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 a correct Python 3 solution for this coding contest problem.
B: Parentheses Number
problem
Define the correct parenthesis string as follows:
* The empty string is the correct parenthesis string
* For the correct parenthesis string S, `(` S `)` is the correct parenthesis string
* For correct parentheses S, T ST is the correct parentheses
Here, the permutations are associated with the correct parentheses according to the following rules.
* When the i-th closing brace corresponds to the j-th opening brace, the i-th value in the sequence is j.
Given a permutation of length n, P = (p_1, p_2, $ \ ldots $, p_n), restore the corresponding parenthesis string.
However, if the parenthesis string corresponding to the permutation does not exist, output `: (`.
Input format
n
p_1 p_2 $ \ ldots $ p_n
The number of permutations n is given in the first row.
Permutations p_1, p_2, $ \ ldots $, p_i, $ \ ldots $, p_n are given in the second row, separated by blanks.
Constraint
* 1 \ leq n \ leq 10 ^ 5
* 1 \ leq p_i \ leq n
* All inputs are integers.
* P = (p_1, p_2, $ \ ldots $, p_n) is a permutation.
Output format
Output the parenthesized string corresponding to the permutation.
Print `: (` if such a parenthesis string does not exist.
Input example 1
2
twenty one
Output example 1
(())
Input example 2
Ten
1 2 3 4 5 6 7 8 9 10
Output example 2
() () () () () () () () () ()
Input example 3
3
3 1 2
Output example 3
:(
Example
Input
2
2 1
Output
(()) | instruction | 0 | 7,501 | 21 | 15,002 |
"Correct Solution:
```
n = int(input())
a = list(map(int,input().split()))
cnt = 0
ans = ''
st = list()
for x in a:
while cnt < x:
ans += '('
cnt += 1
st.append(cnt)
if st[-1] == x:
st.pop()
ans += ')'
else:
ans = ':('
break
print(ans)
``` | output | 1 | 7,501 | 21 | 15,003 |
Provide a correct Python 3 solution for this coding contest problem.
B: Parentheses Number
problem
Define the correct parenthesis string as follows:
* The empty string is the correct parenthesis string
* For the correct parenthesis string S, `(` S `)` is the correct parenthesis string
* For correct parentheses S, T ST is the correct parentheses
Here, the permutations are associated with the correct parentheses according to the following rules.
* When the i-th closing brace corresponds to the j-th opening brace, the i-th value in the sequence is j.
Given a permutation of length n, P = (p_1, p_2, $ \ ldots $, p_n), restore the corresponding parenthesis string.
However, if the parenthesis string corresponding to the permutation does not exist, output `: (`.
Input format
n
p_1 p_2 $ \ ldots $ p_n
The number of permutations n is given in the first row.
Permutations p_1, p_2, $ \ ldots $, p_i, $ \ ldots $, p_n are given in the second row, separated by blanks.
Constraint
* 1 \ leq n \ leq 10 ^ 5
* 1 \ leq p_i \ leq n
* All inputs are integers.
* P = (p_1, p_2, $ \ ldots $, p_n) is a permutation.
Output format
Output the parenthesized string corresponding to the permutation.
Print `: (` if such a parenthesis string does not exist.
Input example 1
2
twenty one
Output example 1
(())
Input example 2
Ten
1 2 3 4 5 6 7 8 9 10
Output example 2
() () () () () () () () () ()
Input example 3
3
3 1 2
Output example 3
:(
Example
Input
2
2 1
Output
(()) | instruction | 0 | 7,502 | 21 | 15,004 |
"Correct Solution:
```
from collections import deque
import sys
sys.setrecursionlimit(10**7)
N = int(input())
P = list(map(int,input().split()))
Q = [-1]*N
for i,p in enumerate(P):
Q[p-1] = i
S = ''
a = 1
def rec(l,r):
global S,a
if l==r:
return
while l < r:
ai = Q[a-1]
if ai >= r:
print(':(')
exit()
a += 1
S += '('
rec(l,ai)
S += ')'
l = ai + 1
rec(0,N)
print(S)
``` | output | 1 | 7,502 | 21 | 15,005 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are fed up with your messy room, so you decided to clean it up.
Your room is a bracket sequence s=s_{1}s_{2}... s_{n} of length n. Each character of this string is either an opening bracket '(' or a closing bracket ')'.
In one operation you can choose any consecutive substring of s and reverse it. In other words, you can choose any substring s[l ... r]=s_l, s_{l+1}, ..., s_r and change the order of elements in it into s_r, s_{r-1}, ..., s_{l}.
For example, if you will decide to reverse substring s[2 ... 4] of string s="((()))" it will be equal to s="()(())".
A regular (aka balanced) bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example, bracket sequences "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.
A prefix of a string s is a substring that starts at position 1. For example, for s="(())()" there are 6 prefixes: "(", "((", "(()", "(())", "(())(" and "(())()".
In your opinion, a neat and clean room s is a bracket sequence that:
* the whole string s is a regular bracket sequence;
* and there are exactly k prefixes of this sequence which are regular (including whole s itself).
For example, if k = 2, then "(())()" is a neat and clean room.
You want to use at most n operations to make your room neat and clean. Operations are applied one after another sequentially.
It is guaranteed that the answer exists. Note that you do not need to minimize the number of operations: find any way to achieve the desired configuration in n or less operations.
Input
The first line contains integer number t (1 β€ t β€ 100) β the number of test cases in the input. Then t test cases follow.
The first line of a test case contains two integers n and k (1 β€ k β€ n/2, 2 β€ n β€ 2000, n is even) β length of s and required number of regular prefixes.
The second line of a test case contains s of length n β the given bracket sequence. It contains only '(' and ')'.
It is guaranteed that there are exactly n/2 characters '(' and exactly n/2 characters ')' in the given string.
The sum of all values n over all the test cases in the input doesn't exceed 2000.
Output
For each test case print an answer.
In the first line print integer m (0 β€ m β€ n) β the number of operations. You do not need to minimize m, any value is suitable.
In the following m lines print description of the operations, each line should contain two integers l,r (1 β€ l β€ r β€ n), representing single reverse operation of s[l ... r]=s_{l}s_{l+1}... s_{r}. Operations are applied one after another sequentially.
