>s2, {s1}s2, [s1]s2, (s1)s2 are also RBS.
For example the string "[[(){}]<>]" is RBS, but the strings "[)()" and "][()()" are not.
Determine the least number of replaces to make the string s RBS.
Input
The only line contains a non empty string s, consisting of only opening and closing brackets of four kinds. The length of s does not exceed 106.
Output
If it's impossible to get RBS from s print Impossible.
Otherwise print the least number of replaces needed to get RBS from s.
Examples
Input
[<}){}
Output
2
Input
{()}[]
Output
0
Input
]]
Output
Impossible | instruction | 0 | 6,228 | 21 | 12,456 |
Tags: data structures, expression parsing, math
Correct Solution:
```
import sys
import bisect
from bisect import bisect_left as lb
input_=lambda: sys.stdin.readline().strip("\r\n")
from math import log
from math import gcd
from math import atan2,acos
from random import randint
sa=lambda :input_()
sb=lambda:int(input_())
sc=lambda:input_().split()
sd=lambda:list(map(int,input_().split()))
se=lambda:float(input_())
sf=lambda:list(input_())
flsh=lambda: sys.stdout.flush()
#sys.setrecursionlimit(10**6)
mod=10**9+7
gp=[]
cost=[]
dp=[]
mx=[]
ans1=[]
ans2=[]
special=[]
specnode=[]
a=0
kthpar=[]
def dfs(root,par):
if par!=-1:
dp[root]=dp[par]+1
for i in range(1,20):
if kthpar[root][i-1]!=-1:
kthpar[root][i]=kthpar[kthpar[root][i-1]][i-1]
for child in gp[root]:
if child==par:continue
kthpar[child][0]=root
dfs(child,root)
d={')':'(',']':'[','}':'{','>':'<'}
op=['(','[','{','<']
def hnbhai():
s=sa()
stck=[]
tot=0
for i in s:
if i in op:
stck.append(i)
elif len(stck)==0:
print("Impossible")
return
else:
if d[i]!=stck[-1]:
tot+=1
stck.pop()
if len(stck)==0:
print(tot)
else:
print("Impossible")
for _ in range(1):
hnbhai()
``` | output | 1 | 6,228 | 21 | 12,457 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given string s consists of opening and closing brackets of four kinds <>, {}, [], (). There are two types of brackets: opening and closing. You can replace any bracket by another of the same type. For example, you can replace < by the bracket {, but you can't replace it by ) or >.
The following definition of a regular bracket sequence is well-known, so you can be familiar with it.
Let's define a regular bracket sequence (RBS). Empty string is RBS. Let s1 and s2 be a RBS then the strings <s1>s2, {s1}s2, [s1]s2, (s1)s2 are also RBS.
For example the string "[[(){}]<>]" is RBS, but the strings "[)()" and "][()()" are not.
Determine the least number of replaces to make the string s RBS.
Input
The only line contains a non empty string s, consisting of only opening and closing brackets of four kinds. The length of s does not exceed 106.
Output
If it's impossible to get RBS from s print Impossible.
Otherwise print the least number of replaces needed to get RBS from s.
Examples
Input
[<}){}
Output
2
Input
{()}[]
Output
0
Input
]]
Output
Impossible
Submitted Solution:
```
st = input()
c = 0
b = r = s = l = 0
for i in st:
if i in [ '[' , '<' , '{' , '(' ]:
c += 1
else:
c -= 1
if c < 0:
break
ans = 0
if c != 0:
print('Impossible')
else:
stack = []
for i in st:
if i in [ '[' , '<' , '{' , '(' ]:
stack.append(i)
else:
if stack[-1] == '(' and i == ')':
stack.pop()
continue
elif stack[-1] == '<' and i == '>':
stack.pop()
continue
elif stack[-1] == '{' and i == '}':
stack.pop()
continue
elif stack[-1] == '[' and i == ']':
stack.pop()
continue
else:
stack.pop()
ans += 1
print(ans)
``` | instruction | 0 | 6,229 | 21 | 12,458 |
Yes | output | 1 | 6,229 | 21 | 12,459 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given string s consists of opening and closing brackets of four kinds <>, {}, [], (). There are two types of brackets: opening and closing. You can replace any bracket by another of the same type. For example, you can replace < by the bracket {, but you can't replace it by ) or >.
