message stringlengths 2 15.4k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 45 107k | cluster float64 21 21 | __index_level_0__ int64 90 214k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya's friends made him a birthday present — a bracket sequence. Petya was quite disappointed with his gift, because he dreamed of correct bracket sequence, yet he told his friends nothing about his dreams and decided to fix present himself.
To make everything right, Petya is going to move at most one bracket from its original place in the sequence to any other position. Reversing the bracket (e.g. turning "(" into ")" or vice versa) isn't allowed.
We remind that bracket sequence s is called correct if:
* s is empty;
* s is equal to "(t)", where t is correct bracket sequence;
* s is equal to t_1 t_2, i.e. concatenation of t_1 and t_2, where t_1 and t_2 are correct bracket sequences.
For example, "(()())", "()" are correct, while ")(" and "())" are not. Help Petya to fix his birthday present and understand whether he can move one bracket so that the sequence becomes correct.
Input
First of line of input contains a single number n (1 ≤ n ≤ 200 000) — length of the sequence which Petya received for his birthday.
Second line of the input contains bracket sequence of length n, containing symbols "(" and ")".
Output
Print "Yes" if Petya can make his sequence correct moving at most one bracket. Otherwise print "No".
Examples
Input
2
)(
Output
Yes
Input
3
(()
Output
No
Input
2
()
Output
Yes
Input
10
)))))(((((
Output
No
Note
In the first example, Petya can move first bracket to the end, thus turning the sequence into "()", which is correct bracket sequence.
In the second example, there is no way to move at most one bracket so that the sequence becomes correct.
In the third example, the sequence is already correct and there's no need to move brackets. | instruction | 0 | 30,150 | 21 | 60,300 |
Tags: data structures, greedy
Correct Solution:
```
ll=lambda:map(int,input().split())
t=lambda:int(input())
ss=lambda:input()
#from math import log10 ,log2,ceil,factorial as f,gcd
#from itertools import combinations_with_replacement as cs
#from functools import reduce
#from bisect import bisect_right as br,bisect_left as bl
from collections import Counter
#for _ in range(t()):
n=t()
a=list(ss())
if len(a)%2==0 :
x=0
j=n-a[::-1].index("(")-1
flag=0
#print(j,a[j])
for i in range(n):
if a[i]=="(":
x+=1
else:
if x>0:
x-=1
else:
if i<j and flag==0:
a.insert(i,"(")
a.pop(j+1)
x+=1
flag=1
else:
x-=1
break
#print(a)
if x==0:
print("YES")
else:
print("NO")
else:
print("NO")
``` | output | 1 | 30,150 | 21 | 60,301 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya's friends made him a birthday present — a bracket sequence. Petya was quite disappointed with his gift, because he dreamed of correct bracket sequence, yet he told his friends nothing about his dreams and decided to fix present himself.
To make everything right, Petya is going to move at most one bracket from its original place in the sequence to any other position. Reversing the bracket (e.g. turning "(" into ")" or vice versa) isn't allowed.
We remind that bracket sequence s is called correct if:
* s is empty;
* s is equal to "(t)", where t is correct bracket sequence;
* s is equal to t_1 t_2, i.e. concatenation of t_1 and t_2, where t_1 and t_2 are correct bracket sequences.
For example, "(()())", "()" are correct, while ")(" and "())" are not. Help Petya to fix his birthday present and understand whether he can move one bracket so that the sequence becomes correct.
Input
First of line of input contains a single number n (1 ≤ n ≤ 200 000) — length of the sequence which Petya received for his birthday.
Second line of the input contains bracket sequence of length n, containing symbols "(" and ")".
Output
Print "Yes" if Petya can make his sequence correct moving at most one bracket. Otherwise print "No".
Examples
Input
2
)(
Output
Yes
Input
3
(()
Output
No
Input
2
()
Output
Yes
Input
10
)))))(((((
Output
No
Note
In the first example, Petya can move first bracket to the end, thus turning the sequence into "()", which is correct bracket sequence.
In the second example, there is no way to move at most one bracket so that the sequence becomes correct.
In the third example, the sequence is already correct and there's no need to move brackets. | instruction | 0 | 30,151 | 21 | 60,302 |
Tags: data structures, greedy
Correct Solution:
```
def solve(s,n):
if len(s)%2:
return "NO"
stack = []
open = s.count("(")
close = s.count(")")
if open != close:
return "NO"
o,c = 0,0
for i in s:
if i == "(":
o += 1
else:
c += 1
#print(c,o)
if c-o>1:
return "NO"
return "YES"
n = int(input())
s = input()
print(solve(s,n))
``` | output | 1 | 30,151 | 21 | 60,303 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya's friends made him a birthday present — a bracket sequence. Petya was quite disappointed with his gift, because he dreamed of correct bracket sequence, yet he told his friends nothing about his dreams and decided to fix present himself.
To make everything right, Petya is going to move at most one bracket from its original place in the sequence to any other position. Reversing the bracket (e.g. turning "(" into ")" or vice versa) isn't allowed.
We remind that bracket sequence s is called correct if:
* s is empty;
* s is equal to "(t)", where t is correct bracket sequence;
* s is equal to t_1 t_2, i.e. concatenation of t_1 and t_2, where t_1 and t_2 are correct bracket sequences.
For example, "(()())", "()" are correct, while ")(" and "())" are not. Help Petya to fix his birthday present and understand whether he can move one bracket so that the sequence becomes correct.
Input
First of line of input contains a single number n (1 ≤ n ≤ 200 000) — length of the sequence which Petya received for his birthday.
Second line of the input contains bracket sequence of length n, containing symbols "(" and ")".
Output
Print "Yes" if Petya can make his sequence correct moving at most one bracket. Otherwise print "No".
Examples
Input
2
)(
Output
Yes
Input
3
(()
Output
No
Input
2
()
Output
Yes
Input
10
)))))(((((
Output
No
Note
In the first example, Petya can move first bracket to the end, thus turning the sequence into "()", which is correct bracket sequence.
In the second example, there is no way to move at most one bracket so that the sequence becomes correct.
In the third example, the sequence is already correct and there's no need to move brackets. | instruction | 0 | 30,152 | 21 | 60,304 |
Tags: data structures, greedy
Correct Solution:
```
import sys
import math
input = sys.stdin.readline
n=int(input())
s=list(input())
if s.count("(")==s.count(")"):
op=0
cl=0
flag=True
for i in range(n):
if s[i]=="(":
op+=1
else:
cl+=1
if cl-op>=2:
print("No")
flag=False
break
if flag:
print("Yes")
else:
print("No")
``` | output | 1 | 30,152 | 21 | 60,305 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya's friends made him a birthday present — a bracket sequence. Petya was quite disappointed with his gift, because he dreamed of correct bracket sequence, yet he told his friends nothing about his dreams and decided to fix present himself.
To make everything right, Petya is going to move at most one bracket from its original place in the sequence to any other position. Reversing the bracket (e.g. turning "(" into ")" or vice versa) isn't allowed.
We remind that bracket sequence s is called correct if:
* s is empty;
* s is equal to "(t)", where t is correct bracket sequence;
* s is equal to t_1 t_2, i.e. concatenation of t_1 and t_2, where t_1 and t_2 are correct bracket sequences.
For example, "(()())", "()" are correct, while ")(" and "())" are not. Help Petya to fix his birthday present and understand whether he can move one bracket so that the sequence becomes correct.
Input
First of line of input contains a single number n (1 ≤ n ≤ 200 000) — length of the sequence which Petya received for his birthday.
Second line of the input contains bracket sequence of length n, containing symbols "(" and ")".
Output
Print "Yes" if Petya can make his sequence correct moving at most one bracket. Otherwise print "No".
Examples
Input
2
)(
Output
Yes
Input
3
(()
Output
No
Input
2
()
Output
Yes
Input
10
)))))(((((
Output
No
Note
In the first example, Petya can move first bracket to the end, thus turning the sequence into "()", which is correct bracket sequence.
In the second example, there is no way to move at most one bracket so that the sequence becomes correct.
In the third example, the sequence is already correct and there's no need to move brackets.
Submitted Solution:
```
# ------------------- fast io --------------------
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")
# ------------------- fast io --------------------
from math import gcd, ceil
def pre(s):
n = len(s)
pi = [0] * n
for i in range(1, n):
j = pi[i - 1]
while j and s[i] != s[j]:
j = pi[j - 1]
if s[i] == s[j]:
j += 1
pi[i] = j
return pi
def prod(a):
ans = 1
for each in a:
ans = (ans * each)
return ans
def lcm(a, b): return a * b // gcd(a, b)
def binary(x, length=16):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
for _ in range(int(input()) if not True else 1):
n = int(input())
#n, k = map(int, input().split())
#a, b = map(int, input().split())
#c, d = map(int, input().split())
#a = list(map(int, input().split()))
#b = list(map(int, input().split()))
s = input()
if s.count("(") != s.count(")"):
print("No")
continue
total = 0
mt = 0
for i in s:
if i == '(':
total += 1
else:
total -= 1
mt = min(total, mt)
print("Yes" if mt >= -1 else "No")
``` | instruction | 0 | 30,153 | 21 | 60,306 |
Yes | output | 1 | 30,153 | 21 | 60,307 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya's friends made him a birthday present — a bracket sequence. Petya was quite disappointed with his gift, because he dreamed of correct bracket sequence, yet he told his friends nothing about his dreams and decided to fix present himself.
To make everything right, Petya is going to move at most one bracket from its original place in the sequence to any other position. Reversing the bracket (e.g. turning "(" into ")" or vice versa) isn't allowed.
We remind that bracket sequence s is called correct if:
* s is empty;
* s is equal to "(t)", where t is correct bracket sequence;
* s is equal to t_1 t_2, i.e. concatenation of t_1 and t_2, where t_1 and t_2 are correct bracket sequences.
For example, "(()())", "()" are correct, while ")(" and "())" are not. Help Petya to fix his birthday present and understand whether he can move one bracket so that the sequence becomes correct.
Input
First of line of input contains a single number n (1 ≤ n ≤ 200 000) — length of the sequence which Petya received for his birthday.
Second line of the input contains bracket sequence of length n, containing symbols "(" and ")".
Output
Print "Yes" if Petya can make his sequence correct moving at most one bracket. Otherwise print "No".
Examples
Input
2
)(
Output
Yes
Input
3
(()
Output
No
Input
2
()
Output
Yes
Input
10
)))))(((((
Output
No
Note
In the first example, Petya can move first bracket to the end, thus turning the sequence into "()", which is correct bracket sequence.
In the second example, there is no way to move at most one bracket so that the sequence becomes correct.
In the third example, the sequence is already correct and there's no need to move brackets.
Submitted Solution:
```
n=int(input())
s=input()
c=0
for i in s:
if i=="(":
c+=1
else:
c-=1
if c<-1:
print("NO")
exit(0)
print("YES" if c==0 else "NO")
``` | instruction | 0 | 30,154 | 21 | 60,308 |
Yes | output | 1 | 30,154 | 21 | 60,309 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya's friends made him a birthday present — a bracket sequence. Petya was quite disappointed with his gift, because he dreamed of correct bracket sequence, yet he told his friends nothing about his dreams and decided to fix present himself.
To make everything right, Petya is going to move at most one bracket from its original place in the sequence to any other position. Reversing the bracket (e.g. turning "(" into ")" or vice versa) isn't allowed.
We remind that bracket sequence s is called correct if:
* s is empty;
* s is equal to "(t)", where t is correct bracket sequence;
* s is equal to t_1 t_2, i.e. concatenation of t_1 and t_2, where t_1 and t_2 are correct bracket sequences.
For example, "(()())", "()" are correct, while ")(" and "())" are not. Help Petya to fix his birthday present and understand whether he can move one bracket so that the sequence becomes correct.
Input
First of line of input contains a single number n (1 ≤ n ≤ 200 000) — length of the sequence which Petya received for his birthday.
Second line of the input contains bracket sequence of length n, containing symbols "(" and ")".
Output
Print "Yes" if Petya can make his sequence correct moving at most one bracket. Otherwise print "No".
Examples
Input
2
)(
Output
Yes
Input
3
(()
Output
No
Input
2
()
Output
Yes
Input
10
)))))(((((
Output
No
Note
In the first example, Petya can move first bracket to the end, thus turning the sequence into "()", which is correct bracket sequence.
In the second example, there is no way to move at most one bracket so that the sequence becomes correct.
In the third example, the sequence is already correct and there's no need to move brackets.
Submitted Solution:
```
n=int(input())
s=input()
f=0
c=0
for i in range(n):
if s[i]==")":
c-=1
else:
c+=1
f=min(f,c)
if f>=-1 and c==0:
print("YES")
else:
print("NO")
``` | instruction | 0 | 30,155 | 21 | 60,310 |
Yes | output | 1 | 30,155 | 21 | 60,311 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya's friends made him a birthday present — a bracket sequence. Petya was quite disappointed with his gift, because he dreamed of correct bracket sequence, yet he told his friends nothing about his dreams and decided to fix present himself.
To make everything right, Petya is going to move at most one bracket from its original place in the sequence to any other position. Reversing the bracket (e.g. turning "(" into ")" or vice versa) isn't allowed.
We remind that bracket sequence s is called correct if:
* s is empty;
* s is equal to "(t)", where t is correct bracket sequence;
* s is equal to t_1 t_2, i.e. concatenation of t_1 and t_2, where t_1 and t_2 are correct bracket sequences.
For example, "(()())", "()" are correct, while ")(" and "())" are not. Help Petya to fix his birthday present and understand whether he can move one bracket so that the sequence becomes correct.
Input
First of line of input contains a single number n (1 ≤ n ≤ 200 000) — length of the sequence which Petya received for his birthday.
Second line of the input contains bracket sequence of length n, containing symbols "(" and ")".
Output
Print "Yes" if Petya can make his sequence correct moving at most one bracket. Otherwise print "No".
Examples
Input
2
)(
Output
Yes
Input
3
(()
Output
No
Input
2
()
Output
Yes
Input
10
)))))(((((
Output
No
Note
In the first example, Petya can move first bracket to the end, thus turning the sequence into "()", which is correct bracket sequence.
In the second example, there is no way to move at most one bracket so that the sequence becomes correct.
In the third example, the sequence is already correct and there's no need to move brackets.
Submitted Solution:
```
import sys, os, io
def rs(): return sys.stdin.readline().rstrip()
def ri(): return int(sys.stdin.readline())
def ria(): return list(map(int, sys.stdin.readline().split()))
def ws(s): sys.stdout.write(s + '\n')
def wi(n): sys.stdout.write(str(n) + '\n')
def wia(a): sys.stdout.write(' '.join([str(x) for x in a]) + '\n')
import math,datetime,functools
from collections import deque,defaultdict,OrderedDict
import collections
def main():
starttime=datetime.datetime.now()
if(os.path.exists('input.txt')):
sys.stdin = open("input.txt","r")
sys.stdout = open("output.txt","w")
#Solving Area Starts-->
for _ in range(1):
n=ri()
s=rs()
ans="No"
if n%2==0:
depth=0
t=0
ans="Yes"
for i in range(n):
if s[i]=="(":
depth+=1
else:
depth-=1
if depth<-1:
ans="No"
break
if depth>1:
ans="No"
print(ans)
#<--Solving Area Ends
endtime=datetime.datetime.now()
time=(endtime-starttime).total_seconds()*1000
if(os.path.exists('input.txt')):
print("Time:",time,"ms")
class FastReader(io.IOBase):
newlines = 0
def __init__(self, fd, chunk_size=1024 * 8):
self._fd = fd
self._chunk_size = chunk_size
self.buffer = io.BytesIO()
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, self._chunk_size))
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, size=-1):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, self._chunk_size if size == -1 else size))
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()
class FastWriter(io.IOBase):
def __init__(self, fd):
self._fd = fd
self.buffer = io.BytesIO()
self.write = self.buffer.write
def flush(self):
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class FastStdin(io.IOBase):
def __init__(self, fd=0):
self.buffer = FastReader(fd)
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
class FastStdout(io.IOBase):
def __init__(self, fd=1):
self.buffer = FastWriter(fd)
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.flush = self.buffer.flush
if __name__ == '__main__':
sys.stdin = FastStdin()
sys.stdout = FastStdout()
main()
``` | instruction | 0 | 30,156 | 21 | 60,312 |
Yes | output | 1 | 30,156 | 21 | 60,313 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya's friends made him a birthday present — a bracket sequence. Petya was quite disappointed with his gift, because he dreamed of correct bracket sequence, yet he told his friends nothing about his dreams and decided to fix present himself.