The final s after all operations should be a regular, also it should be exactly k prefixes (including s) which are regular.
It is guaranteed that the answer exists. If there are several possible answers you can print any.
Example
Input
4
8 2
()(())()
10 3
))()()()((
2 1
()
2 1
)(
Output
4
3 4
1 1
5 8
2 2
3
4 10
1 4
6 7
0
1
1 2
Note
In the first example, the final sequence is "()(()())", where two prefixes are regular, "()" and "()(()())". Note, that all the operations except "5 8" in the example output are useless (they do not change s). | instruction | 0 | 7,671 | 21 | 15,342 |
Tags: constructive algorithms
Correct Solution:
```
for _ in range(int(input())):
n, k = tuple(map(int, input().split()))
s = list(input())
ans = list("()" * (k - 1) + "(" * ((n // 2) - k + 1) + ")" * (n // 2 - k + 1))
ops = []
i = 0
while ans != s and i < n:
# print("----" , i, "----")
if ans[i] != s[i]:
j = s[i:].index(ans[i]) + i
# print(0,"|",j, s[j], s[i])
ops.append(str(i + 1) + " " + str(j + 1))
for k in range(i, (j + i + 1) // 2):
# print(11, "|", j, s[k], s[j + i - k])
(s[k], s[j + i - k]) = (s[j + i - k], s[k])
# print(12, "|", j, s[k], s[j + i - k])
# print(" ".join(s))
# print(" ".join(ans))
# print("|".join(ops))
i += 1
print(len(ops))
if len(ops) != 0:
print("\n".join(ops))
``` | output | 1 | 7,671 | 21 | 15,343 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are fed up with your messy room, so you decided to clean it up.
Your room is a bracket sequence s=s_{1}s_{2}... s_{n} of length n. Each character of this string is either an opening bracket '(' or a closing bracket ')'.
In one operation you can choose any consecutive substring of s and reverse it. In other words, you can choose any substring s[l ... r]=s_l, s_{l+1}, ..., s_r and change the order of elements in it into s_r, s_{r-1}, ..., s_{l}.
For example, if you will decide to reverse substring s[2 ... 4] of string s="((()))" it will be equal to s="()(())".
A regular (aka balanced) bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example, bracket sequences "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.
A prefix of a string s is a substring that starts at position 1. For example, for s="(())()" there are 6 prefixes: "(", "((", "(()", "(())", "(())(" and "(())()".
In your opinion, a neat and clean room s is a bracket sequence that:
* the whole string s is a regular bracket sequence;
* and there are exactly k prefixes of this sequence which are regular (including whole s itself).
For example, if k = 2, then "(())()" is a neat and clean room.
You want to use at most n operations to make your room neat and clean. Operations are applied one after another sequentially.
It is guaranteed that the answer exists. Note that you do not need to minimize the number of operations: find any way to achieve the desired configuration in n or less operations.
Input
The first line contains integer number t (1 β€ t β€ 100) β the number of test cases in the input. Then t test cases follow.
The first line of a test case contains two integers n and k (1 β€ k β€ n/2, 2 β€ n β€ 2000, n is even) β length of s and required number of regular prefixes.
The second line of a test case contains s of length n β the given bracket sequence. It contains only '(' and ')'.
It is guaranteed that there are exactly n/2 characters '(' and exactly n/2 characters ')' in the given string.
The sum of all values n over all the test cases in the input doesn't exceed 2000.
Output
For each test case print an answer.
In the first line print integer m (0 β€ m β€ n) β the number of operations. You do not need to minimize m, any value is suitable.
In the following m lines print description of the operations, each line should contain two integers l,r (1 β€ l β€ r β€ n), representing single reverse operation of s[l ... r]=s_{l}s_{l+1}... s_{r}. Operations are applied one after another sequentially.
The final s after all operations should be a regular, also it should be exactly k prefixes (including s) which are regular.
It is guaranteed that the answer exists. If there are several possible answers you can print any.
Example
Input
4
8 2
()(())()
10 3
))()()()((
2 1
()
2 1
)(
Output
4
3 4
1 1
5 8
2 2
3
4 10
1 4
6 7
0
1
1 2
Note
In the first example, the final sequence is "()(()())", where two prefixes are regular, "()" and "()(()())". Note, that all the operations except "5 8" in the example output are useless (they do not change s). | instruction | 0 | 7,672 | 21 | 15,344 |
Tags: constructive algorithms
Correct Solution:
```
t = int(input())
for test_i in range(t):
n, k = map(int, input().split())
s = list(input())
ans = []
for i in range(k - 1):
if s[2 * i] != '(':
i0 = s.index('(', 2 * i)
ans.append((2 * i + 1, i0 + 1))
s[2 * i], s[i0] = '(', ')'
if s[2 * i + 1] != ')':
i0 = s.index(')', 2 * i + 1)
ans.append((2 * i + 2, i0 + 1))
s[2 * i + 1], s[i0] = ')', '('
for i in range(n // 2 - k + 1):
if s[2 * (k - 1) + i] != '(':
i0 = s.index('(', 2 * (k - 1) + i)
ans.append((2 * (k - 1) + i + 1, i0 + 1))
s[2 * (k - 1) + i], s[i0] = '(', ')'
print(len(ans))
for pair in ans:
print(*pair)
``` | output | 1 | 7,672 | 21 | 15,345 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are fed up with your messy room, so you decided to clean it up.
Your room is a bracket sequence s=s_{1}s_{2}... s_{n} of length n. Each character of this string is either an opening bracket '(' or a closing bracket ')'.
In one operation you can choose any consecutive substring of s and reverse it. In other words, you can choose any substring s[l ... r]=s_l, s_{l+1}, ..., s_r and change the order of elements in it into s_r, s_{r-1}, ..., s_{l}.
For example, if you will decide to reverse substring s[2 ... 4] of string s="((()))" it will be equal to s="()(())".
A regular (aka balanced) bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example, bracket sequences "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.
A prefix of a string s is a substring that starts at position 1. For example, for s="(())()" there are 6 prefixes: "(", "((", "(()", "(())", "(())(" and "(())()".
In your opinion, a neat and clean room s is a bracket sequence that:
* the whole string s is a regular bracket sequence;
* and there are exactly k prefixes of this sequence which are regular (including whole s itself).