The following definition of a regular bracket sequence is well-known, so you can be familiar with it.
Let's define a regular bracket sequence (RBS). Empty string is RBS. Let s1 and s2 be a RBS then the strings <s1>s2, {s1}s2, [s1]s2, (s1)s2 are also RBS.
For example the string "[[(){}]<>]" is RBS, but the strings "[)()" and "][()()" are not.
Determine the least number of replaces to make the string s RBS.
Input
The only line contains a non empty string s, consisting of only opening and closing brackets of four kinds. The length of s does not exceed 106.
Output
If it's impossible to get RBS from s print Impossible.
Otherwise print the least number of replaces needed to get RBS from s.
Examples
Input
[<}){}
Output
2
Input
{()}[]
Output
0
Input
]]
Output
Impossible
Submitted Solution:
```
s = input()
cnt = 0
st = []
for elem in s:
if elem in '([{<':
st.append(elem)
else:
if len(st) == 0:
print('Impossible')
break
elem2 = st.pop()
if elem2 + elem not in '()[]{}<>':
cnt += 1
else:
if len(st) == 0:
print(cnt)
else:
print('Impossible')
``` | instruction | 0 | 6,230 | 21 | 12,460 |
Yes | output | 1 | 6,230 | 21 | 12,461 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given string s consists of opening and closing brackets of four kinds <>, {}, [], (). There are two types of brackets: opening and closing. You can replace any bracket by another of the same type. For example, you can replace < by the bracket {, but you can't replace it by ) or >.
The following definition of a regular bracket sequence is well-known, so you can be familiar with it.
Let's define a regular bracket sequence (RBS). Empty string is RBS. Let s1 and s2 be a RBS then the strings <s1>s2, {s1}s2, [s1]s2, (s1)s2 are also RBS.
For example the string "[[(){}]<>]" is RBS, but the strings "[)()" and "][()()" are not.
Determine the least number of replaces to make the string s RBS.
Input
The only line contains a non empty string s, consisting of only opening and closing brackets of four kinds. The length of s does not exceed 106.
Output
If it's impossible to get RBS from s print Impossible.
Otherwise print the least number of replaces needed to get RBS from s.
Examples
Input
[<}){}
Output
2
Input
{()}[]
Output
0
Input
]]
Output
Impossible
Submitted Solution:
```
'''
Replace To Make Regular Bracket Sequence
'''
S = input()
stack = []
count = 0
length = len(S)
flag = 0
if(length % 2):
flag = 1
else:
for i in range(length):
if S[i] =='<' or S[i] =='(' or S[i] =='{' or S[i] =='[':
stack.append(S[i])
elif stack != []:
#print(S[i])
if S[i] == '>' and stack.pop() != '<':
count += 1
elif S[i] == ')' and stack.pop() != '(':
count += 1
elif S[i] == '}' and stack.pop() != '{':
count += 1
elif S[i] == ']' and stack.pop() != '[':
count += 1
if(flag):
break
else:
flag = 1
break
if flag != 0 or len(stack) != 0:
print("Impossible")
else:
print(count)
``` | instruction | 0 | 6,231 | 21 | 12,462 |
Yes | output | 1 | 6,231 | 21 | 12,463 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given string s consists of opening and closing brackets of four kinds <>, {}, [], (). There are two types of brackets: opening and closing. You can replace any bracket by another of the same type. For example, you can replace < by the bracket {, but you can't replace it by ) or >.
The following definition of a regular bracket sequence is well-known, so you can be familiar with it.
Let's define a regular bracket sequence (RBS). Empty string is RBS. Let s1 and s2 be a RBS then the strings <s1>s2, {s1}s2, [s1]s2, (s1)s2 are also RBS.