To make everything right, Petya is going to move at most one bracket from its original place in the sequence to any other position. Reversing the bracket (e.g. turning "(" into ")" or vice versa) isn't allowed.
We remind that bracket sequence s is called correct if:
* s is empty;
* s is equal to "(t)", where t is correct bracket sequence;
* s is equal to t_1 t_2, i.e. concatenation of t_1 and t_2, where t_1 and t_2 are correct bracket sequences.
For example, "(()())", "()" are correct, while ")(" and "())" are not. Help Petya to fix his birthday present and understand whether he can move one bracket so that the sequence becomes correct.
Input
First of line of input contains a single number n (1 ≤ n ≤ 200 000) — length of the sequence which Petya received for his birthday.
Second line of the input contains bracket sequence of length n, containing symbols "(" and ")".
Output
Print "Yes" if Petya can make his sequence correct moving at most one bracket. Otherwise print "No".
Examples
Input
2
)(
Output
Yes
Input
3
(()
Output
No
Input
2
()
Output
Yes
Input
10
)))))(((((
Output
No
Note
In the first example, Petya can move first bracket to the end, thus turning the sequence into "()", which is correct bracket sequence.
In the second example, there is no way to move at most one bracket so that the sequence becomes correct.
In the third example, the sequence is already correct and there's no need to move brackets.
Submitted Solution:
```
def mm(s):
t=[]
for i in range(len(s)):
if t==[]:
t+=s[i]
elif t[-1]=='(' and s[i]==')':
t.pop()
else:
t+=s[i]
if t==[]:
return True
return False
n=int(input())
s=input()
if n&1:
print("No")
exit(0)
if mm(s):
print("Yes")
else:
print("No")
``` | instruction | 0 | 30,157 | 21 | 60,314 |
No | output | 1 | 30,157 | 21 | 60,315 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya's friends made him a birthday present — a bracket sequence. Petya was quite disappointed with his gift, because he dreamed of correct bracket sequence, yet he told his friends nothing about his dreams and decided to fix present himself.
To make everything right, Petya is going to move at most one bracket from its original place in the sequence to any other position. Reversing the bracket (e.g. turning "(" into ")" or vice versa) isn't allowed.
We remind that bracket sequence s is called correct if:
* s is empty;
* s is equal to "(t)", where t is correct bracket sequence;
* s is equal to t_1 t_2, i.e. concatenation of t_1 and t_2, where t_1 and t_2 are correct bracket sequences.
For example, "(()())", "()" are correct, while ")(" and "())" are not. Help Petya to fix his birthday present and understand whether he can move one bracket so that the sequence becomes correct.
Input
First of line of input contains a single number n (1 ≤ n ≤ 200 000) — length of the sequence which Petya received for his birthday.
Second line of the input contains bracket sequence of length n, containing symbols "(" and ")".
Output
Print "Yes" if Petya can make his sequence correct moving at most one bracket. Otherwise print "No".
Examples
Input
2
)(
Output
Yes
Input
3
(()
Output
No
Input
2
()
Output
Yes
Input
10
)))))(((((
Output
No
Note
In the first example, Petya can move first bracket to the end, thus turning the sequence into "()", which is correct bracket sequence.
In the second example, there is no way to move at most one bracket so that the sequence becomes correct.
In the third example, the sequence is already correct and there's no need to move brackets.
Submitted Solution:
```
n=int(input())
seq=input()
if(n%2):
print("No")
else:
bal=0
err=0
for el in seq:
if el==')':
if(bal==-1 or err):
print("No")
exit()
bal-=1
err+=1
else:
bal+=1
print("Yes")
``` | instruction | 0 | 30,158 | 21 | 60,316 |
No | output | 1 | 30,158 | 21 | 60,317 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya's friends made him a birthday present — a bracket sequence. Petya was quite disappointed with his gift, because he dreamed of correct bracket sequence, yet he told his friends nothing about his dreams and decided to fix present himself.
To make everything right, Petya is going to move at most one bracket from its original place in the sequence to any other position. Reversing the bracket (e.g. turning "(" into ")" or vice versa) isn't allowed.
We remind that bracket sequence s is called correct if:
* s is empty;
* s is equal to "(t)", where t is correct bracket sequence;
* s is equal to t_1 t_2, i.e. concatenation of t_1 and t_2, where t_1 and t_2 are correct bracket sequences.
For example, "(()())", "()" are correct, while ")(" and "())" are not. Help Petya to fix his birthday present and understand whether he can move one bracket so that the sequence becomes correct.
Input
First of line of input contains a single number n (1 ≤ n ≤ 200 000) — length of the sequence which Petya received for his birthday.
Second line of the input contains bracket sequence of length n, containing symbols "(" and ")".
Output
Print "Yes" if Petya can make his sequence correct moving at most one bracket. Otherwise print "No".
Examples
Input
2
)(
Output
Yes
Input
3
(()
Output
No
Input
2
()
Output
Yes
Input
10
)))))(((((
Output
No
Note
In the first example, Petya can move first bracket to the end, thus turning the sequence into "()", which is correct bracket sequence.
In the second example, there is no way to move at most one bracket so that the sequence becomes correct.
In the third example, the sequence is already correct and there's no need to move brackets.
Submitted Solution:
```
n= int(input())
cadena = input()
suma = 0
incrementos = 0
for i in cadena:
if i =='(':
suma+=1
if i==')' and suma <=0:
incrementos+=1
if i==')':
suma-=1
if suma == 0 and incrementos <2:
print("YES")
else:
print("NO")
``` | instruction | 0 | 30,159 | 21 | 60,318 |
No | output | 1 | 30,159 | 21 | 60,319 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya's friends made him a birthday present — a bracket sequence. Petya was quite disappointed with his gift, because he dreamed of correct bracket sequence, yet he told his friends nothing about his dreams and decided to fix present himself.
To make everything right, Petya is going to move at most one bracket from its original place in the sequence to any other position. Reversing the bracket (e.g. turning "(" into ")" or vice versa) isn't allowed.
We remind that bracket sequence s is called correct if:
* s is empty;
* s is equal to "(t)", where t is correct bracket sequence;
* s is equal to t_1 t_2, i.e. concatenation of t_1 and t_2, where t_1 and t_2 are correct bracket sequences.
For example, "(()())", "()" are correct, while ")(" and "())" are not. Help Petya to fix his birthday present and understand whether he can move one bracket so that the sequence becomes correct.
Input
First of line of input contains a single number n (1 ≤ n ≤ 200 000) — length of the sequence which Petya received for his birthday.
Second line of the input contains bracket sequence of length n, containing symbols "(" and ")".
Output
Print "Yes" if Petya can make his sequence correct moving at most one bracket. Otherwise print "No".
Examples
Input
2
)(
Output
Yes
Input
3
(()
Output
No
Input
2
()
Output
Yes
Input
10
)))))(((((
Output
No
Note
In the first example, Petya can move first bracket to the end, thus turning the sequence into "()", which is correct bracket sequence.
In the second example, there is no way to move at most one bracket so that the sequence becomes correct.
In the third example, the sequence is already correct and there's no need to move brackets.
Submitted Solution:
```
import sys, collections, math, itertools, random, bisect
INF = sys.maxsize
def get_ints(): return map(int, sys.stdin.readline().strip().split())
def get_array(): return list(map(int, sys.stdin.readline().strip().split()))
def input(): return sys.stdin.readline().strip()
mod = 1000000007
def check(s):
res = 0
for i in range(len(s)):
if s[i]=='(':
res+=1
if s[i]==')':
res-=1
if res<0:
return 0
return (res>=-1) and (res<=1)
n = int(input())
s = input()
if n%2:
print('No')
else:
if check(s):
print('Yes')
else:
print('No')
``` | instruction | 0 | 30,160 | 21 | 60,320 |
No | output | 1 | 30,160 | 21 | 60,321 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters «+» and «1» into this sequence. For example, sequences «(())()», «()» and «(()(()))» are regular, while «)(», «(()» and «(()))(» are not.
One day Johnny got bracket sequence. He decided to remove some of the brackets from it in order to obtain a regular bracket sequence. What is the maximum length of a regular bracket sequence which can be obtained?
Input
Input consists of a single line with non-empty string of «(» and «)» characters. Its length does not exceed 106.
Output
Output the maximum possible length of a regular bracket sequence.
Examples
Input
(()))(
Output
4
Input
((()())
Output
6 | instruction | 0 | 30,370 | 21 | 60,740 |
Tags: greedy
Correct Solution:
```
#https://codeforces.com/contest/26/problem/B
s=input()
stack=[]
cnt=0
m=0
for i in s:
if i=='(':
stack.append(i)
elif i==')':
if len(stack)!=0:
stack.pop()
cnt+=2
#m=max(m,cnt)
print(cnt)
``` | output | 1 | 30,370 | 21 | 60,741 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters «+» and «1» into this sequence. For example, sequences «(())()», «()» and «(()(()))» are regular, while «)(», «(()» and «(()))(» are not.
One day Johnny got bracket sequence. He decided to remove some of the brackets from it in order to obtain a regular bracket sequence. What is the maximum length of a regular bracket sequence which can be obtained?
Input
Input consists of a single line with non-empty string of «(» and «)» characters. Its length does not exceed 106.
Output
Output the maximum possible length of a regular bracket sequence.
Examples
Input
(()))(
Output
4
Input
((()())
Output
6 | instruction | 0 | 30,371 | 21 | 60,742 |
Tags: greedy
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
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()
# for _ in range(int(ri())):
s = ri()
cnt = 0
ans = 0
for i in range(len(s)):
if s[i] == '(':
cnt+=1
elif s[i] == ')':
if cnt > 0:
cnt-=1
ans+=2
print(ans)
``` | output | 1 | 30,371 | 21 | 60,743 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters «+» and «1» into this sequence. For example, sequences «(())()», «()» and «(()(()))» are regular, while «)(», «(()» and «(()))(» are not.
One day Johnny got bracket sequence. He decided to remove some of the brackets from it in order to obtain a regular bracket sequence. What is the maximum length of a regular bracket sequence which can be obtained?
Input
Input consists of a single line with non-empty string of «(» and «)» characters. Its length does not exceed 106.
Output
Output the maximum possible length of a regular bracket sequence.
Examples
Input
(()))(
Output
4
Input
((()())
Output
6 | instruction | 0 | 30,372 | 21 | 60,744 |
Tags: greedy
Correct Solution:
```
from collections import deque
def solve(s):
d = deque()
n = len(s)
cnt = 0
for i in range(n):
if s[i] == '(':
d.append(s[i])
else:
if s[i] == ')':
if len(d) > 0 and d[-1] == '(':
d.pop()
cnt +=2
else:
continue
return d , cnt
s = input()
r = solve(s)
print(r[1])
``` | output | 1 | 30,372 | 21 | 60,745 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters «+» and «1» into this sequence. For example, sequences «(())()», «()» and «(()(()))» are regular, while «)(», «(()» and «(()))(» are not.
One day Johnny got bracket sequence. He decided to remove some of the brackets from it in order to obtain a regular bracket sequence. What is the maximum length of a regular bracket sequence which can be obtained?
Input
Input consists of a single line with non-empty string of «(» and «)» characters. Its length does not exceed 106.
Output
Output the maximum possible length of a regular bracket sequence.
Examples
Input
(()))(
Output
4
Input
((()())
Output
6 | instruction | 0 | 30,373 | 21 | 60,746 |
Tags: greedy
Correct Solution:
```
ans = cnt = 0
for c in input():
if c == '(':
cnt += 1
elif(c == ')'):
if cnt > 0:
cnt -= 1
ans += 2
print(ans)
``` | output | 1 | 30,373 | 21 | 60,747 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters «+» and «1» into this sequence. For example, sequences «(())()», «()» and «(()(()))» are regular, while «)(», «(()» and «(()))(» are not.
One day Johnny got bracket sequence. He decided to remove some of the brackets from it in order to obtain a regular bracket sequence. What is the maximum length of a regular bracket sequence which can be obtained?
Input
Input consists of a single line with non-empty string of «(» and «)» characters. Its length does not exceed 106.
Output
Output the maximum possible length of a regular bracket sequence.
Examples
Input
(()))(
Output
4
Input
((()())
Output
6 | instruction | 0 | 30,374 | 21 | 60,748 |
Tags: greedy
Correct Solution:
```
b = input()
i = 0
N = 0
x = 0
while(i < len(b)):
if(b[i] == '('):
x = x + 1
else:
x = x - 1
if(x >= 0):
N = N + 2
else:
x = 0
# print(b[i],x,N)
# input()
i = i + 1
print(N)
``` | output | 1 | 30,374 | 21 | 60,749 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters «+» and «1» into this sequence. For example, sequences «(())()», «()» and «(()(()))» are regular, while «)(», «(()» and «(()))(» are not.
One day Johnny got bracket sequence. He decided to remove some of the brackets from it in order to obtain a regular bracket sequence. What is the maximum length of a regular bracket sequence which can be obtained?
Input
Input consists of a single line with non-empty string of «(» and «)» characters. Its length does not exceed 106.
Output
Output the maximum possible length of a regular bracket sequence.
Examples
Input
(()))(
Output
4
Input
((()())
Output
6 | instruction | 0 | 30,375 | 21 | 60,750 |
Tags: greedy
Correct Solution:
```
ch=str(input())
j=0
par=[]
for i in range (len(ch)):
if ch[i]=='(':
par.append(ch[i])
else:
if ch[i]==')' and len(par)!=0:
par.pop()
j+=2
print(j)
``` | output | 1 | 30,375 | 21 | 60,751 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters «+» and «1» into this sequence. For example, sequences «(())()», «()» and «(()(()))» are regular, while «)(», «(()» and «(()))(» are not.
One day Johnny got bracket sequence. He decided to remove some of the brackets from it in order to obtain a regular bracket sequence. What is the maximum length of a regular bracket sequence which can be obtained?
Input
Input consists of a single line with non-empty string of «(» and «)» characters. Its length does not exceed 106.
Output
Output the maximum possible length of a regular bracket sequence.
Examples
Input
(()))(
Output
4
Input
((()())
Output
6 | instruction | 0 | 30,376 | 21 | 60,752 |
Tags: greedy
Correct Solution:
```
k=2
if k!=0:
seq = input()
count = 0
maxn = 0
for i in range(len(seq)):
if(seq[i] == "("):
count += 1
elif(count > 0):
count -= 1
maxn += 2
print(maxn)
``` | output | 1 | 30,376 | 21 | 60,753 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters «+» and «1» into this sequence. For example, sequences «(())()», «()» and «(()(()))» are regular, while «)(», «(()» and «(()))(» are not.