For example, if k = 2, then "(())()" is a neat and clean room.
You want to use at most n operations to make your room neat and clean. Operations are applied one after another sequentially.
It is guaranteed that the answer exists. Note that you do not need to minimize the number of operations: find any way to achieve the desired configuration in n or less operations.
Input
The first line contains integer number t (1 β€ t β€ 100) β the number of test cases in the input. Then t test cases follow.
The first line of a test case contains two integers n and k (1 β€ k β€ n/2, 2 β€ n β€ 2000, n is even) β length of s and required number of regular prefixes.
The second line of a test case contains s of length n β the given bracket sequence. It contains only '(' and ')'.
It is guaranteed that there are exactly n/2 characters '(' and exactly n/2 characters ')' in the given string.
The sum of all values n over all the test cases in the input doesn't exceed 2000.
Output
For each test case print an answer.
In the first line print integer m (0 β€ m β€ n) β the number of operations. You do not need to minimize m, any value is suitable.
In the following m lines print description of the operations, each line should contain two integers l,r (1 β€ l β€ r β€ n), representing single reverse operation of s[l ... r]=s_{l}s_{l+1}... s_{r}. Operations are applied one after another sequentially.
The final s after all operations should be a regular, also it should be exactly k prefixes (including s) which are regular.
It is guaranteed that the answer exists. If there are several possible answers you can print any.
Example
Input
4
8 2
()(())()
10 3
))()()()((
2 1
()
2 1
)(
Output
4
3 4
1 1
5 8
2 2
3
4 10
1 4
6 7
0
1
1 2
Note
In the first example, the final sequence is "()(()())", where two prefixes are regular, "()" and "()(()())". Note, that all the operations except "5 8" in the example output are useless (they do not change s). | instruction | 0 | 7,673 | 21 | 15,346 |
Tags: constructive algorithms
Correct Solution:
```
for _ in range(int(input())):
n, k = map(int, input().split())
s = list(input().strip())
d = '()' * (k-1) + '('*(n//2-k+1) + ')'*(n//2-k+1)
res = []
for i in range(n):
if s[i] != d[i]:
j = s.index(d[i], i)
res.append((i, j))
s[i:j+1] = s[i:j+1][::-1]
print(len(res))
for u, v in res:
print(u+1, v+1)
``` | output | 1 | 7,673 | 21 | 15,347 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are fed up with your messy room, so you decided to clean it up.
Your room is a bracket sequence s=s_{1}s_{2}... s_{n} of length n. Each character of this string is either an opening bracket '(' or a closing bracket ')'.
In one operation you can choose any consecutive substring of s and reverse it. In other words, you can choose any substring s[l ... r]=s_l, s_{l+1}, ..., s_r and change the order of elements in it into s_r, s_{r-1}, ..., s_{l}.
For example, if you will decide to reverse substring s[2 ... 4] of string s="((()))" it will be equal to s="()(())".
A regular (aka balanced) bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example, bracket sequences "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.
A prefix of a string s is a substring that starts at position 1. For example, for s="(())()" there are 6 prefixes: "(", "((", "(()", "(())", "(())(" and "(())()".
In your opinion, a neat and clean room s is a bracket sequence that:
* the whole string s is a regular bracket sequence;
* and there are exactly k prefixes of this sequence which are regular (including whole s itself).
For example, if k = 2, then "(())()" is a neat and clean room.
You want to use at most n operations to make your room neat and clean. Operations are applied one after another sequentially.
It is guaranteed that the answer exists. Note that you do not need to minimize the number of operations: find any way to achieve the desired configuration in n or less operations.
Input
The first line contains integer number t (1 β€ t β€ 100) β the number of test cases in the input. Then t test cases follow.
The first line of a test case contains two integers n and k (1 β€ k β€ n/2, 2 β€ n β€ 2000, n is even) β length of s and required number of regular prefixes.
The second line of a test case contains s of length n β the given bracket sequence. It contains only '(' and ')'.
It is guaranteed that there are exactly n/2 characters '(' and exactly n/2 characters ')' in the given string.
The sum of all values n over all the test cases in the input doesn't exceed 2000.
Output
For each test case print an answer.
In the first line print integer m (0 β€ m β€ n) β the number of operations. You do not need to minimize m, any value is suitable.
In the following m lines print description of the operations, each line should contain two integers l,r (1 β€ l β€ r β€ n), representing single reverse operation of s[l ... r]=s_{l}s_{l+1}... s_{r}. Operations are applied one after another sequentially.
The final s after all operations should be a regular, also it should be exactly k prefixes (including s) which are regular.
It is guaranteed that the answer exists. If there are several possible answers you can print any.
Example
Input
4
8 2
()(())()
10 3
))()()()((
2 1
()
2 1
)(
Output
4
3 4
1 1
5 8
2 2
3
4 10
1 4
6 7
0
1
1 2
Note
In the first example, the final sequence is "()(()())", where two prefixes are regular, "()" and "()(()())". Note, that all the operations except "5 8" in the example output are useless (they do not change s). | instruction | 0 | 7,674 | 21 | 15,348 |
Tags: constructive algorithms
Correct Solution:
```
import sys
t=int(sys.stdin.readline())
def findopen(i,s,n):
for j in range(i,n):
if s[j]=='(':
return j
def findclose(i,s,n):
for j in range(i,n):
if s[j]==')':
return j
for _ in range(t):
n,k=map(int,sys.stdin.readline().split())
s=list(sys.stdin.readline()[:-1])
#print(s,'s')
ans=[]
left,right,i=-1,-1,0
i=0
rem=n//2-(k-1)
while k>1:
a=findopen(i,s,n)
#print(a,'a',i,'i',s,'s')
if a!=i:
ans.append([i+1,a+1])
s[i],s[a]=s[a],s[i]
b=findclose(i+1,s,n)
if b!=i+1:
ans.append([i+2,b+1])
s[i+1],s[b]=s[b],s[i+1]
k-=1
i+=2
#rem=n//2-(k)
#print(rem,'rem')
while rem>0:
a=findopen(i,s,n)
if a!=i:
ans.append([i+1,a+1])
s[i],s[a]=s[a],s[i]
rem-=1
i+=1
#print(s,'s')
m=len(ans)
print(m)
for i in range(m):
print(*ans[i])
``` | output | 1 | 7,674 | 21 | 15,349 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are fed up with your messy room, so you decided to clean it up.