For example the string "[[(){}]<>]" is RBS, but the strings "[)()" and "][()()" are not.
Determine the least number of replaces to make the string s RBS.
Input
The only line contains a non empty string s, consisting of only opening and closing brackets of four kinds. The length of s does not exceed 106.
Output
If it's impossible to get RBS from s print Impossible.
Otherwise print the least number of replaces needed to get RBS from s.
Examples
Input
[<}){}
Output
2
Input
{()}[]
Output
0
Input
]]
Output
Impossible
Submitted 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 *
input=stdin.readline
prin=stdout.write
from random import sample
from collections import Counter,deque
from math import sqrt,ceil,log2,gcd
#dist=[0]*(n+1)
mod=10**9+7
class DisjSet:
def __init__(self, n):
# Constructor to create and
# initialize sets of n items
self.rank = [1] * n
self.parent = [i for i in range(n)]
# Finds set of given item x
def find(self, x):
# Finds the representative of the set
# that x is an element of
if (self.parent[x] != x):
# if x is not the parent of itself
# Then x is not the representative of
# its set,
self.parent[x] = self.find(self.parent[x])
# so we recursively call Find on its parent
# and move i's node directly under the
# representative of this set
return self.parent[x]
# Do union of two sets represented
# by x and y.
def union(self, x, y):
# Find current sets of x and y
xset = self.find(x)
yset = self.find(y)
# If they are already in same set
if xset == yset:
return
# Put smaller ranked item under
# bigger ranked item if ranks are
# different
if self.rank[xset] < self.rank[yset]:
self.parent[xset] = yset
elif self.rank[xset] > self.rank[yset]:
self.parent[yset] = xset
# If ranks are same, then move y under
# x (doesn't matter which one goes where)
# and increment rank of x's tree
else:
self.parent[yset] = xset
self.rank[xset] = self.rank[xset] + 1
# Driver code
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=[];countprev=0;
while n%2==0:
n=n//2
lk+=1
for ps in range(3,ceil(sqrt(n))+1,2):
lk=0
while n%ps==0:
n=n//ps
lk+=1
if n!=1:
lk+=1
return [lk,"NO"]
return [lk,"YES"]
#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 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 outstrlist(array:list)->str:
array=[str(x) for x in array]
return ' '.join(array);
def islist():
return list(map(str,input().split().rstrip()))
def outfast(arr:list)->str:
ss=''
for ip in arr:
ss+=str(ip)+' '
return prin(ss);
###-------------------------CODE STARTS HERE--------------------------------###########
#########################################################################################
#t=int(input())
t=1
for jj in range(t):
aa=input().strip()
kk=[]
ss={'<':0,'{':1,'[':2,'(':3,'>':4,'}':5,']':6,')':7}
cc=0;bb=0
for i in aa:
if ss[i]<4:
kk.append(ss[i])
else:
if len(kk)>0:
if ss[i]-kk[-1]!=4:
cc+=1
kk.pop();
else:
bb=1;break
if bb==1 or len(kk)>0:
print("Impossible")
else:
print(cc)
``` | instruction | 0 | 6,232 | 21 | 12,464 |
Yes | output | 1 | 6,232 | 21 | 12,465 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given string s consists of opening and closing brackets of four kinds <>, {}, [], (). There are two types of brackets: opening and closing. You can replace any bracket by another of the same type. For example, you can replace < by the bracket {, but you can't replace it by ) or >.
The following definition of a regular bracket sequence is well-known, so you can be familiar with it.
Let's define a regular bracket sequence (RBS). Empty string is RBS. Let s1 and s2 be a RBS then the strings <s1>s2, {s1}s2, [s1]s2, (s1)s2 are also RBS.
For example the string "[[(){}]<>]" is RBS, but the strings "[)()" and "][()()" are not.
Determine the least number of replaces to make the string s RBS.
Input
The only line contains a non empty string s, consisting of only opening and closing brackets of four kinds. The length of s does not exceed 106.
Output
If it's impossible to get RBS from s print Impossible.