One day Johnny got bracket sequence. He decided to remove some of the brackets from it in order to obtain a regular bracket sequence. What is the maximum length of a regular bracket sequence which can be obtained?
Input
Input consists of a single line with non-empty string of «(» and «)» characters. Its length does not exceed 106.
Output
Output the maximum possible length of a regular bracket sequence.
Examples
Input
(()))(
Output
4
Input
((()())
Output
6 | instruction | 0 | 30,377 | 21 | 60,754 |
Tags: greedy
Correct Solution:
```
def check(n):
e = 0
b = 0
for i in n:
if i == '(':
b += 1
if i == ')' and b > 0:
e += 1
b -= 1
return e
n = input()
print(check(n)*2)
``` | output | 1 | 30,377 | 21 | 60,755 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters «+» and «1» into this sequence. For example, sequences «(())()», «()» and «(()(()))» are regular, while «)(», «(()» and «(()))(» are not.
One day Johnny got bracket sequence. He decided to remove some of the brackets from it in order to obtain a regular bracket sequence. What is the maximum length of a regular bracket sequence which can be obtained?
Input
Input consists of a single line with non-empty string of «(» and «)» characters. Its length does not exceed 106.
Output
Output the maximum possible length of a regular bracket sequence.
Examples
Input
(()))(
Output
4
Input
((()())
Output
6
Submitted Solution:
```
a,b = 0,0
for c in input():
if c=="(": b += 1
else: b-=1
if b<0: b=0
else: a+=1
print(a-b)
``` | instruction | 0 | 30,378 | 21 | 60,756 |
Yes | output | 1 | 30,378 | 21 | 60,757 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters «+» and «1» into this sequence. For example, sequences «(())()», «()» and «(()(()))» are regular, while «)(», «(()» and «(()))(» are not.
One day Johnny got bracket sequence. He decided to remove some of the brackets from it in order to obtain a regular bracket sequence. What is the maximum length of a regular bracket sequence which can be obtained?
Input
Input consists of a single line with non-empty string of «(» and «)» characters. Its length does not exceed 106.
Output
Output the maximum possible length of a regular bracket sequence.
Examples
Input
(()))(
Output
4
Input
((()())
Output
6
Submitted Solution:
```
s ,l ,r = input(), 0, 0
for i in s :
if i == '(' : l += 1
elif r < l : r += 1
print(2*r)
``` | instruction | 0 | 30,379 | 21 | 60,758 |
Yes | output | 1 | 30,379 | 21 | 60,759 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters «+» and «1» into this sequence. For example, sequences «(())()», «()» and «(()(()))» are regular, while «)(», «(()» and «(()))(» are not.
One day Johnny got bracket sequence. He decided to remove some of the brackets from it in order to obtain a regular bracket sequence. What is the maximum length of a regular bracket sequence which can be obtained?
Input
Input consists of a single line with non-empty string of «(» and «)» characters. Its length does not exceed 106.
Output
Output the maximum possible length of a regular bracket sequence.
Examples
Input
(()))(
Output
4
Input
((()())
Output
6
Submitted Solution:
```
line = list(input())
op = cl = i = 0
while i < len(line):
if line[i] == '(':
op+=1
elif cl != op:
cl +=1
i+=1
print(min(op, cl)*2)
``` | instruction | 0 | 30,380 | 21 | 60,760 |
Yes | output | 1 | 30,380 | 21 | 60,761 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters «+» and «1» into this sequence. For example, sequences «(())()», «()» and «(()(()))» are regular, while «)(», «(()» and «(()))(» are not.
One day Johnny got bracket sequence. He decided to remove some of the brackets from it in order to obtain a regular bracket sequence. What is the maximum length of a regular bracket sequence which can be obtained?
Input
Input consists of a single line with non-empty string of «(» and «)» characters. Its length does not exceed 106.
Output
Output the maximum possible length of a regular bracket sequence.
Examples
Input
(()))(
Output
4
Input
((()())
Output
6
Submitted Solution:
```
from collections import *
s, ans, l, r = input(), 0, 0, 0
arr = deque(s)
# print(arr)
for i in arr:
if i == '(':
l += 1
else:
if l > 0:
l-=1
ans+=2
print(ans)
``` | instruction | 0 | 30,381 | 21 | 60,762 |
Yes | output | 1 | 30,381 | 21 | 60,763 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters «+» and «1» into this sequence. For example, sequences «(())()», «()» and «(()(()))» are regular, while «)(», «(()» and «(()))(» are not.
One day Johnny got bracket sequence. He decided to remove some of the brackets from it in order to obtain a regular bracket sequence. What is the maximum length of a regular bracket sequence which can be obtained?
Input
Input consists of a single line with non-empty string of «(» and «)» characters. Its length does not exceed 106.
Output
Output the maximum possible length of a regular bracket sequence.
Examples
Input
(()))(
Output
4
Input
((()())
Output
6
Submitted Solution:
```
line = input()
open_b = 0
closed_b = 0
i = 0
good_lens = []
while closed_b <= open_b and i < len(line):
if line[i] == '(':
open_b += 1
else:
closed_b += 1
i += 1
if open_b == closed_b:
good_lens.append(i)
if len(good_lens) == 0:
print(min([open_b, closed_b])*2)
else:
print(max(good_lens))
``` | instruction | 0 | 30,382 | 21 | 60,764 |
No | output | 1 | 30,382 | 21 | 60,765 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters «+» and «1» into this sequence. For example, sequences «(())()», «()» and «(()(()))» are regular, while «)(», «(()» and «(()))(» are not.
One day Johnny got bracket sequence. He decided to remove some of the brackets from it in order to obtain a regular bracket sequence. What is the maximum length of a regular bracket sequence which can be obtained?
Input
Input consists of a single line with non-empty string of «(» and «)» characters. Its length does not exceed 106.
Output
Output the maximum possible length of a regular bracket sequence.
Examples
Input
(()))(
Output
4
Input
((()())
Output
6
Submitted Solution:
```
def Alfa(a):
n0 = a.find('()')
b = list(a)
N = 0
while(n0 != -1):
del b[n0],b[n0]
N = N + 1
a = ''.join(b)
n0 = a.find('()')
if(a == ''):
a = '|'
return[a,N]
a = input()
a = a[a.find("("):(len(a) - a[::-1].find(")")):1]
n = 0
N1 = 0
b1 = ''
N = []
b = []
#c = [0]
i = a.find(')(',n)
while(i != -1):
[b1,N1] = Alfa(a[n:i+1:1])
# c = c +[len(b1) + c[len(c)-1]]
N = N + [N1]
b = b + [b1]
n = i + 1
i = a.find(')(',n)
i = len(a)
[b1,N1] = Alfa(a[n:i:1])
N = N + [N1]
b = b + [b1]
#print(b,N)
#input()
a = ''.join(b)
while(a.find('()') != -1):
i = 0
while(i < (len(b) -1)):
m = 1
if((b[i].find('(') != -1)&(b[i+1].find(')') != -1)):
if(len(b[i]) > len(b[i+1])):
b[i] = '('*(len(b[i]) - len(b[i+1]))
# c[i] = len(b[i]) + c[len(c)-1]
N[i] = N[i] + N[i+1] + len(b[i+1])
del N[i+1]
del b[i+1]
elif(len(b[i]) < len(b[i+1])):
b[i] = ')'*(len(b[i+1]) - len(b[i]))
N[i] = N[i] + N[i+1] + len(b[i])
del N[i+1]
del b[i+1]
else:
N[i] = N[i] + N[i+1] + len(b[i])
b[i] = '|'
del N[i+1]
del b[i+1]
i = i - 1
i = i + 1
a = ''.join(b)
# print(b,N)
i = 0
while(i < (len(b) -1)):
if((b[i].find('(') != -1)&(b[i+1] == '|')):
j = i + 2
while(j < (len(b) -1)):
if(b[j] != '|'):
break
j = j + 1
if(b[j].find(')') != -1):
while(i < (j-1) ):
N[i] = N[i] + N[i+1]
del b[i+1]
del N[i+1]
i = i + 1
i = i + 1
# print(b,N)
# input()
a = ''.join(b)
i = 0
while(i < (len(b) -1)):
if((b[i] == '|')&(b[i+1] == '|')):
N[i] = N[i] + N[i+1]
del b[i+1]
del N[i+1]
i = i - 1
i = i + 1
print(sum(N)*2)
``` | instruction | 0 | 30,383 | 21 | 60,766 |
No | output | 1 | 30,383 | 21 | 60,767 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters «+» and «1» into this sequence. For example, sequences «(())()», «()» and «(()(()))» are regular, while «)(», «(()» and «(()))(» are not.
One day Johnny got bracket sequence. He decided to remove some of the brackets from it in order to obtain a regular bracket sequence. What is the maximum length of a regular bracket sequence which can be obtained?
Input
Input consists of a single line with non-empty string of «(» and «)» characters. Its length does not exceed 106.
Output
Output the maximum possible length of a regular bracket sequence.
Examples
Input
(()))(
Output
4
Input
((()())
Output
6
Submitted Solution:
```
def solution():
s = input().strip()
pos = 0
total = 0
for c in s:
if c == '(':
pos += 1
else:
pos -= 1
if pos >= 0:
total += 2
print(total)
solution()
``` | instruction | 0 | 30,384 | 21 | 60,768 |
No | output | 1 | 30,384 | 21 | 60,769 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters «+» and «1» into this sequence. For example, sequences «(())()», «()» and «(()(()))» are regular, while «)(», «(()» and «(()))(» are not.
One day Johnny got bracket sequence. He decided to remove some of the brackets from it in order to obtain a regular bracket sequence. What is the maximum length of a regular bracket sequence which can be obtained?
Input
Input consists of a single line with non-empty string of «(» and «)» characters. Its length does not exceed 106.
Output
Output the maximum possible length of a regular bracket sequence.
Examples
Input
(()))(
Output
4
Input
((()())
Output
6
Submitted Solution:
```
s = input()
balance = 0;
ans = ''
for i in range(len(s)):
if s[i] == '(':
balance += 1
ans += '('
if s[i] == ')' and balance > 0:
balance -= 1
ans += ')'
while len(ans) and ans[-1] == '(':
balance -= 1
ans = ans[:-1]
print(ans[balance:])
``` | instruction | 0 | 30,385 | 21 | 60,770 |
No | output | 1 | 30,385 | 21 | 60,771 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Implication is a function of two logical arguments, its value is false if and only if the value of the first argument is true and the value of the second argument is false.
Implication is written by using character '<image>', and the arguments and the result of the implication are written as '0' (false) and '1' (true). According to the definition of the implication:
<image>
<image>
<image>
<image>
When a logical expression contains multiple implications, then when there are no brackets, it will be calculated from left to fight. For example,
<image>.
When there are brackets, we first calculate the expression in brackets. For example,
<image>.
For the given logical expression <image> determine if it is possible to place there brackets so that the value of a logical expression is false. If it is possible, your task is to find such an arrangement of brackets.
Input
The first line contains integer n (1 ≤ n ≤ 100 000) — the number of arguments in a logical expression.
The second line contains n numbers a1, a2, ..., an (<image>), which means the values of arguments in the expression in the order they occur.
Output
Print "NO" (without the quotes), if it is impossible to place brackets in the expression so that its value was equal to 0.
Otherwise, print "YES" in the first line and the logical expression with the required arrangement of brackets in the second line.
The expression should only contain characters '0', '1', '-' (character with ASCII code 45), '>' (character with ASCII code 62), '(' and ')'. Characters '-' and '>' can occur in an expression only paired like that: ("->") and represent implication. The total number of logical arguments (i.e. digits '0' and '1') in the expression must be equal to n. The order in which the digits follow in the expression from left to right must coincide with a1, a2, ..., an.
The expression should be correct. More formally, a correct expression is determined as follows:
* Expressions "0", "1" (without the quotes) are correct.
* If v1, v2 are correct, then v1->v2 is a correct expression.
* If v is a correct expression, then (v) is a correct expression.
The total number of characters in the resulting expression mustn't exceed 106.
If there are multiple possible answers, you are allowed to print any of them.
Examples
Input
4
0 1 1 0
Output
YES
(((0)->1)->(1->0))
Input
2
1 1
Output
NO
Input
1
0
Output
YES
0 | instruction | 0 | 32,940 | 21 | 65,880 |
Tags: constructive algorithms, greedy, implementation, math
Correct Solution:
```
x = int(input())
seq = list(map(int, input().split(' ')))
if seq == [0]:
print("YES")
print(0)
elif seq == [0, 0]:
print("NO")
elif seq == [1, 0]:
print("YES")
print('1->0')
elif seq == [0, 0, 0]:
print("YES")
print("(0->0)->0")
elif seq == [1, 0, 0]:
print("NO")
elif seq[x-1] == 1:
print("NO")
#ENDS IN 1
elif seq[x-2] == 1:
print("YES")
print('->'.join([str(x) for x in seq]))
#ENDS IN 10
elif seq == [1] * (x-2) + [0, 0]:
print("NO")
#000 BELOW
elif seq[x-3] == 0:
a = ('->'.join([str(x) for x in seq][0:x-3]))
print("YES")
print(a + '->(0->0)->0')
#100
else:
last = 0
for i in range(x-1):
if seq[i] == 0 and seq[i+1] == 1:
last = i
seq[last] = '(0'
seq[last+1] = '(1'
seq[x-2] = '0))'
print("YES")
print('->'.join([str(x) for x in seq]))
``` | output | 1 | 32,940 | 21 | 65,881 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Implication is a function of two logical arguments, its value is false if and only if the value of the first argument is true and the value of the second argument is false.
Implication is written by using character '<image>', and the arguments and the result of the implication are written as '0' (false) and '1' (true). According to the definition of the implication:
<image>
<image>
<image>
<image>
When a logical expression contains multiple implications, then when there are no brackets, it will be calculated from left to fight. For example,
<image>.
When there are brackets, we first calculate the expression in brackets. For example,
<image>.
For the given logical expression <image> determine if it is possible to place there brackets so that the value of a logical expression is false. If it is possible, your task is to find such an arrangement of brackets.
Input
The first line contains integer n (1 ≤ n ≤ 100 000) — the number of arguments in a logical expression.
The second line contains n numbers a1, a2, ..., an (<image>), which means the values of arguments in the expression in the order they occur.
Output
Print "NO" (without the quotes), if it is impossible to place brackets in the expression so that its value was equal to 0.
Otherwise, print "YES" in the first line and the logical expression with the required arrangement of brackets in the second line.
The expression should only contain characters '0', '1', '-' (character with ASCII code 45), '>' (character with ASCII code 62), '(' and ')'. Characters '-' and '>' can occur in an expression only paired like that: ("->") and represent implication. The total number of logical arguments (i.e. digits '0' and '1') in the expression must be equal to n. The order in which the digits follow in the expression from left to right must coincide with a1, a2, ..., an.
The expression should be correct. More formally, a correct expression is determined as follows:
* Expressions "0", "1" (without the quotes) are correct.
* If v1, v2 are correct, then v1->v2 is a correct expression.
* If v is a correct expression, then (v) is a correct expression.
The total number of characters in the resulting expression mustn't exceed 106.
If there are multiple possible answers, you are allowed to print any of them.