Your room is a bracket sequence s=s_{1}s_{2}... s_{n} of length n. Each character of this string is either an opening bracket '(' or a closing bracket ')'.
In one operation you can choose any consecutive substring of s and reverse it. In other words, you can choose any substring s[l ... r]=s_l, s_{l+1}, ..., s_r and change the order of elements in it into s_r, s_{r-1}, ..., s_{l}.
For example, if you will decide to reverse substring s[2 ... 4] of string s="((()))" it will be equal to s="()(())".
A regular (aka balanced) bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example, bracket sequences "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.
A prefix of a string s is a substring that starts at position 1. For example, for s="(())()" there are 6 prefixes: "(", "((", "(()", "(())", "(())(" and "(())()".
In your opinion, a neat and clean room s is a bracket sequence that:
* the whole string s is a regular bracket sequence;
* and there are exactly k prefixes of this sequence which are regular (including whole s itself).
For example, if k = 2, then "(())()" is a neat and clean room.
You want to use at most n operations to make your room neat and clean. Operations are applied one after another sequentially.
It is guaranteed that the answer exists. Note that you do not need to minimize the number of operations: find any way to achieve the desired configuration in n or less operations.
Input
The first line contains integer number t (1 β€ t β€ 100) β the number of test cases in the input. Then t test cases follow.
The first line of a test case contains two integers n and k (1 β€ k β€ n/2, 2 β€ n β€ 2000, n is even) β length of s and required number of regular prefixes.
The second line of a test case contains s of length n β the given bracket sequence. It contains only '(' and ')'.
It is guaranteed that there are exactly n/2 characters '(' and exactly n/2 characters ')' in the given string.
The sum of all values n over all the test cases in the input doesn't exceed 2000.
Output
For each test case print an answer.
In the first line print integer m (0 β€ m β€ n) β the number of operations. You do not need to minimize m, any value is suitable.
In the following m lines print description of the operations, each line should contain two integers l,r (1 β€ l β€ r β€ n), representing single reverse operation of s[l ... r]=s_{l}s_{l+1}... s_{r}. Operations are applied one after another sequentially.
The final s after all operations should be a regular, also it should be exactly k prefixes (including s) which are regular.
It is guaranteed that the answer exists. If there are several possible answers you can print any.
Example
Input
4
8 2
()(())()
10 3
))()()()((
2 1
()
2 1
)(
Output
4
3 4
1 1
5 8
2 2
3
4 10
1 4
6 7
0
1
1 2
Note
In the first example, the final sequence is "()(()())", where two prefixes are regular, "()" and "()(()())". Note, that all the operations except "5 8" in the example output are useless (they do not change s). | instruction | 0 | 7,675 | 21 | 15,350 |
Tags: constructive algorithms
Correct Solution:
```
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
from fractions import Fraction
from collections import defaultdict
from itertools import permutations
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")
#-------------------game starts now-----------------------------------------------------
for _ in range (int(input())):
n,k=map(int,input().split())
s=list(input())
r=list()
cnt=0
ind=-1
for i in range (n):
if cnt==k-1:
ind=i
break
if i%2==0:
if s[i]=='(':
continue
j=i+1
for p in range (i+1,n):
if s[p]=='(':
j=p
break
temp=s[i:j+1]
r.append((i+1,j+1))
p=len(temp)-1
for j in range (i,j+1):
s[j]=temp[p]
p-=1
else:
cnt+=1
if s[i]==')':
continue
j=i+1
for p in range (i+1,n):
if s[p]==')':
j=p
break
temp=s[i:j+1]
r.append((i+1,j+1))
p=len(temp)-1
for j in range (i,j+1):
s[j]=temp[p]
p-=1
for i in range (ind,n):
if i<=(n+ind-1)//2:
if s[i]=='(':
continue
j=i+1
for p in range (i+1,n):
if s[p]=='(':
j=p
break
temp=s[i:j+1]
r.append((i+1,j+1))
p=len(temp)-1
for j in range (i,j+1):
s[j]=temp[p]
p-=1
else:
cnt+=1
if s[i]==')':
continue
j=i+1
for p in range (i+1,n):
if s[p]==')':
j=p
break
temp=s[i:j+1]
r.append((i+1,j+1))
p=len(temp)-1
for j in range (i,j+1):
s[j]=temp[p]
p-=1
#print(s)
print(len(r))
for i in range (len(r)):
print(*r[i])
``` | output | 1 | 7,675 | 21 | 15,351 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are fed up with your messy room, so you decided to clean it up.
Your room is a bracket sequence s=s_{1}s_{2}... s_{n} of length n. Each character of this string is either an opening bracket '(' or a closing bracket ')'.
In one operation you can choose any consecutive substring of s and reverse it. In other words, you can choose any substring s[l ... r]=s_l, s_{l+1}, ..., s_r and change the order of elements in it into s_r, s_{r-1}, ..., s_{l}.
For example, if you will decide to reverse substring s[2 ... 4] of string s="((()))" it will be equal to s="()(())".
A regular (aka balanced) bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example, bracket sequences "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.
A prefix of a string s is a substring that starts at position 1. For example, for s="(())()" there are 6 prefixes: "(", "((", "(()", "(())", "(())(" and "(())()".
In your opinion, a neat and clean room s is a bracket sequence that:
* the whole string s is a regular bracket sequence;
* and there are exactly k prefixes of this sequence which are regular (including whole s itself).
For example, if k = 2, then "(())()" is a neat and clean room.
You want to use at most n operations to make your room neat and clean. Operations are applied one after another sequentially.
It is guaranteed that the answer exists. Note that you do not need to minimize the number of operations: find any way to achieve the desired configuration in n or less operations.
Input
The first line contains integer number t (1 β€ t β€ 100) β the number of test cases in the input. Then t test cases follow.
The first line of a test case contains two integers n and k (1 β€ k β€ n/2, 2 β€ n β€ 2000, n is even) β length of s and required number of regular prefixes.
The second line of a test case contains s of length n β the given bracket sequence. It contains only '(' and ')'.
It is guaranteed that there are exactly n/2 characters '(' and exactly n/2 characters ')' in the given string.