Otherwise print the least number of replaces needed to get RBS from s.
Examples
Input
[<}){}
Output
2
Input
{()}[]
Output
0
Input
]]
Output
Impossible
Submitted Solution:
```
from sys import stdin, stdout
class SOLVE:
def solve(self):
R = stdin.readline
#f = open('input.txt');R = f.readline
W = stdout.write
s = R()[:-1]
brackets, cnt = [], 0
for i in range(len(s)):
if s[i] in ['(', '<', '{', '[']:
brackets.append(s[i])
else:
if len(brackets) == 0:
W("Impossible\n")
return 0
if s[i] == ')':
if brackets[-1] != '(':
cnt += 1
elif s[i] == '>':
if brackets[-1] != '<':
cnt += 1
elif s[i] == '}':
if brackets[-1] != '{':
cnt += 1
elif s[i] == ']':
if brackets[-1] != '[':
cnt += 1
brackets.pop()
W('%d\n' % cnt)
return 0
def main():
s = SOLVE()
s.solve()
main()
``` | instruction | 0 | 6,233 | 21 | 12,466 |
No | output | 1 | 6,233 | 21 | 12,467 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given string s consists of opening and closing brackets of four kinds <>, {}, [], (). There are two types of brackets: opening and closing. You can replace any bracket by another of the same type. For example, you can replace < by the bracket {, but you can't replace it by ) or >.
The following definition of a regular bracket sequence is well-known, so you can be familiar with it.
Let's define a regular bracket sequence (RBS). Empty string is RBS. Let s1 and s2 be a RBS then the strings <s1>s2, {s1}s2, [s1]s2, (s1)s2 are also RBS.
For example the string "[[(){}]<>]" is RBS, but the strings "[)()" and "][()()" are not.
Determine the least number of replaces to make the string s RBS.
Input
The only line contains a non empty string s, consisting of only opening and closing brackets of four kinds. The length of s does not exceed 106.
Output
If it's impossible to get RBS from s print Impossible.
Otherwise print the least number of replaces needed to get RBS from s.
Examples
Input
[<}){}
Output
2
Input
{()}[]
Output
0
Input
]]
Output
Impossible
Submitted Solution:
```
s=list(input())
o,cl=0,0
a,b,c,d=0,0,0,0
for i in s:
if i=="(":
a+=1
o+=1
elif i==")":
a-=1
cl+=1
elif i=="{":
b+=1
o+=1
elif i=="}":
b-=1
cl+=1
elif i=="[":
c+=1
o+=1
elif i=="]":
c-=1
cl+=1
elif i=="<":
d+=1
o+=1
else:
d-=1
cl+=1
z=abs(a)+abs(b)+abs(c)+abs(d)
if o==cl and z==0:
print(0)
elif o==cl and z!=0:
print(z//2)
elif o!=c:
print("impossible")
``` | instruction | 0 | 6,234 | 21 | 12,468 |
No | output | 1 | 6,234 | 21 | 12,469 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given string s consists of opening and closing brackets of four kinds <>, {}, [], (). There are two types of brackets: opening and closing. You can replace any bracket by another of the same type. For example, you can replace < by the bracket {, but you can't replace it by ) or >.
The following definition of a regular bracket sequence is well-known, so you can be familiar with it.
Let's define a regular bracket sequence (RBS). Empty string is RBS. Let s1 and s2 be a RBS then the strings <s1>s2, {s1}s2, [s1]s2, (s1)s2 are also RBS.
For example the string "[[(){}]<>]" is RBS, but the strings "[)()" and "][()()" are not.
Determine the least number of replaces to make the string s RBS.
Input
The only line contains a non empty string s, consisting of only opening and closing brackets of four kinds. The length of s does not exceed 106.
Output
If it's impossible to get RBS from s print Impossible.
Otherwise print the least number of replaces needed to get RBS from s.