Examples
Input
4
0 1 1 0
Output
YES
(((0)->1)->(1->0))
Input
2
1 1
Output
NO
Input
1
0
Output
YES
0 | instruction | 0 | 32,941 | 21 | 65,882 |
Tags: constructive algorithms, greedy, implementation, math
Correct Solution:
```
# https://codeforces.com/contest/550/problem/E
# TLE
from collections import deque
def solve():
n = int(input())
a = list(map(int, input().split()))
S = n - sum(a)
if a[-1] != 0:
return 'NO', None
if S == 1 and n == 1:
return 'YES', '0'
if S == 2 and (a[-1] == a[-2] and a[-1] == 0):
return 'NO', None
return 'YES', make(a)
def insert_head(x, cur):
cur.appendleft('>')
cur.appendleft('-')
cur.appendleft(str(x))
cur.appendleft('(')
cur.append(')')
def insert_tail(x, cur):
cur.append('-')
cur.append('>')
cur.append(x)
def make(a):
cur = None
for x in a[::-1][1:]:
if cur is None:
cur = deque([str(x)])
else:
insert_head(x, cur)
insert_tail(str(a[-1]), cur)
return ''.join(cur)
ans, s = solve()
if ans == 'NO':
print(ans)
else:
print(ans)
print(s)
``` | output | 1 | 32,941 | 21 | 65,883 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Implication is a function of two logical arguments, its value is false if and only if the value of the first argument is true and the value of the second argument is false.
Implication is written by using character '<image>', and the arguments and the result of the implication are written as '0' (false) and '1' (true). According to the definition of the implication:
<image>
<image>
<image>
<image>
When a logical expression contains multiple implications, then when there are no brackets, it will be calculated from left to fight. For example,
<image>.
When there are brackets, we first calculate the expression in brackets. For example,
<image>.
For the given logical expression <image> determine if it is possible to place there brackets so that the value of a logical expression is false. If it is possible, your task is to find such an arrangement of brackets.
Input
The first line contains integer n (1 ≤ n ≤ 100 000) — the number of arguments in a logical expression.
The second line contains n numbers a1, a2, ..., an (<image>), which means the values of arguments in the expression in the order they occur.
Output
Print "NO" (without the quotes), if it is impossible to place brackets in the expression so that its value was equal to 0.
Otherwise, print "YES" in the first line and the logical expression with the required arrangement of brackets in the second line.
The expression should only contain characters '0', '1', '-' (character with ASCII code 45), '>' (character with ASCII code 62), '(' and ')'. Characters '-' and '>' can occur in an expression only paired like that: ("->") and represent implication. The total number of logical arguments (i.e. digits '0' and '1') in the expression must be equal to n. The order in which the digits follow in the expression from left to right must coincide with a1, a2, ..., an.
The expression should be correct. More formally, a correct expression is determined as follows:
* Expressions "0", "1" (without the quotes) are correct.
* If v1, v2 are correct, then v1->v2 is a correct expression.
* If v is a correct expression, then (v) is a correct expression.
The total number of characters in the resulting expression mustn't exceed 106.
If there are multiple possible answers, you are allowed to print any of them.
Examples
Input
4
0 1 1 0
Output
YES
(((0)->1)->(1->0))
Input
2
1 1
Output
NO
Input
1
0
Output
YES
0 | instruction | 0 | 32,942 | 21 | 65,884 |
Tags: constructive algorithms, greedy, implementation, math
Correct Solution:
```
n=int(input())
a=list(input().split())
if (a[-1]>'0') or (len(a)>1 and a.count('0')==2 and a[-1]<'1' and a[-2]<'1'): print("NO")
else:
if len(a)>1:
if (a[-2]<'1'):
a[-2]+='))'
b=a[:-2]
b.reverse()
p=len(a)-3-b.index('0')
a[p]='('+a[p]
a[p+1]='('+a[p+1]
print("YES")
print("->".join(a))
``` | output | 1 | 32,942 | 21 | 65,885 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Implication is a function of two logical arguments, its value is false if and only if the value of the first argument is true and the value of the second argument is false.
Implication is written by using character '<image>', and the arguments and the result of the implication are written as '0' (false) and '1' (true). According to the definition of the implication:
<image>
<image>
<image>
<image>
When a logical expression contains multiple implications, then when there are no brackets, it will be calculated from left to fight. For example,
<image>.
When there are brackets, we first calculate the expression in brackets. For example,
<image>.
For the given logical expression <image> determine if it is possible to place there brackets so that the value of a logical expression is false. If it is possible, your task is to find such an arrangement of brackets.
Input
The first line contains integer n (1 ≤ n ≤ 100 000) — the number of arguments in a logical expression.
The second line contains n numbers a1, a2, ..., an (<image>), which means the values of arguments in the expression in the order they occur.
Output
Print "NO" (without the quotes), if it is impossible to place brackets in the expression so that its value was equal to 0.
Otherwise, print "YES" in the first line and the logical expression with the required arrangement of brackets in the second line.
The expression should only contain characters '0', '1', '-' (character with ASCII code 45), '>' (character with ASCII code 62), '(' and ')'. Characters '-' and '>' can occur in an expression only paired like that: ("->") and represent implication. The total number of logical arguments (i.e. digits '0' and '1') in the expression must be equal to n. The order in which the digits follow in the expression from left to right must coincide with a1, a2, ..., an.
The expression should be correct. More formally, a correct expression is determined as follows:
* Expressions "0", "1" (without the quotes) are correct.
* If v1, v2 are correct, then v1->v2 is a correct expression.
* If v is a correct expression, then (v) is a correct expression.
The total number of characters in the resulting expression mustn't exceed 106.
If there are multiple possible answers, you are allowed to print any of them.
Examples
Input
4
0 1 1 0
Output
YES
(((0)->1)->(1->0))
Input
2
1 1
Output
NO
Input
1
0
Output
YES
0 | instruction | 0 | 32,943 | 21 | 65,886 |
Tags: constructive algorithms, greedy, implementation, math
Correct Solution:
```
n=int(input())
s=input().split()
t=[]
j=-1
for i in range(n):
t+=[int(s[i])]
if int(s[i])==1:j=i
if t[n-1]==1:print("NO")
elif j==-1:
if n==1:
print("YES")
print(0)
elif n==2:
print("NO")
else:
print("YES")
for i in range(n-3):
print(s[i]+"->",end="")
print("(0->0)->0")
else:
if j<n-3:
print("YES")
for i in range(n-3):
print(s[i]+"->",end="")
print("(0->0)->0")
elif j==n-2:
print("YES")
for i in range(n-1):
print(s[i]+"->",end="")
print("0")
else:
k=-1
for i in range(n-2):
if t[i]==0:k=i
if k==-1:print("NO")
else:
print("YES")
for i in range(k):
print(s[i]+"->",end="")
print("(0->(",end="")
for i in range(n-k-3):
print("(",end="")
for i in range(n-k-3):
print("1)->",end="")
print("0))->0")
``` | output | 1 | 32,943 | 21 | 65,887 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Implication is a function of two logical arguments, its value is false if and only if the value of the first argument is true and the value of the second argument is false.
Implication is written by using character '<image>', and the arguments and the result of the implication are written as '0' (false) and '1' (true). According to the definition of the implication:
<image>
<image>
<image>
<image>
When a logical expression contains multiple implications, then when there are no brackets, it will be calculated from left to fight. For example,
<image>.
When there are brackets, we first calculate the expression in brackets. For example,
<image>.
For the given logical expression <image> determine if it is possible to place there brackets so that the value of a logical expression is false. If it is possible, your task is to find such an arrangement of brackets.
Input
The first line contains integer n (1 ≤ n ≤ 100 000) — the number of arguments in a logical expression.
The second line contains n numbers a1, a2, ..., an (<image>), which means the values of arguments in the expression in the order they occur.
Output
Print "NO" (without the quotes), if it is impossible to place brackets in the expression so that its value was equal to 0.
Otherwise, print "YES" in the first line and the logical expression with the required arrangement of brackets in the second line.
The expression should only contain characters '0', '1', '-' (character with ASCII code 45), '>' (character with ASCII code 62), '(' and ')'. Characters '-' and '>' can occur in an expression only paired like that: ("->") and represent implication. The total number of logical arguments (i.e. digits '0' and '1') in the expression must be equal to n. The order in which the digits follow in the expression from left to right must coincide with a1, a2, ..., an.
The expression should be correct. More formally, a correct expression is determined as follows:
* Expressions "0", "1" (without the quotes) are correct.
* If v1, v2 are correct, then v1->v2 is a correct expression.
* If v is a correct expression, then (v) is a correct expression.
The total number of characters in the resulting expression mustn't exceed 106.
If there are multiple possible answers, you are allowed to print any of them.
Examples
Input
4
0 1 1 0
Output
YES
(((0)->1)->(1->0))
Input
2
1 1
Output
NO
Input
1
0
Output
YES
0 | instruction | 0 | 32,944 | 21 | 65,888 |
Tags: constructive algorithms, greedy, implementation, math
Correct Solution:
```
def f(a):
if len(a) == 1:
if a[0] == 0:
print("YES\n0")
return
else:
print("NO")
return
if a[-1] == 1:
print("NO")
return
if a[-2] == 1:
print("YES")
print("->".join(str(x) for x in a))
return
elif len(a) == 2:
print("NO")
return
elif len(a) >= 3 and a[-3] == 0:
a[-3] = '(0'
a[-2] = '0)'
print("YES\n" + "->".join(str(x) for x in a))
return
for i in range(len(a) - 3, -1, -1):
if a[i] == 0:
a[i] = '(' + str(a[i])
a[i+1] = '(' + str(a[i+1])
a[-2] = '0))'
print("YES\n" + "->".join(str(x) for x in a))
return
print("NO")
return
n = int(input())
a = list(int(x) for x in input().split())
f(a)
``` | output | 1 | 32,944 | 21 | 65,889 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Implication is a function of two logical arguments, its value is false if and only if the value of the first argument is true and the value of the second argument is false.
Implication is written by using character '<image>', and the arguments and the result of the implication are written as '0' (false) and '1' (true). According to the definition of the implication:
<image>
<image>
<image>
<image>
When a logical expression contains multiple implications, then when there are no brackets, it will be calculated from left to fight. For example,
<image>.
When there are brackets, we first calculate the expression in brackets. For example,
<image>.
For the given logical expression <image> determine if it is possible to place there brackets so that the value of a logical expression is false. If it is possible, your task is to find such an arrangement of brackets.
Input
The first line contains integer n (1 ≤ n ≤ 100 000) — the number of arguments in a logical expression.
The second line contains n numbers a1, a2, ..., an (<image>), which means the values of arguments in the expression in the order they occur.
Output
Print "NO" (without the quotes), if it is impossible to place brackets in the expression so that its value was equal to 0.
Otherwise, print "YES" in the first line and the logical expression with the required arrangement of brackets in the second line.
The expression should only contain characters '0', '1', '-' (character with ASCII code 45), '>' (character with ASCII code 62), '(' and ')'. Characters '-' and '>' can occur in an expression only paired like that: ("->") and represent implication. The total number of logical arguments (i.e. digits '0' and '1') in the expression must be equal to n. The order in which the digits follow in the expression from left to right must coincide with a1, a2, ..., an.
The expression should be correct. More formally, a correct expression is determined as follows:
* Expressions "0", "1" (without the quotes) are correct.
* If v1, v2 are correct, then v1->v2 is a correct expression.
* If v is a correct expression, then (v) is a correct expression.
The total number of characters in the resulting expression mustn't exceed 106.
If there are multiple possible answers, you are allowed to print any of them.
Examples
Input
4
0 1 1 0
Output
YES
(((0)->1)->(1->0))
Input
2
1 1
Output
NO
Input
1
0
Output
YES
0 | instruction | 0 | 32,945 | 21 | 65,890 |
Tags: constructive algorithms, greedy, implementation, math
Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect
from itertools import chain, dropwhile, permutations, combinations
from collections import defaultdict, deque
def VI(): return list(map(int,input().split()))
def ret_yes(x):
print("YES")
print(x)
def pr(a):
print(a[0],end="")
for x in a[1:]:
print("->{}".format(x),end="")
print()
def pr2(a):
p = 0
for i,x in enumerate(a[:-2]):
if x==0:# and a[i+1]==1:
print("{}->(".format(x),end="")
p += 1
else:
print("{}->".format(x),end="")
print(a[-2],end="")
for i in range(p):
print(")",end="")
print("->{}".format(a[-1]))
def main(n,a):
if a[-1] != 0:
print("NO")
elif n==1:
print("YES")
print("0")
elif n==2 and a[0]==0:
print("NO")
elif a[-2]==0 and 0 not in set(a[:-2]):
print("NO")
elif a[-2]==1:
print("YES")
pr(a)
else: # ..-> 0 -> .. 1 -> 0 -> 0 -- needs to add parentheses
print("YES")
pr2(a)
def main_input(info=0):
n = int(input())
a = VI()
main(n,a)
if __name__ == "__main__":
main_input()
``` | output | 1 | 32,945 | 21 | 65,891 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Implication is a function of two logical arguments, its value is false if and only if the value of the first argument is true and the value of the second argument is false.
Implication is written by using character '<image>', and the arguments and the result of the implication are written as '0' (false) and '1' (true). According to the definition of the implication:
<image>
<image>
<image>
<image>
When a logical expression contains multiple implications, then when there are no brackets, it will be calculated from left to fight. For example,
<image>.
When there are brackets, we first calculate the expression in brackets. For example,
<image>.
For the given logical expression <image> determine if it is possible to place there brackets so that the value of a logical expression is false. If it is possible, your task is to find such an arrangement of brackets.
Input
The first line contains integer n (1 ≤ n ≤ 100 000) — the number of arguments in a logical expression.
The second line contains n numbers a1, a2, ..., an (<image>), which means the values of arguments in the expression in the order they occur.
Output
Print "NO" (without the quotes), if it is impossible to place brackets in the expression so that its value was equal to 0.
Otherwise, print "YES" in the first line and the logical expression with the required arrangement of brackets in the second line.
The expression should only contain characters '0', '1', '-' (character with ASCII code 45), '>' (character with ASCII code 62), '(' and ')'. Characters '-' and '>' can occur in an expression only paired like that: ("->") and represent implication. The total number of logical arguments (i.e. digits '0' and '1') in the expression must be equal to n. The order in which the digits follow in the expression from left to right must coincide with a1, a2, ..., an.
The expression should be correct. More formally, a correct expression is determined as follows:
* Expressions "0", "1" (without the quotes) are correct.
* If v1, v2 are correct, then v1->v2 is a correct expression.
* If v is a correct expression, then (v) is a correct expression.
The total number of characters in the resulting expression mustn't exceed 106.
If there are multiple possible answers, you are allowed to print any of them.
Examples
Input
4
0 1 1 0
Output
YES
(((0)->1)->(1->0))
Input
2
1 1
Output
NO
Input
1
0
Output
YES
0 | instruction | 0 | 32,946 | 21 | 65,892 |
Tags: constructive algorithms, greedy, implementation, math
Correct Solution:
```
n = int(input())
v = input().split()
if v[-1] != '0':
print("NO")
else:
if n >= 2 and v[-2] == "0":
for i in range(n-3, -1, -1):
if v[i] == "0":
v[i] = "(" + v[i]
v[i+1] = "(" + v[i+1]
v[-2] = v[-2] + "))"
break
else:
print("NO")
exit()
print("YES")
print("->".join(v))
``` | output | 1 | 32,946 | 21 | 65,893 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Implication is a function of two logical arguments, its value is false if and only if the value of the first argument is true and the value of the second argument is false.
Implication is written by using character '<image>', and the arguments and the result of the implication are written as '0' (false) and '1' (true). According to the definition of the implication:
<image>
<image>
<image>
<image>
When a logical expression contains multiple implications, then when there are no brackets, it will be calculated from left to fight. For example,
<image>.
When there are brackets, we first calculate the expression in brackets. For example,
<image>.
For the given logical expression <image> determine if it is possible to place there brackets so that the value of a logical expression is false. If it is possible, your task is to find such an arrangement of brackets.