The sum of all values n over all the test cases in the input doesn't exceed 2000.
Output
For each test case print an answer.
In the first line print integer m (0 β€ m β€ n) β the number of operations. You do not need to minimize m, any value is suitable.
In the following m lines print description of the operations, each line should contain two integers l,r (1 β€ l β€ r β€ n), representing single reverse operation of s[l ... r]=s_{l}s_{l+1}... s_{r}. Operations are applied one after another sequentially.
The final s after all operations should be a regular, also it should be exactly k prefixes (including s) which are regular.
It is guaranteed that the answer exists. If there are several possible answers you can print any.
Example
Input
4
8 2
()(())()
10 3
))()()()((
2 1
()
2 1
)(
Output
4
3 4
1 1
5 8
2 2
3
4 10
1 4
6 7
0
1
1 2
Note
In the first example, the final sequence is "()(()())", where two prefixes are regular, "()" and "()(()())". Note, that all the operations except "5 8" in the example output are useless (they do not change s). | instruction | 0 | 7,676 | 21 | 15,352 |
Tags: constructive algorithms
Correct Solution:
```
def openBracket(i):
global firstOpen, ans
ind = index[0][firstOpen]
a = s[i: ind + 1]
a.reverse()
#print(i + 1, ind + 1)
s[i: ind + 1] = a
ans += [[i + 1, ind + 1]]
firstOpen += 1
def closeBracket(i):
global firstClose, ans
ind = index[1][firstClose]
a = s[i: ind + 1]
a.reverse()
#print(i + 1, ind + 1)
ans += [[i + 1, ind + 1]]
s[i: ind + 1] = a
firstClose += 1
t = int(input())
for h in range(t):
n, k = map(int, input().split())
s = list(input())
ans = []
fl = 0
index = [[], []]
firstOpen = 0
firstClose = 0
for i in range(n):
if s[i] == "(":
index[0] += [i];
else:
index[1] += [i];
for i in range(2 * k - 2):
if fl == 0:
if s[i] != "(":
openBracket(i)
else:
firstOpen += 1
elif fl == 1:
if s[i] != ")":
closeBracket(i)
else:
firstClose += 1
fl = abs(fl - 1)
fl = 0
for i in range(2 * k - 2, n):
if fl == 0:
if s[i] != "(":
openBracket(i)
else:
firstOpen += 1
elif fl == 1:
if s[i] != ")":
closeBracket(i)
else:
firstClose += 1
if i == n // 2 - k + 2 * k - 2:
fl = 1
print(len(ans))
[print(*i) for i in ans]
``` | output | 1 | 7,676 | 21 | 15,353 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are fed up with your messy room, so you decided to clean it up.
Your room is a bracket sequence s=s_{1}s_{2}... s_{n} of length n. Each character of this string is either an opening bracket '(' or a closing bracket ')'.
In one operation you can choose any consecutive substring of s and reverse it. In other words, you can choose any substring s[l ... r]=s_l, s_{l+1}, ..., s_r and change the order of elements in it into s_r, s_{r-1}, ..., s_{l}.
For example, if you will decide to reverse substring s[2 ... 4] of string s="((()))" it will be equal to s="()(())".
A regular (aka balanced) bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example, bracket sequences "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.
A prefix of a string s is a substring that starts at position 1. For example, for s="(())()" there are 6 prefixes: "(", "((", "(()", "(())", "(())(" and "(())()".
In your opinion, a neat and clean room s is a bracket sequence that:
* the whole string s is a regular bracket sequence;
* and there are exactly k prefixes of this sequence which are regular (including whole s itself).
For example, if k = 2, then "(())()" is a neat and clean room.
You want to use at most n operations to make your room neat and clean. Operations are applied one after another sequentially.
It is guaranteed that the answer exists. Note that you do not need to minimize the number of operations: find any way to achieve the desired configuration in n or less operations.
Input
The first line contains integer number t (1 β€ t β€ 100) β the number of test cases in the input. Then t test cases follow.
The first line of a test case contains two integers n and k (1 β€ k β€ n/2, 2 β€ n β€ 2000, n is even) β length of s and required number of regular prefixes.
The second line of a test case contains s of length n β the given bracket sequence. It contains only '(' and ')'.
It is guaranteed that there are exactly n/2 characters '(' and exactly n/2 characters ')' in the given string.
The sum of all values n over all the test cases in the input doesn't exceed 2000.
Output
For each test case print an answer.
In the first line print integer m (0 β€ m β€ n) β the number of operations. You do not need to minimize m, any value is suitable.
In the following m lines print description of the operations, each line should contain two integers l,r (1 β€ l β€ r β€ n), representing single reverse operation of s[l ... r]=s_{l}s_{l+1}... s_{r}. Operations are applied one after another sequentially.
The final s after all operations should be a regular, also it should be exactly k prefixes (including s) which are regular.
It is guaranteed that the answer exists. If there are several possible answers you can print any.
Example
Input
4
8 2
()(())()
10 3
))()()()((
2 1
()
2 1
)(
Output
4
3 4
1 1
5 8
2 2
3
4 10
1 4
6 7
0
1
1 2
Note
In the first example, the final sequence is "()(()())", where two prefixes are regular, "()" and "()(()())". Note, that all the operations except "5 8" in the example output are useless (they do not change s). | instruction | 0 | 7,677 | 21 | 15,354 |
Tags: constructive algorithms
Correct Solution:
```
t = int(input())
for request in range(t):
n, k = map(int, input().split())
box = list(input())
pattern = '()' * (k - 1) + '(' + ('()' * ((n - (k) * 2) // 2) ) + ')'
changes = []
for i in range(n):
if box[i] != pattern[i]:
for j in range(i + 1, n):
if box[j] == pattern[i]:
for z in range((j - i + 1) // 2):
box[i + z], box[j - z] = box[j - z], box[i + z]
changes.append((i + 1, j + 1))
break
print(len(changes))
for i in range(len(changes)):
print(*changes[i])
``` | output | 1 | 7,677 | 21 | 15,355 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are fed up with your messy room, so you decided to clean it up.
Your room is a bracket sequence s=s_{1}s_{2}... s_{n} of length n. Each character of this string is either an opening bracket '(' or a closing bracket ')'.