Examples
Input
[<}){}
Output
2
Input
{()}[]
Output
0
Input
]]
Output
Impossible
Submitted Solution:
```
s = input()
brackets = {
'<': '>',
'{': '}',
'[': ']',
'(': ')',
}
OPENING = "<{[("
CLOSING = ">}])"
def main(s):
stack = []
res = 0
for c in s:
if c in OPENING:
stack.append(c)
else:
if stack:
top = stack.pop()
if top in OPENING:
if brackets[top] != c:
res += 1
continue
return 'Impossible'
return res
print(main(s))
``` | instruction | 0 | 6,235 | 21 | 12,470 |
No | output | 1 | 6,235 | 21 | 12,471 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given string s consists of opening and closing brackets of four kinds <>, {}, [], (). There are two types of brackets: opening and closing. You can replace any bracket by another of the same type. For example, you can replace < by the bracket {, but you can't replace it by ) or >.
The following definition of a regular bracket sequence is well-known, so you can be familiar with it.
Let's define a regular bracket sequence (RBS). Empty string is RBS. Let s1 and s2 be a RBS then the strings <s1>s2, {s1}s2, [s1]s2, (s1)s2 are also RBS.
For example the string "[[(){}]<>]" is RBS, but the strings "[)()" and "][()()" are not.
Determine the least number of replaces to make the string s RBS.
Input
The only line contains a non empty string s, consisting of only opening and closing brackets of four kinds. The length of s does not exceed 106.
Output
If it's impossible to get RBS from s print Impossible.
Otherwise print the least number of replaces needed to get RBS from s.
Examples
Input
[<}){}
Output
2
Input
{()}[]
Output
0
Input
]]
Output
Impossible
Submitted Solution:
```
import sys
s = input()
stack = []
ans = 0
brackets = {
'>': '<',
'}': '{',
']': '[',
')': '('
}
for c in s:
if c in '>}])':
if not stack:
print('Impossible')
exit()
if stack[-1] != brackets[c]:
ans += 1
stack.pop()
else:
stack.append(c)
print(ans)
``` | instruction | 0 | 6,236 | 21 | 12,472 |
No | output | 1 | 6,236 | 21 | 12,473 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarpus has a finite sequence of opening and closing brackets. In order not to fall asleep in a lecture, Polycarpus is having fun with his sequence. He is able to perform two operations:
* adding any bracket in any position (in the beginning, the end, or between any two existing brackets);
* cyclic shift — moving the last bracket from the end of the sequence to the beginning.
Polycarpus can apply any number of operations to his sequence and adding a cyclic shift in any order. As a result, he wants to get the correct bracket sequence of the minimum possible length. If there are several such sequences, Polycarpus is interested in the lexicographically smallest one. Help him find such a sequence.
Acorrect bracket sequence is a sequence of opening and closing brackets, from which you can get a correct arithmetic expression by adding characters "1" and "+" . Each opening bracket must correspond to a closed one. For example, the sequences "(())()", "()", "(()(()))" are correct and ")(", "(()" and "(()))(" are not.
The sequence a1 a2... an is lexicographically smaller than sequence b1 b2... bn, if there is such number i from 1 to n, thatak = bk for 1 ≤ k < i and ai < bi. Consider that "(" < ")".
Input
The first line contains Polycarpus's sequence consisting of characters "(" and ")". The length of a line is from 1 to 1 000 000.
Output
Print a correct bracket sequence of the minimum length that Polycarpus can obtain by his operations. If there are multiple such sequences, print the lexicographically minimum one.
Examples
Input
()(())
Output
(())()
Input
()(
Output
(())
Note
The sequence in the first example is already correct, but to get the lexicographically minimum answer, you need to perform four cyclic shift operations. In the second example you need to add a closing parenthesis between the second and third brackets and make a cyclic shift. You can first make the shift, and then add the bracket at the end.