Input
The first line contains integer n (1 ≤ n ≤ 100 000) — the number of arguments in a logical expression.
The second line contains n numbers a1, a2, ..., an (<image>), which means the values of arguments in the expression in the order they occur.
Output
Print "NO" (without the quotes), if it is impossible to place brackets in the expression so that its value was equal to 0.
Otherwise, print "YES" in the first line and the logical expression with the required arrangement of brackets in the second line.
The expression should only contain characters '0', '1', '-' (character with ASCII code 45), '>' (character with ASCII code 62), '(' and ')'. Characters '-' and '>' can occur in an expression only paired like that: ("->") and represent implication. The total number of logical arguments (i.e. digits '0' and '1') in the expression must be equal to n. The order in which the digits follow in the expression from left to right must coincide with a1, a2, ..., an.
The expression should be correct. More formally, a correct expression is determined as follows:
* Expressions "0", "1" (without the quotes) are correct.
* If v1, v2 are correct, then v1->v2 is a correct expression.
* If v is a correct expression, then (v) is a correct expression.
The total number of characters in the resulting expression mustn't exceed 106.
If there are multiple possible answers, you are allowed to print any of them.
Examples
Input
4
0 1 1 0
Output
YES
(((0)->1)->(1->0))
Input
2
1 1
Output
NO
Input
1
0
Output
YES
0 | instruction | 0 | 32,947 | 21 | 65,894 |
Tags: constructive algorithms, greedy, implementation, math
Correct Solution:
```
import sys
input = []
input_index = 0
def next(type, number = None):
def next():
global input, input_index
while input_index == len(input):
if sys.stdin:
input = sys.stdin.readline().split()
input_index = 0
else:
raise Exception()
input_index += 1
return input[input_index - 1]
if number is None:
result = type(next())
else:
result = [type(next()) for _ in range(number)]
return result
n = next(int)
ks = next(str, n)
if ks[-1] == "0":
if n == 1:
print("YES")
print("0")
elif "0" in ks[:-2] or ks[-2] != "0":
print("YES")
for k in ks[:-2]:
sys.stdout.write(k + "->(")
sys.stdout.write(ks[-2])
for _ in range(len(ks) - 2):
sys.stdout.write(")")
sys.stdout.write("->0\n")
sys.stdout.flush()
else:
print("NO")
else:
print("NO")
``` | output | 1 | 32,947 | 21 | 65,895 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Implication is a function of two logical arguments, its value is false if and only if the value of the first argument is true and the value of the second argument is false.
Implication is written by using character '<image>', and the arguments and the result of the implication are written as '0' (false) and '1' (true). According to the definition of the implication:
<image>
<image>
<image>
<image>
When a logical expression contains multiple implications, then when there are no brackets, it will be calculated from left to fight. For example,
<image>.
When there are brackets, we first calculate the expression in brackets. For example,
<image>.
For the given logical expression <image> determine if it is possible to place there brackets so that the value of a logical expression is false. If it is possible, your task is to find such an arrangement of brackets.
Input
The first line contains integer n (1 ≤ n ≤ 100 000) — the number of arguments in a logical expression.
The second line contains n numbers a1, a2, ..., an (<image>), which means the values of arguments in the expression in the order they occur.
Output
Print "NO" (without the quotes), if it is impossible to place brackets in the expression so that its value was equal to 0.
Otherwise, print "YES" in the first line and the logical expression with the required arrangement of brackets in the second line.
The expression should only contain characters '0', '1', '-' (character with ASCII code 45), '>' (character with ASCII code 62), '(' and ')'. Characters '-' and '>' can occur in an expression only paired like that: ("->") and represent implication. The total number of logical arguments (i.e. digits '0' and '1') in the expression must be equal to n. The order in which the digits follow in the expression from left to right must coincide with a1, a2, ..., an.
The expression should be correct. More formally, a correct expression is determined as follows:
* Expressions "0", "1" (without the quotes) are correct.
* If v1, v2 are correct, then v1->v2 is a correct expression.
* If v is a correct expression, then (v) is a correct expression.
The total number of characters in the resulting expression mustn't exceed 106.
If there are multiple possible answers, you are allowed to print any of them.
Examples
Input
4
0 1 1 0
Output
YES
(((0)->1)->(1->0))
Input
2
1 1
Output
NO
Input
1
0
Output
YES
0
Submitted Solution:
```
n = int(input())
c = list(map(int, input().split()))
idx = [i for i in range(n) if c[i] == 0]
one = [i for i in range(n) if c[i] == 1]
if len(idx) >= 1 and idx[-1] == n-1:
if len(idx) >= 2 and idx[-2] == n-2:
if len(idx) >= 3 and idx[-3] <= n-3:
if idx[0] == 0:
print("YES")
r = ""
for i in range(n):
if i == 1:
r += "("
r += str(c[i])
if i == n-2:
r += ")"
if i != n-1:
r += "->"
print(r)
else:
print("YES")
r = ""
if idx[-2]-1 == 0:
for i in range(n):
if i == 0:
r += "("
r += str(c[i])
if i == idx[-2]-1:
r += ")"
if i != n-1:
r += "->"
else:
for i in range(n):
if i == 0:
r += "("
elif i == idx[0]:
r += "("
elif i == idx[0] + 1:
r += "("
r += str(c[i])
if i == idx[-1] - 1:
r += "))"
elif i == idx[0]-1:
r += ")"
if i != n-1:
r += "->"
print(r)
else:
print("NO")
else:
print("YES")
print("->".join(map(str, c)))
else:
print("NO")
``` | instruction | 0 | 32,948 | 21 | 65,896 |
Yes | output | 1 | 32,948 | 21 | 65,897 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Implication is a function of two logical arguments, its value is false if and only if the value of the first argument is true and the value of the second argument is false.
Implication is written by using character '<image>', and the arguments and the result of the implication are written as '0' (false) and '1' (true). According to the definition of the implication:
<image>
<image>
<image>
<image>
When a logical expression contains multiple implications, then when there are no brackets, it will be calculated from left to fight. For example,
<image>.
When there are brackets, we first calculate the expression in brackets. For example,
<image>.
For the given logical expression <image> determine if it is possible to place there brackets so that the value of a logical expression is false. If it is possible, your task is to find such an arrangement of brackets.
Input
The first line contains integer n (1 ≤ n ≤ 100 000) — the number of arguments in a logical expression.
The second line contains n numbers a1, a2, ..., an (<image>), which means the values of arguments in the expression in the order they occur.
Output
Print "NO" (without the quotes), if it is impossible to place brackets in the expression so that its value was equal to 0.
Otherwise, print "YES" in the first line and the logical expression with the required arrangement of brackets in the second line.
The expression should only contain characters '0', '1', '-' (character with ASCII code 45), '>' (character with ASCII code 62), '(' and ')'. Characters '-' and '>' can occur in an expression only paired like that: ("->") and represent implication. The total number of logical arguments (i.e. digits '0' and '1') in the expression must be equal to n. The order in which the digits follow in the expression from left to right must coincide with a1, a2, ..., an.
The expression should be correct. More formally, a correct expression is determined as follows:
* Expressions "0", "1" (without the quotes) are correct.
* If v1, v2 are correct, then v1->v2 is a correct expression.
* If v is a correct expression, then (v) is a correct expression.
The total number of characters in the resulting expression mustn't exceed 106.
If there are multiple possible answers, you are allowed to print any of them.
Examples
Input
4
0 1 1 0
Output
YES
(((0)->1)->(1->0))
Input
2
1 1
Output
NO
Input
1
0
Output
YES
0
Submitted Solution:
```
#!/usr/bin/env python3
n = int(input())
arr = list(map(int, input().strip().split(' ')))
class MyExcept(BaseException):
pass
def left_rec(lst, L, R):
k = R - L + 1
ans = "(" * k + ")->".join(map(str, lst[L:R+1])) + ")"
return ans
def solve():
if arr[-2] == 1:
return "(" + left_rec(arr, 0, len(arr) - 2) + "->0)"
else: # 00
if n == 3:
if arr[-3] == 0:
return "((0->0)->0)"
else:
raise MyExcept()
else: #XX00
if arr[-3] == 0: #X000
return "((" + left_rec(arr, 0, len(arr) - 4) + "->(0->0))->0)"
else: #X100
k = 0
for i in range(0, len(arr) - 3):
if arr[i] == 0:
k = i
break
else:
raise MyExcept()
return "((" + left_rec(arr, 0, k) + "->" + left_rec(arr, k + 1, len(arr) - 2) + ")->0)"
def main():
if n == 1:
if arr[0] == 0:
print("YES")
print("0")
else:
print("NO")
elif n == 2:
if arr[0] == 1 and arr[1] == 0:
print("YES")
print("(1->0)")
else:
print("NO")
elif arr[-1] == 1:
print("NO")
else:
try:
s = solve()
print("YES")
print(s)
except MyExcept:
print("NO")
main()
``` | instruction | 0 | 32,949 | 21 | 65,898 |
Yes | output | 1 | 32,949 | 21 | 65,899 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Implication is a function of two logical arguments, its value is false if and only if the value of the first argument is true and the value of the second argument is false.
Implication is written by using character '<image>', and the arguments and the result of the implication are written as '0' (false) and '1' (true). According to the definition of the implication:
<image>
<image>
<image>
<image>
When a logical expression contains multiple implications, then when there are no brackets, it will be calculated from left to fight. For example,
<image>.
When there are brackets, we first calculate the expression in brackets. For example,
<image>.
For the given logical expression <image> determine if it is possible to place there brackets so that the value of a logical expression is false. If it is possible, your task is to find such an arrangement of brackets.
Input
The first line contains integer n (1 ≤ n ≤ 100 000) — the number of arguments in a logical expression.
The second line contains n numbers a1, a2, ..., an (<image>), which means the values of arguments in the expression in the order they occur.
Output
Print "NO" (without the quotes), if it is impossible to place brackets in the expression so that its value was equal to 0.
Otherwise, print "YES" in the first line and the logical expression with the required arrangement of brackets in the second line.
The expression should only contain characters '0', '1', '-' (character with ASCII code 45), '>' (character with ASCII code 62), '(' and ')'. Characters '-' and '>' can occur in an expression only paired like that: ("->") and represent implication. The total number of logical arguments (i.e. digits '0' and '1') in the expression must be equal to n. The order in which the digits follow in the expression from left to right must coincide with a1, a2, ..., an.
The expression should be correct. More formally, a correct expression is determined as follows:
* Expressions "0", "1" (without the quotes) are correct.
* If v1, v2 are correct, then v1->v2 is a correct expression.
* If v is a correct expression, then (v) is a correct expression.
The total number of characters in the resulting expression mustn't exceed 106.
If there are multiple possible answers, you are allowed to print any of them.
Examples
Input
4
0 1 1 0
Output
YES
(((0)->1)->(1->0))
Input
2
1 1
Output
NO
Input
1
0
Output
YES
0
Submitted Solution:
```
def constructExpression( a , l , r ):
#print("construct:",a)
#print("l:",l,"r:",r)
res = "("
for i in range(l,r):
res += (a[i]+("->" if i != r-1 else ")"))
#print(res)
return str(res)
def allOnes( a , l , r ):
for i in range(l,r):
if a[i] != "1":
return False
return True
if __name__ == "__main__":
N = int(input())
a = [x for x in input().split()]
if len(a) == 1:
#1
if a[0] == "1":print("NO")
#0
else:print("YES");print("0")
elif len(a) == 2:
#1 0
if a[0] == "1" and a[1] == "0":print("YES");print("1->0")
#0 0
#0 1
#1 1
else:print("NO")
else:
#... 1
if a[len(a)-1] == "1":print("NO")
#... 0
else:
#... 1 0
if a[len(a)-2] == "1":
print("YES")
print( constructExpression(a,0,len(a)) )
#... 0 0
else:
#1 1 ... 1 0 0
if allOnes(a , 0 , len(a)-2 ):
print("NO")
# ... 0 ... 0 0
else:
index = len(a)-3
for i in range(len(a)-3,-1,-1):
if a[i] == "0":
index = i
break
#print("index:",index)
a = a[:index+1] + [constructExpression(a,index+1,len(a)-1)] + a[len(a)-1:]
#print("1) ",a)
a = a[:len(a)-3] + [constructExpression(a,len(a)-3,len(a)-1)] + a[len(a)-1:]
#print("2) ",a)
print("YES")
print( constructExpression(a,0,len(a)) )
``` | instruction | 0 | 32,950 | 21 | 65,900 |
Yes | output | 1 | 32,950 | 21 | 65,901 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Implication is a function of two logical arguments, its value is false if and only if the value of the first argument is true and the value of the second argument is false.
Implication is written by using character '<image>', and the arguments and the result of the implication are written as '0' (false) and '1' (true). According to the definition of the implication:
<image>
<image>
<image>
<image>
When a logical expression contains multiple implications, then when there are no brackets, it will be calculated from left to fight. For example,
<image>.
When there are brackets, we first calculate the expression in brackets. For example,
<image>.
For the given logical expression <image> determine if it is possible to place there brackets so that the value of a logical expression is false. If it is possible, your task is to find such an arrangement of brackets.
Input
The first line contains integer n (1 ≤ n ≤ 100 000) — the number of arguments in a logical expression.
The second line contains n numbers a1, a2, ..., an (<image>), which means the values of arguments in the expression in the order they occur.
Output
Print "NO" (without the quotes), if it is impossible to place brackets in the expression so that its value was equal to 0.
Otherwise, print "YES" in the first line and the logical expression with the required arrangement of brackets in the second line.
The expression should only contain characters '0', '1', '-' (character with ASCII code 45), '>' (character with ASCII code 62), '(' and ')'. Characters '-' and '>' can occur in an expression only paired like that: ("->") and represent implication. The total number of logical arguments (i.e. digits '0' and '1') in the expression must be equal to n. The order in which the digits follow in the expression from left to right must coincide with a1, a2, ..., an.
The expression should be correct. More formally, a correct expression is determined as follows:
* Expressions "0", "1" (without the quotes) are correct.
* If v1, v2 are correct, then v1->v2 is a correct expression.
* If v is a correct expression, then (v) is a correct expression.
The total number of characters in the resulting expression mustn't exceed 106.
If there are multiple possible answers, you are allowed to print any of them.
Examples
Input
4
0 1 1 0
Output
YES
(((0)->1)->(1->0))
Input
2
1 1
Output
NO
Input
1
0
Output
YES
0
Submitted Solution:
```
n = int(input())
s = "".join(input().split())
if s == "0":
print("YES")
print("0")
elif s[-1] == '1':
print("NO")
elif s[-2:] == "10":
print("YES")
print("->".join(s))
else:
# s[-2:] == "00"
pos = s.rfind('0', 0, -2)
if pos != -1:
print("YES")
if pos != 0:
print("->".join(s[:pos]), end="->")
print("(0->(%s))->0" % "->".join(s[pos + 1:-1]))
else:
print("NO")
``` | instruction | 0 | 32,951 | 21 | 65,902 |
Yes | output | 1 | 32,951 | 21 | 65,903 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Implication is a function of two logical arguments, its value is false if and only if the value of the first argument is true and the value of the second argument is false.
Implication is written by using character '<image>', and the arguments and the result of the implication are written as '0' (false) and '1' (true). According to the definition of the implication:
<image>
<image>
<image>
<image>
When a logical expression contains multiple implications, then when there are no brackets, it will be calculated from left to fight. For example,
<image>.
When there are brackets, we first calculate the expression in brackets. For example,
<image>.
For the given logical expression <image> determine if it is possible to place there brackets so that the value of a logical expression is false. If it is possible, your task is to find such an arrangement of brackets.