In one operation you can choose any consecutive substring of s and reverse it. In other words, you can choose any substring s[l ... r]=s_l, s_{l+1}, ..., s_r and change the order of elements in it into s_r, s_{r-1}, ..., s_{l}.
For example, if you will decide to reverse substring s[2 ... 4] of string s="((()))" it will be equal to s="()(())".
A regular (aka balanced) bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example, bracket sequences "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.
A prefix of a string s is a substring that starts at position 1. For example, for s="(())()" there are 6 prefixes: "(", "((", "(()", "(())", "(())(" and "(())()".
In your opinion, a neat and clean room s is a bracket sequence that:
* the whole string s is a regular bracket sequence;
* and there are exactly k prefixes of this sequence which are regular (including whole s itself).
For example, if k = 2, then "(())()" is a neat and clean room.
You want to use at most n operations to make your room neat and clean. Operations are applied one after another sequentially.
It is guaranteed that the answer exists. Note that you do not need to minimize the number of operations: find any way to achieve the desired configuration in n or less operations.
Input
The first line contains integer number t (1 β€ t β€ 100) β the number of test cases in the input. Then t test cases follow.
The first line of a test case contains two integers n and k (1 β€ k β€ n/2, 2 β€ n β€ 2000, n is even) β length of s and required number of regular prefixes.
The second line of a test case contains s of length n β the given bracket sequence. It contains only '(' and ')'.
It is guaranteed that there are exactly n/2 characters '(' and exactly n/2 characters ')' in the given string.
The sum of all values n over all the test cases in the input doesn't exceed 2000.
Output
For each test case print an answer.
In the first line print integer m (0 β€ m β€ n) β the number of operations. You do not need to minimize m, any value is suitable.
In the following m lines print description of the operations, each line should contain two integers l,r (1 β€ l β€ r β€ n), representing single reverse operation of s[l ... r]=s_{l}s_{l+1}... s_{r}. Operations are applied one after another sequentially.
The final s after all operations should be a regular, also it should be exactly k prefixes (including s) which are regular.
It is guaranteed that the answer exists. If there are several possible answers you can print any.
Example
Input
4
8 2
()(())()
10 3
))()()()((
2 1
()
2 1
)(
Output
4
3 4
1 1
5 8
2 2
3
4 10
1 4
6 7
0
1
1 2
Note
In the first example, the final sequence is "()(()())", where two prefixes are regular, "()" and "()(()())". Note, that all the operations except "5 8" in the example output are useless (they do not change s). | instruction | 0 | 7,678 | 21 | 15,356 |
Tags: constructive algorithms
Correct Solution:
```
def newstroka(f,a):
pp = f
new = []
ss = 0
for i in range(0,a,2):
if f==0:
ss=i
break
else:
f-=1
new.append("(")
new.append(")")
if pp+1!=a//2+1:
for i in range(ss,ss+((a-ss)//2)):
new.append("(")
for j in range(i+1,a):
new.append(")")
return new
for _ in range(int(input())):
a,b = map(int,input().split())
c = list(input())
f = b-1
newstr = newstroka(f,a)
ansi = 0
ans = []
for i in range(a):
if c[i]!=newstr[i]:
ansi+=1
j = i+1
while c[i]==c[j]:
j+=1
ans.append((i+1,j+1))
if i == 0: c = c[j::-1]+c[j+1:]
else: c = c[0:i]+c[j:i-1:-1]+c[j+1:]
print(ansi)
if ansi!=0:
for i in range(ansi): print(*ans[i])
``` | output | 1 | 7,678 | 21 | 15,357 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are fed up with your messy room, so you decided to clean it up.
Your room is a bracket sequence s=s_{1}s_{2}... s_{n} of length n. Each character of this string is either an opening bracket '(' or a closing bracket ')'.
In one operation you can choose any consecutive substring of s and reverse it. In other words, you can choose any substring s[l ... r]=s_l, s_{l+1}, ..., s_r and change the order of elements in it into s_r, s_{r-1}, ..., s_{l}.
For example, if you will decide to reverse substring s[2 ... 4] of string s="((()))" it will be equal to s="()(())".
A regular (aka balanced) bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example, bracket sequences "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.
A prefix of a string s is a substring that starts at position 1. For example, for s="(())()" there are 6 prefixes: "(", "((", "(()", "(())", "(())(" and "(())()".
In your opinion, a neat and clean room s is a bracket sequence that:
* the whole string s is a regular bracket sequence;
* and there are exactly k prefixes of this sequence which are regular (including whole s itself).
For example, if k = 2, then "(())()" is a neat and clean room.
You want to use at most n operations to make your room neat and clean. Operations are applied one after another sequentially.
It is guaranteed that the answer exists. Note that you do not need to minimize the number of operations: find any way to achieve the desired configuration in n or less operations.
Input
The first line contains integer number t (1 β€ t β€ 100) β the number of test cases in the input. Then t test cases follow.
The first line of a test case contains two integers n and k (1 β€ k β€ n/2, 2 β€ n β€ 2000, n is even) β length of s and required number of regular prefixes.
The second line of a test case contains s of length n β the given bracket sequence. It contains only '(' and ')'.
It is guaranteed that there are exactly n/2 characters '(' and exactly n/2 characters ')' in the given string.
The sum of all values n over all the test cases in the input doesn't exceed 2000.
Output
For each test case print an answer.
In the first line print integer m (0 β€ m β€ n) β the number of operations. You do not need to minimize m, any value is suitable.
In the following m lines print description of the operations, each line should contain two integers l,r (1 β€ l β€ r β€ n), representing single reverse operation of s[l ... r]=s_{l}s_{l+1}... s_{r}. Operations are applied one after another sequentially.
The final s after all operations should be a regular, also it should be exactly k prefixes (including s) which are regular.
It is guaranteed that the answer exists. If there are several possible answers you can print any.
Example
Input
4
8 2
()(())()
10 3
))()()()((
2 1
()
2 1
)(
Output
4
3 4
1 1
5 8
2 2
3
4 10
1 4
6 7
0
1
1 2
Note
In the first example, the final sequence is "()(()())", where two prefixes are regular, "()" and "()(()())". Note, that all the operations except "5 8" in the example output are useless (they do not change s).