Submitted Solution:
```
import sys
line = sys.stdin.readline()
counter = 0
max_counter = 0
longest_seq_end = 0
if line[0] == '(':
i = 1
while line[i] == '(':
i += 1
line = line[i:-1] + line[:i]
for i in range(1, len(line)):
if line[i] == '(':
counter += 1
else:
if counter > max_counter:
longest_seq_end = i
max_counter = counter
counter = 0
elif counter == max_counter:
for j in range(1, len(line) - 1):
if line[(longest_seq_end + j) % len(line)] == '(' and line[(i + j) % len(line)] == ')':
break
elif line[(longest_seq_end + j) % len(line)] == ')' and line[(i + j) % len(line)] == '(':
longest_seq_end = i
line = line[longest_seq_end - max_counter:longest_seq_end] + line[longest_seq_end:] + line[:longest_seq_end - max_counter]
addition = str()
for i in range(1, len(line) - line.count('(')):
addition += ')'
line = line + addition
else:
addition = str()
for i in range(0, len(line)):
addition += '('
line = addition + line
print(line)
``` | instruction | 0 | 7,074 | 21 | 14,148 |
No | output | 1 | 7,074 | 21 | 14,149 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarpus has a finite sequence of opening and closing brackets. In order not to fall asleep in a lecture, Polycarpus is having fun with his sequence. He is able to perform two operations:
* adding any bracket in any position (in the beginning, the end, or between any two existing brackets);
* cyclic shift — moving the last bracket from the end of the sequence to the beginning.
Polycarpus can apply any number of operations to his sequence and adding a cyclic shift in any order. As a result, he wants to get the correct bracket sequence of the minimum possible length. If there are several such sequences, Polycarpus is interested in the lexicographically smallest one. Help him find such a sequence.
Acorrect bracket sequence is a sequence of opening and closing brackets, from which you can get a correct arithmetic expression by adding characters "1" and "+" . Each opening bracket must correspond to a closed one. For example, the sequences "(())()", "()", "(()(()))" are correct and ")(", "(()" and "(()))(" are not.
The sequence a1 a2... an is lexicographically smaller than sequence b1 b2... bn, if there is such number i from 1 to n, thatak = bk for 1 ≤ k < i and ai < bi. Consider that "(" < ")".
Input
The first line contains Polycarpus's sequence consisting of characters "(" and ")". The length of a line is from 1 to 1 000 000.
Output
Print a correct bracket sequence of the minimum length that Polycarpus can obtain by his operations. If there are multiple such sequences, print the lexicographically minimum one.
Examples
Input
()(())
Output
(())()
Input
()(
Output
(())
Note
The sequence in the first example is already correct, but to get the lexicographically minimum answer, you need to perform four cyclic shift operations. In the second example you need to add a closing parenthesis between the second and third brackets and make a cyclic shift. You can first make the shift, and then add the bracket at the end.
Submitted Solution:
```
def check(s):
level = 0
for i in s:
if i == '(':
level+=1
if i == ')':
level-=1
if level<0:
return level
return level
def moved(s, times = 1):
buff = ''
n = 0
for i in range(-times, len(s)-times):
buff+=s[i]
return buff
def number(s):
n = 0
for i in s:
if i == '(':
n+=1
else:
return n
s = input()
if not check(s) == 0:
#print('Inn')
if check(s)>0:
for i in range(check(s)):
s=s+')'
if check(s)<0:
for i in range(check(s), 0):
s=s+'('
#print(s)
arr = []
for i in range(len(s)-1):
string = moved(s, i)
if check(string):
arr.append(number(string))
else:
arr.append(-300)
m = 0
b = 0
for i in arr:
if i > b:
b = arr.index(i)
if i < m:
m = arr.index(i)
print(moved(s, b))
``` | instruction | 0 | 7,075 | 21 | 14,150 |
No | output | 1 | 7,075 | 21 | 14,151 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarpus has a finite sequence of opening and closing brackets. In order not to fall asleep in a lecture, Polycarpus is having fun with his sequence. He is able to perform two operations:
* adding any bracket in any position (in the beginning, the end, or between any two existing brackets);
* cyclic shift — moving the last bracket from the end of the sequence to the beginning.