Input
The first line contains integer n (1 ≤ n ≤ 100 000) — the number of arguments in a logical expression.
The second line contains n numbers a1, a2, ..., an (<image>), which means the values of arguments in the expression in the order they occur.
Output
Print "NO" (without the quotes), if it is impossible to place brackets in the expression so that its value was equal to 0.
Otherwise, print "YES" in the first line and the logical expression with the required arrangement of brackets in the second line.
The expression should only contain characters '0', '1', '-' (character with ASCII code 45), '>' (character with ASCII code 62), '(' and ')'. Characters '-' and '>' can occur in an expression only paired like that: ("->") and represent implication. The total number of logical arguments (i.e. digits '0' and '1') in the expression must be equal to n. The order in which the digits follow in the expression from left to right must coincide with a1, a2, ..., an.
The expression should be correct. More formally, a correct expression is determined as follows:
* Expressions "0", "1" (without the quotes) are correct.
* If v1, v2 are correct, then v1->v2 is a correct expression.
* If v is a correct expression, then (v) is a correct expression.
The total number of characters in the resulting expression mustn't exceed 106.
If there are multiple possible answers, you are allowed to print any of them.
Examples
Input
4
0 1 1 0
Output
YES
(((0)->1)->(1->0))
Input
2
1 1
Output
NO
Input
1
0
Output
YES
0
Submitted Solution:
```
from sys import stdin
input = stdin.readline
cache = {}
def implication(a, b):
if cache.get((a, b)):
return cache[(a, b)]
if len(a) == 1:
cache[(a, b)] = a[0]
return int(a[0]) == b
for i in range(1, len(a)):
print(a, a[:i], a[i:], b)
if not b:
x1 = implication(a[:i], True)
x2 = implication(a[i:], False)
if x1 and x2:
cache[(a, b)] = '(%s)->(%s)' % (cache[(a[:i], True)], cache[(a[i:], False)])
return True
elif b:
p1 = implication(a[:i], False)
p2 = implication(a[i:], False)
if p1 and p2:
cache[(a, b)] = '(%s)->(%s)' % (cache[(a[:i], False)], cache[(a[i:], False)])
return True
p3 = implication(a[i:], True)
if p1 and p3:
cache[(a, b)] = '(%s)->(%s)' % (cache[(a[:i], False)], cache[(a[i:], True)])
return True
p4 = implication(a[:i], True)
if p4 and p3:
cache[(a, b)] = '(%s)->(%s)' % (cache[(a[:i], True)], cache[(a[i:], True)])
return True
cache[(a, b)] = False
return False
def int_main(n):
a = tuple(map(int, input().split()))
if not implication(a, False):
return 'NO'
return 'YES\n%s' % (cache[(a, False)])
print(int_main(int(input())))
``` | instruction | 0 | 32,952 | 21 | 65,904 |
No | output | 1 | 32,952 | 21 | 65,905 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Implication is a function of two logical arguments, its value is false if and only if the value of the first argument is true and the value of the second argument is false.
Implication is written by using character '<image>', and the arguments and the result of the implication are written as '0' (false) and '1' (true). According to the definition of the implication:
<image>
<image>
<image>
<image>
When a logical expression contains multiple implications, then when there are no brackets, it will be calculated from left to fight. For example,
<image>.
When there are brackets, we first calculate the expression in brackets. For example,
<image>.
For the given logical expression <image> determine if it is possible to place there brackets so that the value of a logical expression is false. If it is possible, your task is to find such an arrangement of brackets.
Input
The first line contains integer n (1 ≤ n ≤ 100 000) — the number of arguments in a logical expression.
The second line contains n numbers a1, a2, ..., an (<image>), which means the values of arguments in the expression in the order they occur.
Output
Print "NO" (without the quotes), if it is impossible to place brackets in the expression so that its value was equal to 0.
Otherwise, print "YES" in the first line and the logical expression with the required arrangement of brackets in the second line.
The expression should only contain characters '0', '1', '-' (character with ASCII code 45), '>' (character with ASCII code 62), '(' and ')'. Characters '-' and '>' can occur in an expression only paired like that: ("->") and represent implication. The total number of logical arguments (i.e. digits '0' and '1') in the expression must be equal to n. The order in which the digits follow in the expression from left to right must coincide with a1, a2, ..., an.
The expression should be correct. More formally, a correct expression is determined as follows:
* Expressions "0", "1" (without the quotes) are correct.
* If v1, v2 are correct, then v1->v2 is a correct expression.
* If v is a correct expression, then (v) is a correct expression.
The total number of characters in the resulting expression mustn't exceed 106.
If there are multiple possible answers, you are allowed to print any of them.
Examples
Input
4
0 1 1 0
Output
YES
(((0)->1)->(1->0))
Input
2
1 1
Output
NO
Input
1
0
Output
YES
0
Submitted Solution:
```
n=int(input())
s=input().split()
t=[]
j=-1
for i in range(n):
t+=[int(s[i])]
if int(s[i])==1:j=i
if t[n-1]==1:print("NO")
elif j==-1:
if n==1:
print("YES")
print(0)
elif n==2:
print("NO")
else:
print("YES")
for i in range(n-3):
print(s[i]+"->",end="")
print("(0->0)->0")
else:
if j<n-3:
print("YES")
for i in range(n-3):
print(s[i]+"->",end="")
print("(0->0)->0")
elif j==n-2:
print("YES")
for i in range(n-1):
print(s[i]+"->",end="")
print("0")
else:
k=-1
for i in range(n-2):
if t[i]==0:k=i
if k==-1:print("NO")
else:
print("YES")
for i in range(k):
print(s[i]+"->",end="")
print("(0->",end="")
for i in range(n-3):
print("(",end="")
for i in range(n-3):
print("1)->",end="")
print("0))->0")
``` | instruction | 0 | 32,953 | 21 | 65,906 |
No | output | 1 | 32,953 | 21 | 65,907 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Implication is a function of two logical arguments, its value is false if and only if the value of the first argument is true and the value of the second argument is false.
Implication is written by using character '<image>', and the arguments and the result of the implication are written as '0' (false) and '1' (true). According to the definition of the implication:
<image>
<image>
<image>
<image>
When a logical expression contains multiple implications, then when there are no brackets, it will be calculated from left to fight. For example,
<image>.
When there are brackets, we first calculate the expression in brackets. For example,
<image>.
For the given logical expression <image> determine if it is possible to place there brackets so that the value of a logical expression is false. If it is possible, your task is to find such an arrangement of brackets.
Input
The first line contains integer n (1 ≤ n ≤ 100 000) — the number of arguments in a logical expression.
The second line contains n numbers a1, a2, ..., an (<image>), which means the values of arguments in the expression in the order they occur.
Output
Print "NO" (without the quotes), if it is impossible to place brackets in the expression so that its value was equal to 0.
Otherwise, print "YES" in the first line and the logical expression with the required arrangement of brackets in the second line.
The expression should only contain characters '0', '1', '-' (character with ASCII code 45), '>' (character with ASCII code 62), '(' and ')'. Characters '-' and '>' can occur in an expression only paired like that: ("->") and represent implication. The total number of logical arguments (i.e. digits '0' and '1') in the expression must be equal to n. The order in which the digits follow in the expression from left to right must coincide with a1, a2, ..., an.
The expression should be correct. More formally, a correct expression is determined as follows:
* Expressions "0", "1" (without the quotes) are correct.
* If v1, v2 are correct, then v1->v2 is a correct expression.
* If v is a correct expression, then (v) is a correct expression.
The total number of characters in the resulting expression mustn't exceed 106.
If there are multiple possible answers, you are allowed to print any of them.
Examples
Input
4
0 1 1 0
Output
YES
(((0)->1)->(1->0))
Input
2
1 1
Output
NO
Input
1
0
Output
YES
0
Submitted Solution:
```
#!/usr/bin/env python3
n = int(input())
arr = list(map(int, input().strip().split(' ')))
class MyExcept(BaseException):
pass
def left_rec(lst, L, R):
k = R - L + 1
ans = "(" * (k - 1) + ")->".join(map(str, lst[L:R+1]))
return ans
def solve():
if arr[-2] == 1:
return "(" + left_rec(arr, 0, len(arr) - 2) + "->0)"
else: # 00
if n == 3:
if arr[-3] == 0:
return "((0->0)->0)"
else:
raise 0
else: #XX00
if arr[-3] == 0: #X000
return "((" + left_rec(arr, 0, len(arr) - 4) + "->(0->0))->0)"
else: #X100
k = 0
for i in range(0, len(arr) - 3):
if arr[i] == 0:
k = i
break
else:
print("MEOW")
raise MyExcept()
return "((" + left_rec(arr, 0, k) + "->" + left_rec(arr, k + 1, len(arr) - 2) + ")->0)"
def main():
if n == 1:
if arr[0] == 0:
print("YES")
print("0")
else:
print("NO")
elif n == 2:
if arr[0] == 1 and arr[1] == 0:
print("YES")
print("(1->0)")
else:
print("NO")
elif arr[-1] == 1:
if arr[0:8] == [1, 0, 1, 0, 0, 0, 0, 0]:
print("MEOW")
print("NO")
else:
try:
s = solve()
print("YES")
print(s)
except MyExcept:
print("NO")
main()
``` | instruction | 0 | 32,954 | 21 | 65,908 |
No | output | 1 | 32,954 | 21 | 65,909 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Implication is a function of two logical arguments, its value is false if and only if the value of the first argument is true and the value of the second argument is false.
Implication is written by using character '<image>', and the arguments and the result of the implication are written as '0' (false) and '1' (true). According to the definition of the implication:
<image>
<image>
<image>
<image>
When a logical expression contains multiple implications, then when there are no brackets, it will be calculated from left to fight. For example,
<image>.
When there are brackets, we first calculate the expression in brackets. For example,
<image>.
For the given logical expression <image> determine if it is possible to place there brackets so that the value of a logical expression is false. If it is possible, your task is to find such an arrangement of brackets.
Input
The first line contains integer n (1 ≤ n ≤ 100 000) — the number of arguments in a logical expression.
The second line contains n numbers a1, a2, ..., an (<image>), which means the values of arguments in the expression in the order they occur.
Output
Print "NO" (without the quotes), if it is impossible to place brackets in the expression so that its value was equal to 0.
Otherwise, print "YES" in the first line and the logical expression with the required arrangement of brackets in the second line.
The expression should only contain characters '0', '1', '-' (character with ASCII code 45), '>' (character with ASCII code 62), '(' and ')'. Characters '-' and '>' can occur in an expression only paired like that: ("->") and represent implication. The total number of logical arguments (i.e. digits '0' and '1') in the expression must be equal to n. The order in which the digits follow in the expression from left to right must coincide with a1, a2, ..., an.
The expression should be correct. More formally, a correct expression is determined as follows:
* Expressions "0", "1" (without the quotes) are correct.
* If v1, v2 are correct, then v1->v2 is a correct expression.
* If v is a correct expression, then (v) is a correct expression.
The total number of characters in the resulting expression mustn't exceed 106.
If there are multiple possible answers, you are allowed to print any of them.
Examples
Input
4
0 1 1 0
Output
YES
(((0)->1)->(1->0))
Input
2
1 1
Output
NO
Input
1
0
Output
YES
0
Submitted Solution:
```
n = int(input())
c = list(map(int, input().split()))
idx = [i for i in range(n) if c[i] == 0]
if len(idx) >= 1 and idx[-1] == n-1:
if len(idx) >= 2 and idx[-2] == n-2:
if len(idx) >= 3 and idx[-3] <= n-3:
if idx[0] == 0:
print("YES")
r = ""
for i in range(n):
if i == 1:
r += "("
r += str(c[i])
if i == n-2:
r += ")"
if i != n-1:
r += "->"
print(r)
else:
print("YES")
r = ""
for i in range(n):
if i == idx[-3]:
r += "("
r += str(c[i])
if i == idx[-2]:
r += ")"
if i != n-1:
r += "->"
print(r)
else:
print("NO")
else:
print("YES")
print("->".join(map(str, c)))
else:
print("NO")
``` | instruction | 0 | 32,955 | 21 | 65,910 |
No | output | 1 | 32,955 | 21 | 65,911 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The development of a text editor is a hard problem. You need to implement an extra module for brackets coloring in text.
Your editor consists of a line with infinite length and cursor, which points to the current character. Please note that it points to only one of the characters (and not between a pair of characters). Thus, it points to an index character. The user can move the cursor left or right one position. If the cursor is already at the first (leftmost) position, then it does not move left.
Initially, the cursor is in the first (leftmost) character.
Also, the user can write a letter or brackets (either (, or )) to the position that the cursor is currently pointing at. A new character always overwrites the old value at that position.
Your editor must check, whether the current line is the correct text. Text is correct if the brackets in them form the correct bracket sequence.
Formally, correct text (CT) must satisfy the following rules:
* any line without brackets is CT (the line can contain whitespaces);
* If the first character of the string — is (, the last — is ), and all the rest form a CT, then the whole line is a CT;
* two consecutively written CT is also CT.
Examples of correct texts: hello(codeforces), round, ((i)(write))edi(tor)s, ( me). Examples of incorrect texts: hello)oops(, round), ((me).
The user uses special commands to work with your editor. Each command has its symbol, which must be written to execute this command.
The correspondence of commands and characters is as follows:
* L — move the cursor one character to the left (remains in place if it already points to the first character);
* R — move the cursor one character to the right;
* any lowercase Latin letter or bracket (( or )) — write the entered character to the position where the cursor is now.
For a complete understanding, take a look at the first example and its illustrations in the note below.
You are given a string containing the characters that the user entered. For the brackets coloring module's work, after each command you need to:
* check if the current text in the editor is a correct text;
* if it is, print the least number of colors that required, to color all brackets.
If two pairs of brackets are nested (the first in the second or vice versa), then these pairs of brackets should be painted in different colors. If two pairs of brackets are not nested, then they can be painted in different or the same colors. For example, for the bracket sequence ()(())()() the least number of colors is 2, and for the bracket sequence (()(()())())(()) — is 3.
Write a program that prints the minimal number of colors after processing each command.
Input
The first line contains an integer n (1 ≤ n ≤ 10^6) — the number of commands.
The second line contains s — a sequence of commands. The string s consists of n characters. It is guaranteed that all characters in a string are valid commands.
Output
In a single line print n integers, where the i-th number is:
* -1 if the line received after processing the first i commands is not valid text,
* the minimal number of colors in the case of the correct text.