Submitted Solution:
```
t = int(input())
for i in range(t):
n, k = map(int, input().split())
*s, = input()
operations = []
best = (['('] + [')']) * (k - 1) + (['('] * (n // 2 - k + 1) + [')'] * (n // 2 - k + 1))
for startx_pos in range((k - 1) * 2):
try:
if s[startx_pos - 1] == ')' or startx_pos == 0:
end_pos = s.index('(', startx_pos)
else:
end_pos = s.index(')', startx_pos)
except ValueError:
continue
if startx_pos == end_pos:
continue
if startx_pos == 0:
s = s[:startx_pos] + s[end_pos::-1] + s[end_pos + 1:]
else:
s = s[:startx_pos] + s[end_pos:startx_pos - 1:-1] + s[end_pos + 1:]
operations.append(f'{startx_pos + 1} {end_pos + 1}')
for startx_pos in range((k - 1) * 2, (k - 1) * 2 + (n // 2 - k + 1)):
try:
end_pos = s.index('(', startx_pos)
except ValueError:
continue
if startx_pos == end_pos:
continue
if startx_pos == 0:
s = s[:startx_pos] + s[end_pos::-1] + s[end_pos + 1:]
else:
s = s[:startx_pos] + s[end_pos:startx_pos - 1:-1] + s[end_pos + 1:]
operations.append(f'{startx_pos + 1} {end_pos + 1}')
print(len(operations))
if len(operations):
print(*operations, sep='\n')
``` | instruction | 0 | 7,679 | 21 | 15,358 |
Yes | output | 1 | 7,679 | 21 | 15,359 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are fed up with your messy room, so you decided to clean it up.
Your room is a bracket sequence s=s_{1}s_{2}... s_{n} of length n. Each character of this string is either an opening bracket '(' or a closing bracket ')'.
In one operation you can choose any consecutive substring of s and reverse it. In other words, you can choose any substring s[l ... r]=s_l, s_{l+1}, ..., s_r and change the order of elements in it into s_r, s_{r-1}, ..., s_{l}.
For example, if you will decide to reverse substring s[2 ... 4] of string s="((()))" it will be equal to s="()(())".
A regular (aka balanced) bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example, bracket sequences "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.
A prefix of a string s is a substring that starts at position 1. For example, for s="(())()" there are 6 prefixes: "(", "((", "(()", "(())", "(())(" and "(())()".
In your opinion, a neat and clean room s is a bracket sequence that:
* the whole string s is a regular bracket sequence;
* and there are exactly k prefixes of this sequence which are regular (including whole s itself).
For example, if k = 2, then "(())()" is a neat and clean room.
You want to use at most n operations to make your room neat and clean. Operations are applied one after another sequentially.
It is guaranteed that the answer exists. Note that you do not need to minimize the number of operations: find any way to achieve the desired configuration in n or less operations.
Input
The first line contains integer number t (1 β€ t β€ 100) β the number of test cases in the input. Then t test cases follow.
The first line of a test case contains two integers n and k (1 β€ k β€ n/2, 2 β€ n β€ 2000, n is even) β length of s and required number of regular prefixes.
The second line of a test case contains s of length n β the given bracket sequence. It contains only '(' and ')'.
It is guaranteed that there are exactly n/2 characters '(' and exactly n/2 characters ')' in the given string.
The sum of all values n over all the test cases in the input doesn't exceed 2000.
Output
For each test case print an answer.
In the first line print integer m (0 β€ m β€ n) β the number of operations. You do not need to minimize m, any value is suitable.
In the following m lines print description of the operations, each line should contain two integers l,r (1 β€ l β€ r β€ n), representing single reverse operation of s[l ... r]=s_{l}s_{l+1}... s_{r}. Operations are applied one after another sequentially.
The final s after all operations should be a regular, also it should be exactly k prefixes (including s) which are regular.
It is guaranteed that the answer exists. If there are several possible answers you can print any.
Example
Input
4
8 2
()(())()
10 3
))()()()((
2 1
()
2 1
)(
Output
4
3 4
1 1
5 8
2 2
3
4 10
1 4
6 7
0
1
1 2
Note
In the first example, the final sequence is "()(()())", where two prefixes are regular, "()" and "()(()())". Note, that all the operations except "5 8" in the example output are useless (they do not change s).
Submitted Solution:
```
t = int(input())
def conv1(v) :
global z
index, q = 0, 0
for i in range(len(v)) :
if v[i] == '(' : q += 1
else : q -= 1
if q == 0 and v[i] == '(' :
if i != len(v) : v = v[:index] + list(reversed(v[index:i+1])) + v[i+1:]
else : v = v[:index] + list(reversed(v[index:i+1]))
z.append([index+1, i+1])
index = i+1
elif q == 0 : index = i+1
return v
def count(v) :
q, k = 0, 0
for i in v :
if i == '(' : q += 1
else : q -= 1
if q == 0 : k += 1
return k
def conv_min(v, k, n) :
global z
q = 0
for i in range(0, len(v)) :
if k == n : return v
if v[i] == '(' : q += 1
else : q -= 1
if q == 0 :
z.append([i+1, i+2])
n -= 1
def conv_max(v, k, n) :
global z
q = 0
for i in range(0, len(v)) :
if k == n : return v
if v[i] == '(' : q += 1
else :
if q == 2 :
v[i-1], v[i] = v[i], v[i-1]
q = 1
z.append([i, i+1])
n += 1
elif q > 2 :
v[i-q+1], v[i] = v[i], v[i-q+1]
z.append([i-q+1, i+1])
z.append([i-q+1, i-q+2])
q -= 1
n += 1
else : q = 0
if 1 == 2 :
s = list('()(())')
z = []
print(''.join(conv_max(s, 3, 2)))
raise SystemExit
for _ in range(t) :
_, k = map(lambda x : int(x), input().split())
s = list(input())
z = []
s = conv1(s)
ct = count(s)
if ct >= k : conv_min(s, k, ct)
else : conv_max(s, k, ct)
print(len(z))
print('\n'.join(list(map(lambda x : str(x[0])+' '+str(x[1]), z))))
``` | instruction | 0 | 7,680 | 21 | 15,360 |
Yes | output | 1 | 7,680 | 21 | 15,361 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are fed up with your messy room, so you decided to clean it up.