Polycarpus can apply any number of operations to his sequence and adding a cyclic shift in any order. As a result, he wants to get the correct bracket sequence of the minimum possible length. If there are several such sequences, Polycarpus is interested in the lexicographically smallest one. Help him find such a sequence.
Acorrect bracket sequence is a sequence of opening and closing brackets, from which you can get a correct arithmetic expression by adding characters "1" and "+" . Each opening bracket must correspond to a closed one. For example, the sequences "(())()", "()", "(()(()))" are correct and ")(", "(()" and "(()))(" are not.
The sequence a1 a2... an is lexicographically smaller than sequence b1 b2... bn, if there is such number i from 1 to n, thatak = bk for 1 ≤ k < i and ai < bi. Consider that "(" < ")".
Input
The first line contains Polycarpus's sequence consisting of characters "(" and ")". The length of a line is from 1 to 1 000 000.
Output
Print a correct bracket sequence of the minimum length that Polycarpus can obtain by his operations. If there are multiple such sequences, print the lexicographically minimum one.
Examples
Input
()(())
Output
(())()
Input
()(
Output
(())
Note
The sequence in the first example is already correct, but to get the lexicographically minimum answer, you need to perform four cyclic shift operations. In the second example you need to add a closing parenthesis between the second and third brackets and make a cyclic shift. You can first make the shift, and then add the bracket at the end.
Submitted Solution:
```
123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123
``` | instruction | 0 | 7,076 | 21 | 14,152 |
No | output | 1 | 7,076 | 21 | 14,153 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarpus has a finite sequence of opening and closing brackets. In order not to fall asleep in a lecture, Polycarpus is having fun with his sequence. He is able to perform two operations:
* adding any bracket in any position (in the beginning, the end, or between any two existing brackets);
* cyclic shift — moving the last bracket from the end of the sequence to the beginning.
Polycarpus can apply any number of operations to his sequence and adding a cyclic shift in any order. As a result, he wants to get the correct bracket sequence of the minimum possible length. If there are several such sequences, Polycarpus is interested in the lexicographically smallest one. Help him find such a sequence.
Acorrect bracket sequence is a sequence of opening and closing brackets, from which you can get a correct arithmetic expression by adding characters "1" and "+" . Each opening bracket must correspond to a closed one. For example, the sequences "(())()", "()", "(()(()))" are correct and ")(", "(()" and "(()))(" are not.
The sequence a1 a2... an is lexicographically smaller than sequence b1 b2... bn, if there is such number i from 1 to n, thatak = bk for 1 ≤ k < i and ai < bi. Consider that "(" < ")".
Input
The first line contains Polycarpus's sequence consisting of characters "(" and ")". The length of a line is from 1 to 1 000 000.
Output
Print a correct bracket sequence of the minimum length that Polycarpus can obtain by his operations. If there are multiple such sequences, print the lexicographically minimum one.
Examples
Input
()(())
Output
(())()
Input
()(
Output
(())
Note
The sequence in the first example is already correct, but to get the lexicographically minimum answer, you need to perform four cyclic shift operations. In the second example you need to add a closing parenthesis between the second and third brackets and make a cyclic shift. You can first make the shift, and then add the bracket at the end.
Submitted Solution:
```
per = input()
t = len(per)
hach = 0
cou = 0
for i in range(t-1,-1,-1):
if per[i] == '(':
hach += 2**cou
cou +=1
else:
cou+=1
answer = hach
for i in range(t-1,-1,-1):
s = (hach - (0 if per[i] == ')' else 1))//2
s += (0 if per[i] == ')' else 2**(t-1))
answer = max(answer, s)
hach = s
c = bin(answer)
cou = 0
for i in range(t):
if per[i] == '(':
cou -=1
else:
cou +=1
if cou >= 0:
print(cou * '(' , end = '')
for i in range(2,t+2):
if c[i] == '0':
print(')', end = '')
else:
print('(', end = '')
if cou < 0:
print(-cou * ')')
``` | instruction | 0 | 7,077 | 21 | 14,154 |
No | output | 1 | 7,077 | 21 | 14,155 |
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