Examples
Input
11
(RaRbR)L)L(
Output
-1 -1 -1 -1 -1 -1 1 1 -1 -1 2
Input
11
(R)R(R)Ra)c
Output
-1 -1 1 1 -1 -1 1 1 1 -1 1
Note
In the first example, the text in the editor will take the following form:
1. (
^
2. (
^
3. (a
^
4. (a
^
5. (ab
^
6. (ab
^
7. (ab)
^
8. (ab)
^
9. (a))
^
10. (a))
^
11. (())
^ | instruction | 0 | 33,452 | 21 | 66,904 |
Tags: data structures, implementation
Correct Solution:
```
import os, sys, math
def solve(seq):
result = []
# +d, min +d, max +d
max_depth_allowed = min(20, int(math.log2(max(1, seq.count('R')))) + 1)
dbase_delta_d = [ 0 ] * (2 ** (max_depth_allowed + 1))
dbase_min_d = [ 0 ] * (2 ** (max_depth_allowed + 1))
dbase_max_d = [ 0 ] * (2 ** (max_depth_allowed + 1))
element_offset = 2 ** max_depth_allowed
cursor_position = 0
def print_db():
index = 1
depth = 0
while index < len(dbase):
size = 2 ** depth
res = []
for x in range(index, index + size):
res.append(','.join(map(str, dbase[x])))
#res2 = []
#for q in dbase[x]:
# res2.append(str(q))
#res.append(','.join(res2))
txt = ' '.join(res)
print(f'{depth} {txt}')
index += size
depth += 1
def update_dbase(d):
db_index = element_offset + cursor_position
if dbase_delta_d[db_index] != d:
dbase_delta_d[db_index] = d
dbase_min_d[db_index] = min(0, d)
dbase_max_d[db_index] = max(0, d)
while True:
parent_db_index = db_index // 2
if parent_db_index == 0:
break
first_child_db_index = parent_db_index * 2
d1 = dbase_delta_d[first_child_db_index]
min_d1 = dbase_min_d[first_child_db_index]
max_d1 = dbase_max_d[first_child_db_index]
first_child_db_index += 1
d2 = dbase_delta_d[first_child_db_index]
min_d2 = dbase_min_d[first_child_db_index]
max_d2 = dbase_max_d[first_child_db_index]
new_d = d1 + d2
new_min_d = min(min_d1, d1 + min_d2)
new_max_d = max(max_d1, d1 + max_d2)
if new_d == dbase_delta_d[parent_db_index] and new_min_d == dbase_min_d[parent_db_index] and new_max_d == dbase_max_d[parent_db_index]:
break
dbase_delta_d[parent_db_index] = new_d
dbase_min_d[parent_db_index] = new_min_d
dbase_max_d[parent_db_index] = new_max_d
db_index = parent_db_index
for s in seq:
if 'a' <= s <= 'z':
update_dbase(0)
elif s == 'L':
if cursor_position > 0:
cursor_position -= 1
elif s == 'R':
cursor_position += 1
elif s == '(':
update_dbase(1)
elif s == ')':
update_dbase(-1)
final_d = dbase_delta_d[1]
min_d = dbase_min_d[1]
max_d = dbase_max_d[1]
colors = max_d if min_d == 0 and final_d == 0 else -1
result.append(colors)
return result
#res = solve('(R' + ('(R)R' * 2) + ')')
if os.path.exists('testing'):
name = os.path.basename(__file__)
if name.endswith('.py'):
name = name[:-3]
src = open(name + '.in.txt', encoding='utf8')
input = src.readline
num = int(input().strip())
seq = input().strip()
res = solve(seq)
print(' '.join(map(str, res)))
``` | output | 1 | 33,452 | 21 | 66,905 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The development of a text editor is a hard problem. You need to implement an extra module for brackets coloring in text.
Your editor consists of a line with infinite length and cursor, which points to the current character. Please note that it points to only one of the characters (and not between a pair of characters). Thus, it points to an index character. The user can move the cursor left or right one position. If the cursor is already at the first (leftmost) position, then it does not move left.
Initially, the cursor is in the first (leftmost) character.
Also, the user can write a letter or brackets (either (, or )) to the position that the cursor is currently pointing at. A new character always overwrites the old value at that position.
Your editor must check, whether the current line is the correct text. Text is correct if the brackets in them form the correct bracket sequence.
Formally, correct text (CT) must satisfy the following rules:
* any line without brackets is CT (the line can contain whitespaces);
* If the first character of the string — is (, the last — is ), and all the rest form a CT, then the whole line is a CT;
* two consecutively written CT is also CT.
Examples of correct texts: hello(codeforces), round, ((i)(write))edi(tor)s, ( me). Examples of incorrect texts: hello)oops(, round), ((me).
The user uses special commands to work with your editor. Each command has its symbol, which must be written to execute this command.
The correspondence of commands and characters is as follows:
* L — move the cursor one character to the left (remains in place if it already points to the first character);
* R — move the cursor one character to the right;
* any lowercase Latin letter or bracket (( or )) — write the entered character to the position where the cursor is now.
For a complete understanding, take a look at the first example and its illustrations in the note below.
You are given a string containing the characters that the user entered. For the brackets coloring module's work, after each command you need to:
* check if the current text in the editor is a correct text;
* if it is, print the least number of colors that required, to color all brackets.
If two pairs of brackets are nested (the first in the second or vice versa), then these pairs of brackets should be painted in different colors. If two pairs of brackets are not nested, then they can be painted in different or the same colors. For example, for the bracket sequence ()(())()() the least number of colors is 2, and for the bracket sequence (()(()())())(()) — is 3.
Write a program that prints the minimal number of colors after processing each command.
Input
The first line contains an integer n (1 ≤ n ≤ 10^6) — the number of commands.
The second line contains s — a sequence of commands. The string s consists of n characters. It is guaranteed that all characters in a string are valid commands.
Output
In a single line print n integers, where the i-th number is:
* -1 if the line received after processing the first i commands is not valid text,
* the minimal number of colors in the case of the correct text.
Examples
Input
11
(RaRbR)L)L(
Output
-1 -1 -1 -1 -1 -1 1 1 -1 -1 2
Input
11
(R)R(R)Ra)c
Output
-1 -1 1 1 -1 -1 1 1 1 -1 1
Note
In the first example, the text in the editor will take the following form:
1. (
^
2. (
^
3. (a
^
4. (a
^
5. (ab
^
6. (ab
^
7. (ab)
^
8. (ab)
^
9. (a))
^
10. (a))
^
11. (())
^ | instruction | 0 | 33,453 | 21 | 66,906 |
Tags: data structures, implementation
Correct Solution:
```
'''input
11
(R)R(R)Ra)c
'''
# https://codeforces.com/blog/entry/18051
# Min = minimum prefix sum instead of minimum value
class SegmentTree:
def __init__(self, n, arr=[]):
self.n = n
self.tsum = [0] * (2 * n)
self.tmin = [0] * (2 * n)
self.tmax = [0] * (2 * n)
if arr:
for i in range(len(arr)):
self.tsum[n + i] = arr[i];
for i in range(len(arr) - 1, 0, -1) :
self.tsum[i] = self.tsum[i<<1] + self.tsum[i<<1|1];
def update(self, p, val):
p += self.n;
self.tsum[p] = val;
self.tmin[p] = val;
self.tmax[p] = val;
i = p;
while i > 1:
par = i>>1 # parent
if i & 1: # i is a right child (odd index)
self.tsum[par] = self.tsum[i] + self.tsum[i^1];
self.tmin[par] = min(self.tmin[i^1], self.tmin[i] + self.tsum[i^1]) # i^1 = other child of i's parent
self.tmax[par] = max(self.tmax[i^1], self.tmax[i] + self.tsum[i^1])
else: # i is a left child
self.tsum[par] = self.tsum[i] + self.tsum[i^1];
self.tmin[par] = min(self.tmin[i], self.tmin[i^1] + self.tsum[i])
self.tmax[par] = max(self.tmax[i], self.tmax[i^1] + self.tsum[i])
i >>= 1;
'''
int query(int l, int r) { // sum on interval [l, r)
int res = 0;
for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
if (l&1) res += t[l++];
if (r&1) res += t[--r];
}
return res;
}
'''
'''
import math
array = [1,3,5,7,9,11]
n = 2 ** math.ceil(math.log(len(array), 2))
st = SegmentTree(n, array)
st.update(0, 2)
'''
from sys import stdin
import math
def input():
return stdin.readline()[:-1]
def main():
n = int(input())
s = input()
#n = 2 ** math.ceil(math.log(n, 2))
n = 1048576
st = SegmentTree(n)
maxit = -1
currentit = 0
output = []
for c in s:
if c == 'L':
currentit = max(0, currentit - 1)
elif c == 'R':
currentit += 1
else:
maxit = max(maxit, currentit)
if c == '(':
st.update(currentit, 1)
elif c == ')':
st.update(currentit, -1)
else:
st.update(currentit, 0)
vmax = st.tmax[1]
vmin = st.tmin[1]
vsum = st.tsum[1]
if vmin >= 0 and vsum == 0:
output.append(vmax)
else:
output.append(-1)
print(' '.join(map(str, output)))
if __name__ == '__main__':
main()
``` | output | 1 | 33,453 | 21 | 66,907 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The development of a text editor is a hard problem. You need to implement an extra module for brackets coloring in text.
Your editor consists of a line with infinite length and cursor, which points to the current character. Please note that it points to only one of the characters (and not between a pair of characters). Thus, it points to an index character. The user can move the cursor left or right one position. If the cursor is already at the first (leftmost) position, then it does not move left.
Initially, the cursor is in the first (leftmost) character.
Also, the user can write a letter or brackets (either (, or )) to the position that the cursor is currently pointing at. A new character always overwrites the old value at that position.
Your editor must check, whether the current line is the correct text. Text is correct if the brackets in them form the correct bracket sequence.
Formally, correct text (CT) must satisfy the following rules:
* any line without brackets is CT (the line can contain whitespaces);
* If the first character of the string — is (, the last — is ), and all the rest form a CT, then the whole line is a CT;
* two consecutively written CT is also CT.
Examples of correct texts: hello(codeforces), round, ((i)(write))edi(tor)s, ( me). Examples of incorrect texts: hello)oops(, round), ((me).
The user uses special commands to work with your editor. Each command has its symbol, which must be written to execute this command.
The correspondence of commands and characters is as follows:
* L — move the cursor one character to the left (remains in place if it already points to the first character);
* R — move the cursor one character to the right;
* any lowercase Latin letter or bracket (( or )) — write the entered character to the position where the cursor is now.
For a complete understanding, take a look at the first example and its illustrations in the note below.
You are given a string containing the characters that the user entered. For the brackets coloring module's work, after each command you need to:
* check if the current text in the editor is a correct text;
* if it is, print the least number of colors that required, to color all brackets.
If two pairs of brackets are nested (the first in the second or vice versa), then these pairs of brackets should be painted in different colors. If two pairs of brackets are not nested, then they can be painted in different or the same colors. For example, for the bracket sequence ()(())()() the least number of colors is 2, and for the bracket sequence (()(()())())(()) — is 3.
Write a program that prints the minimal number of colors after processing each command.
Input
The first line contains an integer n (1 ≤ n ≤ 10^6) — the number of commands.
The second line contains s — a sequence of commands. The string s consists of n characters. It is guaranteed that all characters in a string are valid commands.
Output
In a single line print n integers, where the i-th number is:
* -1 if the line received after processing the first i commands is not valid text,
* the minimal number of colors in the case of the correct text.
Examples
Input
11
(RaRbR)L)L(
Output
-1 -1 -1 -1 -1 -1 1 1 -1 -1 2
Input
11
(R)R(R)Ra)c
Output
-1 -1 1 1 -1 -1 1 1 1 -1 1
Note
In the first example, the text in the editor will take the following form:
1. (
^
2. (
^
3. (a
^
4. (a
^
5. (ab
^
6. (ab
^
7. (ab)
^
8. (ab)
^
9. (a))
^
10. (a))
^
11. (())
^ | instruction | 0 | 33,454 | 21 | 66,908 |
Tags: data structures, implementation
Correct Solution:
```
# reproduction of solution 66039386 by @windhunterSB
import sys
# inf = open('input.txt', 'r')
# reader = (line.rstrip() for line in inf)
reader = (s.rstrip() for s in sys.stdin)
n = int(next(reader))
operations = next(reader)
# inf.close()
left_sum = [0 for i in range(n + 2)]
right_sum = [0 for i in range(n + 2)]
left_min = [0 for i in range(n + 2)]
right_min = [0 for i in range(n + 2)]
left_max = [0 for i in range(n + 2)]
right_max = [0 for i in range(n + 2)]
# left_sum = [0] * (n + 2) # for s[:i+1]
# right_sum = [0] * (n + 2) # for s[i:][::-1]
# left_min = [0] * (n + 2) # min - validation of bracket sequence
# right_min = [0] * (n + 2)
# left_max = [0] * (n + 2) # max - depth of bracket sequence
# right_max = [0] * (n + 2)
text = [0] * (n + 2) # entered text, letters marked as 0
op_map = {'(': 1,
')': -1}
ans = []
i = 1 # cursor loc i >= 1
for op in operations:
if op == 'L':
i = max(1, i - 1)
elif op == 'R':
i += 1
else:
text[i] = op_map.get(op, 0)
left_sum[i] = left_sum[i - 1] + text[i]
left_min[i] = min(left_min[i - 1], left_sum[i])
left_max[i] = max(left_max[i - 1], left_sum[i])
right_sum[i] = right_sum[i + 1] - text[i] # -text[i] cause of symmetry
right_min[i] = min(right_min[i + 1], right_sum[i])
right_max[i] = max(right_max[i + 1], right_sum[i])
correct = left_min[i] >= 0 and right_min[i + 1] >= 0 and left_sum[i] == right_sum[i + 1]
status = max(left_max[i], right_max[i + 1]) if correct else -1
ans.append(status)
print(*ans)
``` | output | 1 | 33,454 | 21 | 66,909 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The development of a text editor is a hard problem. You need to implement an extra module for brackets coloring in text.
Your editor consists of a line with infinite length and cursor, which points to the current character. Please note that it points to only one of the characters (and not between a pair of characters). Thus, it points to an index character. The user can move the cursor left or right one position. If the cursor is already at the first (leftmost) position, then it does not move left.
Initially, the cursor is in the first (leftmost) character.
Also, the user can write a letter or brackets (either (, or )) to the position that the cursor is currently pointing at. A new character always overwrites the old value at that position.
Your editor must check, whether the current line is the correct text. Text is correct if the brackets in them form the correct bracket sequence.
Formally, correct text (CT) must satisfy the following rules:
* any line without brackets is CT (the line can contain whitespaces);
* If the first character of the string — is (, the last — is ), and all the rest form a CT, then the whole line is a CT;
* two consecutively written CT is also CT.
Examples of correct texts: hello(codeforces), round, ((i)(write))edi(tor)s, ( me). Examples of incorrect texts: hello)oops(, round), ((me).
The user uses special commands to work with your editor. Each command has its symbol, which must be written to execute this command.
The correspondence of commands and characters is as follows:
* L — move the cursor one character to the left (remains in place if it already points to the first character);
* R — move the cursor one character to the right;
* any lowercase Latin letter or bracket (( or )) — write the entered character to the position where the cursor is now.
For a complete understanding, take a look at the first example and its illustrations in the note below.
You are given a string containing the characters that the user entered. For the brackets coloring module's work, after each command you need to:
* check if the current text in the editor is a correct text;
* if it is, print the least number of colors that required, to color all brackets.
If two pairs of brackets are nested (the first in the second or vice versa), then these pairs of brackets should be painted in different colors. If two pairs of brackets are not nested, then they can be painted in different or the same colors. For example, for the bracket sequence ()(())()() the least number of colors is 2, and for the bracket sequence (()(()())())(()) — is 3.
Write a program that prints the minimal number of colors after processing each command.
Input
The first line contains an integer n (1 ≤ n ≤ 10^6) — the number of commands.
The second line contains s — a sequence of commands. The string s consists of n characters. It is guaranteed that all characters in a string are valid commands.
Output
In a single line print n integers, where the i-th number is:
* -1 if the line received after processing the first i commands is not valid text,
* the minimal number of colors in the case of the correct text.
Examples
Input
11
(RaRbR)L)L(
Output
-1 -1 -1 -1 -1 -1 1 1 -1 -1 2
Input
11
(R)R(R)Ra)c
Output
-1 -1 1 1 -1 -1 1 1 1 -1 1
Note
In the first example, the text in the editor will take the following form:
1. (
^
2. (
^
3. (a
^
4. (a
^
5. (ab
^
6. (ab
^
7. (ab)
^
8. (ab)
^
9. (a))
^
10. (a))
^
11. (())
^ | instruction | 0 | 33,455 | 21 | 66,910 |
Tags: data structures, implementation
Correct Solution:
```
# reproduction of solution № 66039386 by @windhunterSB
import sys
# inf = open('input.txt', 'r')
# reader = (line.rstrip() for line in inf)
reader = (s.rstrip() for s in sys.stdin)
n = int(next(reader))
operations = next(reader)
# inf.close()
left_sum = [0] * (n + 2) # for s[:i+1]
right_sum = [0] * (n + 2) # for s[i:][::-1]
left_min = [0] * (n + 2) # min - validation of bracket sequence
right_min = [0] * (n + 2)
left_max = [0] * (n + 2) # max - depth of bracket sequence
right_max = [0] * (n + 2)
text = [0] * (n + 2) # entered text, letters marked as 0
op_map = {'(': 1,
')': -1}
ans = []
i = 1 # cursor loc i >= 1
for op in operations:
if op == 'L':
i = max(1, i - 1)
elif op == 'R':
i += 1
else:
text[i] = op_map.get(op, 0)
left_sum[i] = left_sum[i - 1] + text[i]
left_min[i] = min(left_min[i - 1], left_sum[i])
left_max[i] = max(left_max[i - 1], left_sum[i])
right_sum[i] = right_sum[i + 1] - text[i] # -text[i] cause of symmetry
right_min[i] = min(right_min[i + 1], right_sum[i])
right_max[i] = max(right_max[i + 1], right_sum[i])
correct = left_min[i] >= 0 and right_min[i + 1] >= 0 and left_sum[i] == right_sum[i + 1]
status = max(left_max[i], right_max[i + 1]) if correct else -1
ans.append(status)
print(' '.join(map(str, ans)))
``` | output | 1 | 33,455 | 21 | 66,911 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The development of a text editor is a hard problem. You need to implement an extra module for brackets coloring in text.
Your editor consists of a line with infinite length and cursor, which points to the current character. Please note that it points to only one of the characters (and not between a pair of characters). Thus, it points to an index character. The user can move the cursor left or right one position. If the cursor is already at the first (leftmost) position, then it does not move left.
Initially, the cursor is in the first (leftmost) character.
Also, the user can write a letter or brackets (either (, or )) to the position that the cursor is currently pointing at. A new character always overwrites the old value at that position.
Your editor must check, whether the current line is the correct text. Text is correct if the brackets in them form the correct bracket sequence.
Formally, correct text (CT) must satisfy the following rules:
* any line without brackets is CT (the line can contain whitespaces);
* If the first character of the string — is (, the last — is ), and all the rest form a CT, then the whole line is a CT;
* two consecutively written CT is also CT.
Examples of correct texts: hello(codeforces), round, ((i)(write))edi(tor)s, ( me). Examples of incorrect texts: hello)oops(, round), ((me).
The user uses special commands to work with your editor. Each command has its symbol, which must be written to execute this command.
The correspondence of commands and characters is as follows:
* L — move the cursor one character to the left (remains in place if it already points to the first character);
* R — move the cursor one character to the right;
* any lowercase Latin letter or bracket (( or )) — write the entered character to the position where the cursor is now.
For a complete understanding, take a look at the first example and its illustrations in the note below.
You are given a string containing the characters that the user entered. For the brackets coloring module's work, after each command you need to:
* check if the current text in the editor is a correct text;
* if it is, print the least number of colors that required, to color all brackets.
If two pairs of brackets are nested (the first in the second or vice versa), then these pairs of brackets should be painted in different colors. If two pairs of brackets are not nested, then they can be painted in different or the same colors. For example, for the bracket sequence ()(())()() the least number of colors is 2, and for the bracket sequence (()(()())())(()) — is 3.
Write a program that prints the minimal number of colors after processing each command.
Input
The first line contains an integer n (1 ≤ n ≤ 10^6) — the number of commands.
The second line contains s — a sequence of commands. The string s consists of n characters. It is guaranteed that all characters in a string are valid commands.
Output
In a single line print n integers, where the i-th number is:
* -1 if the line received after processing the first i commands is not valid text,
* the minimal number of colors in the case of the correct text.
Examples
Input
11
(RaRbR)L)L(
Output
-1 -1 -1 -1 -1 -1 1 1 -1 -1 2
Input
11
(R)R(R)Ra)c
Output
-1 -1 1 1 -1 -1 1 1 1 -1 1
Note
In the first example, the text in the editor will take the following form:
1. (
^
2. (
^
3. (a
^
4. (a
^
5. (ab
^
6. (ab
^
7. (ab)
^
8. (ab)
^
9. (a))
^
10. (a))
^
11. (())
^ | instruction | 0 | 33,456 | 21 | 66,912 |
Tags: data structures, implementation
Correct Solution:
```
# reproduction of solution № 66039386 by @windhunterSB
import sys
# inf = open('input.txt', 'r')
# reader = (line.rstrip() for line in inf)
reader = (s.rstrip() for s in sys.stdin)
n = int(next(reader))
operations = next(reader)
# inf.close()
left_sum = [0] * (n + 2) # for s[:i+1]
right_sum = [0] * (n + 2) # for s[i:][::-1]
left_min = [0] * (n + 2) # min - validation of bracket sequence
right_min = [0] * (n + 2)
left_max = [0] * (n + 2) # max - depth of bracket sequence
right_max = [0] * (n + 2)
text = [0] * (n + 2) # entered text, letters marked as 0
op_map = {'(': 1,
')': -1}
ans = []
i = 1 # cursor loc i >= 1
for op in operations:
if op == 'L':
i = max(1, i - 1)
elif op == 'R':
i += 1
else:
text[i] = op_map.get(op, 0)
left_sum[i] = left_sum[i - 1] + text[i]
left_min[i] = min(left_min[i - 1], left_sum[i])
left_max[i] = max(left_max[i - 1], left_sum[i])
right_sum[i] = right_sum[i + 1] - text[i] # -text[i] cause of symmetry
right_min[i] = min(right_min[i + 1], right_sum[i])
right_max[i] = max(right_max[i + 1], right_sum[i])
correct = left_min[i] >= 0 and right_min[i + 1] >= 0 and left_sum[i] == right_sum[i + 1]
status = max(left_max[i], right_max[i + 1]) if correct else -1
ans.append(status)
print(*ans)
``` | output | 1 | 33,456 | 21 | 66,913 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The development of a text editor is a hard problem. You need to implement an extra module for brackets coloring in text.
Your editor consists of a line with infinite length and cursor, which points to the current character. Please note that it points to only one of the characters (and not between a pair of characters). Thus, it points to an index character. The user can move the cursor left or right one position. If the cursor is already at the first (leftmost) position, then it does not move left.
Initially, the cursor is in the first (leftmost) character.
Also, the user can write a letter or brackets (either (, or )) to the position that the cursor is currently pointing at. A new character always overwrites the old value at that position.
Your editor must check, whether the current line is the correct text. Text is correct if the brackets in them form the correct bracket sequence.
Formally, correct text (CT) must satisfy the following rules:
* any line without brackets is CT (the line can contain whitespaces);
* If the first character of the string — is (, the last — is ), and all the rest form a CT, then the whole line is a CT;
* two consecutively written CT is also CT.
Examples of correct texts: hello(codeforces), round, ((i)(write))edi(tor)s, ( me). Examples of incorrect texts: hello)oops(, round), ((me).
The user uses special commands to work with your editor. Each command has its symbol, which must be written to execute this command.
The correspondence of commands and characters is as follows:
* L — move the cursor one character to the left (remains in place if it already points to the first character);
* R — move the cursor one character to the right;
* any lowercase Latin letter or bracket (( or )) — write the entered character to the position where the cursor is now.
For a complete understanding, take a look at the first example and its illustrations in the note below.
You are given a string containing the characters that the user entered. For the brackets coloring module's work, after each command you need to:
* check if the current text in the editor is a correct text;
* if it is, print the least number of colors that required, to color all brackets.
If two pairs of brackets are nested (the first in the second or vice versa), then these pairs of brackets should be painted in different colors. If two pairs of brackets are not nested, then they can be painted in different or the same colors. For example, for the bracket sequence ()(())()() the least number of colors is 2, and for the bracket sequence (()(()())())(()) — is 3.
Write a program that prints the minimal number of colors after processing each command.
Input
The first line contains an integer n (1 ≤ n ≤ 10^6) — the number of commands.
The second line contains s — a sequence of commands. The string s consists of n characters. It is guaranteed that all characters in a string are valid commands.
Output
In a single line print n integers, where the i-th number is:
* -1 if the line received after processing the first i commands is not valid text,
* the minimal number of colors in the case of the correct text.
Examples
Input
11
(RaRbR)L)L(
Output
-1 -1 -1 -1 -1 -1 1 1 -1 -1 2
Input
11
(R)R(R)Ra)c
Output
-1 -1 1 1 -1 -1 1 1 1 -1 1
Note
In the first example, the text in the editor will take the following form:
1. (
^
2. (
^
3. (a
^
4. (a
^
5. (ab
^
6. (ab
^
7. (ab)
^
8. (ab)
^
9. (a))
^
10. (a))
^
11. (())
^ | instruction | 0 | 33,457 | 21 | 66,914 |
Tags: data structures, implementation
Correct Solution:
```
'''input
11
(R)R(R)Ra)c
'''
# MODIFIED SEGMENT TREE (MIN = MINIMUM PREFIX SUM instead of MINIMUM ELEMENT IN PREFIX)
class SegmentTree:
def __init__(self, n, arr=[]):
self.n = n
self.tsum = [0] * (2 * n)
self.tmin = [0] * (2 * n)
self.tmax = [0] * (2 * n)
if arr:
for i in range(len(arr)):
self.tsum[n + i] = arr[i];
for i in range(len(arr) - 1, 0, -1) :
self.tsum[i] = self.tsum[i << 1] + self.tsum[i << 1 | 1];
def update(self, p, val):
p += self.n;
self.tsum[p] = val;
self.tmin[p] = val;
self.tmax[p] = val;
i = p;
while i > 1:
self.tsum[i >> 1] = self.tsum[i] + self.tsum[i ^ 1];
self.tmin[i >> 1] = min(self.tmin[i], self.tmin[i ^ 1] + self.tsum[i]) if i%2==0 else min(self.tmin[i^1], self.tmin[i] + self.tsum[i^1])
self.tmax[i >> 1] = max(self.tmax[i], self.tmax[i ^ 1] + self.tsum[i]) if i%2==0 else max(self.tmax[i^1], self.tmax[i] + self.tsum[i^1])
i >>= 1;
'''
import math
array = [1,3,5,7,9,11]
n = 2 ** math.ceil(math.log(len(array), 2))
st = SegmentTree(n, array)
st.update(0, 2)
'''
from sys import stdin
import math
def input():
return stdin.readline()[:-1]
n = int(input())
s = input()
#n = 2 ** math.ceil(math.log(n, 2))
n = 1048576
st = SegmentTree(n)
maxit = -1
currentit = 0
output = []
for c in s:
if c == 'L':
currentit = max(0, currentit - 1)
elif c == 'R':
currentit += 1
else:
maxit = max(maxit, currentit)
if c == '(':
st.update(currentit, 1)
elif c == ')':
st.update(currentit, -1)
else:
st.update(currentit, 0)
vmax = st.tmax[1]
vmin = st.tmin[1]
vsum = st.tsum[1]
if vmin >= 0 and vsum == 0:
output.append(vmax)
else:
output.append(-1)
print(' '.join(map(str, output)))
``` | output | 1 | 33,457 | 21 | 66,915 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The development of a text editor is a hard problem. You need to implement an extra module for brackets coloring in text.
Your editor consists of a line with infinite length and cursor, which points to the current character. Please note that it points to only one of the characters (and not between a pair of characters). Thus, it points to an index character. The user can move the cursor left or right one position. If the cursor is already at the first (leftmost) position, then it does not move left.
Initially, the cursor is in the first (leftmost) character.
Also, the user can write a letter or brackets (either (, or )) to the position that the cursor is currently pointing at. A new character always overwrites the old value at that position.
Your editor must check, whether the current line is the correct text. Text is correct if the brackets in them form the correct bracket sequence.
Formally, correct text (CT) must satisfy the following rules:
* any line without brackets is CT (the line can contain whitespaces);
* If the first character of the string — is (, the last — is ), and all the rest form a CT, then the whole line is a CT;
* two consecutively written CT is also CT.
Examples of correct texts: hello(codeforces), round, ((i)(write))edi(tor)s, ( me). Examples of incorrect texts: hello)oops(, round), ((me).
The user uses special commands to work with your editor. Each command has its symbol, which must be written to execute this command.
The correspondence of commands and characters is as follows:
* L — move the cursor one character to the left (remains in place if it already points to the first character);
* R — move the cursor one character to the right;
* any lowercase Latin letter or bracket (( or )) — write the entered character to the position where the cursor is now.
For a complete understanding, take a look at the first example and its illustrations in the note below.
You are given a string containing the characters that the user entered. For the brackets coloring module's work, after each command you need to:
* check if the current text in the editor is a correct text;
* if it is, print the least number of colors that required, to color all brackets.
If two pairs of brackets are nested (the first in the second or vice versa), then these pairs of brackets should be painted in different colors. If two pairs of brackets are not nested, then they can be painted in different or the same colors. For example, for the bracket sequence ()(())()() the least number of colors is 2, and for the bracket sequence (()(()())())(()) — is 3.
Write a program that prints the minimal number of colors after processing each command.
Input
The first line contains an integer n (1 ≤ n ≤ 10^6) — the number of commands.
The second line contains s — a sequence of commands. The string s consists of n characters. It is guaranteed that all characters in a string are valid commands.
Output
In a single line print n integers, where the i-th number is:
* -1 if the line received after processing the first i commands is not valid text,
* the minimal number of colors in the case of the correct text.
Examples
Input
11
(RaRbR)L)L(
Output
-1 -1 -1 -1 -1 -1 1 1 -1 -1 2
Input
11
(R)R(R)Ra)c
Output
-1 -1 1 1 -1 -1 1 1 1 -1 1
Note
In the first example, the text in the editor will take the following form:
1. (
^
2. (
^
3. (a
^
4. (a
^
5. (ab
^
6. (ab
^
7. (ab)
^
8. (ab)
^
9. (a))
^
10. (a))
^
11. (())
^ | instruction | 0 | 33,458 | 21 | 66,916 |
Tags: data structures, implementation
Correct Solution:
```
import traceback
def push(stack, delta):
if stack:
_, d, mn, mx = stack[-1]
else:
d, mn, mx = 0, 0, 0
stack.append((delta, d + delta, min(d + delta, mn), max(d + delta, mx)))
def main():
n = int(input())
ss = input()
left = [(0, 0,0,0)]
right = [(0, 0,0,0)] * n
res = []
try:
for s in ss:
if s == 'R':
delta = right.pop()[0]
push(left, -delta)
elif s == 'L':
if len(left) > 1:
delta = left.pop()[0]
push(right, -delta)
else:
if left:
left.pop()
if s == '(':
delta = 1
elif s == ')':
delta = -1
else:
delta = 0
push(left, delta)
_, ld, lmn, lmx = left[-1]
_, rd, rmn, rmx = right[-1]
if ld == rd and lmn >= 0 and rmn >= 0:
res.append(max(ld, lmx, rmx))
else:
res.append(-1)
except Exception as e:
print(e)
traceback.print_exc()
print("size left %d size right %d"%(len(left), len(right)))
print(' '.join(map(str, res)))
if __name__ == "__main__":
main()
``` | output | 1 | 33,458 | 21 | 66,917 |
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