Your room is a bracket sequence s=s_{1}s_{2}... s_{n} of length n. Each character of this string is either an opening bracket '(' or a closing bracket ')'.
In one operation you can choose any consecutive substring of s and reverse it. In other words, you can choose any substring s[l ... r]=s_l, s_{l+1}, ..., s_r and change the order of elements in it into s_r, s_{r-1}, ..., s_{l}.
For example, if you will decide to reverse substring s[2 ... 4] of string s="((()))" it will be equal to s="()(())".
A regular (aka balanced) bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example, bracket sequences "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.
A prefix of a string s is a substring that starts at position 1. For example, for s="(())()" there are 6 prefixes: "(", "((", "(()", "(())", "(())(" and "(())()".
In your opinion, a neat and clean room s is a bracket sequence that:
* the whole string s is a regular bracket sequence;
* and there are exactly k prefixes of this sequence which are regular (including whole s itself).
For example, if k = 2, then "(())()" is a neat and clean room.
You want to use at most n operations to make your room neat and clean. Operations are applied one after another sequentially.
It is guaranteed that the answer exists. Note that you do not need to minimize the number of operations: find any way to achieve the desired configuration in n or less operations.
Input
The first line contains integer number t (1 β€ t β€ 100) β the number of test cases in the input. Then t test cases follow.
The first line of a test case contains two integers n and k (1 β€ k β€ n/2, 2 β€ n β€ 2000, n is even) β length of s and required number of regular prefixes.
The second line of a test case contains s of length n β the given bracket sequence. It contains only '(' and ')'.
It is guaranteed that there are exactly n/2 characters '(' and exactly n/2 characters ')' in the given string.
The sum of all values n over all the test cases in the input doesn't exceed 2000.
Output
For each test case print an answer.
In the first line print integer m (0 β€ m β€ n) β the number of operations. You do not need to minimize m, any value is suitable.
In the following m lines print description of the operations, each line should contain two integers l,r (1 β€ l β€ r β€ n), representing single reverse operation of s[l ... r]=s_{l}s_{l+1}... s_{r}. Operations are applied one after another sequentially.
The final s after all operations should be a regular, also it should be exactly k prefixes (including s) which are regular.
It is guaranteed that the answer exists. If there are several possible answers you can print any.
Example
Input
4
8 2
()(())()
10 3
))()()()((
2 1
()
2 1
)(
Output
4
3 4
1 1
5 8
2 2
3
4 10
1 4
6 7
0
1
1 2
Note
In the first example, the final sequence is "()(()())", where two prefixes are regular, "()" and "()(()())". Note, that all the operations except "5 8" in the example output are useless (they do not change s).
Submitted Solution:
```
import sys
input = sys.stdin.readline
T = int(input())
for _ in range(T):
N, K = map(int, input().split())
# N γ―εΆζ°
S = list(input()[:-1])
bl = []
for i in range(K-1):
bl.extend(["(", ")"])
for i in range(N//2-K+1):
bl.append("(")
for i in range(N//2-K+1):
bl.append(")")
ans = []
for i in range(N):
if S[i] != bl[i]:
for j in range(i+1, N):
if S[j] == bl[i]:
ans.append(f"{i+1} {j+1}")
S[i:j+1] = reversed(S[i:j+1])
break
else:
assert False, (S, bl)
print(len(ans))
if ans:
print("\n".join(ans))
``` | instruction | 0 | 7,681 | 21 | 15,362 |
Yes | output | 1 | 7,681 | 21 | 15,363 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are fed up with your messy room, so you decided to clean it up.
Your room is a bracket sequence s=s_{1}s_{2}... s_{n} of length n. Each character of this string is either an opening bracket '(' or a closing bracket ')'.
In one operation you can choose any consecutive substring of s and reverse it. In other words, you can choose any substring s[l ... r]=s_l, s_{l+1}, ..., s_r and change the order of elements in it into s_r, s_{r-1}, ..., s_{l}.
For example, if you will decide to reverse substring s[2 ... 4] of string s="((()))" it will be equal to s="()(())".
A regular (aka balanced) bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example, bracket sequences "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.
A prefix of a string s is a substring that starts at position 1. For example, for s="(())()" there are 6 prefixes: "(", "((", "(()", "(())", "(())(" and "(())()".
In your opinion, a neat and clean room s is a bracket sequence that:
* the whole string s is a regular bracket sequence;
* and there are exactly k prefixes of this sequence which are regular (including whole s itself).
For example, if k = 2, then "(())()" is a neat and clean room.
You want to use at most n operations to make your room neat and clean. Operations are applied one after another sequentially.
It is guaranteed that the answer exists. Note that you do not need to minimize the number of operations: find any way to achieve the desired configuration in n or less operations.
Input
The first line contains integer number t (1 β€ t β€ 100) β the number of test cases in the input. Then t test cases follow.
The first line of a test case contains two integers n and k (1 β€ k β€ n/2, 2 β€ n β€ 2000, n is even) β length of s and required number of regular prefixes.
The second line of a test case contains s of length n β the given bracket sequence. It contains only '(' and ')'.
It is guaranteed that there are exactly n/2 characters '(' and exactly n/2 characters ')' in the given string.
The sum of all values n over all the test cases in the input doesn't exceed 2000.
Output
For each test case print an answer.
In the first line print integer m (0 β€ m β€ n) β the number of operations. You do not need to minimize m, any value is suitable.
In the following m lines print description of the operations, each line should contain two integers l,r (1 β€ l β€ r β€ n), representing single reverse operation of s[l ... r]=s_{l}s_{l+1}... s_{r}. Operations are applied one after another sequentially.
The final s after all operations should be a regular, also it should be exactly k prefixes (including s) which are regular.
It is guaranteed that the answer exists. If there are several possible answers you can print any.
Example
Input
4
8 2
()(())()
10 3
))()()()((
2 1
()
2 1
)(
Output
4
3 4
1 1
5 8
2 2
3
4 10
1 4
6 7
0
1
1 2
Note
In the first example, the final sequence is "()(()())", where two prefixes are regular, "()" and "()(()())". Note, that all the operations except "5 8" in the example output are useless (they do not change s).
Submitted Solution:
```
from math import *
from collections import *
import sys
sys.setrecursionlimit(10**9)
test = int(input())
for y in range(test):
n,k = map(int,input().split())
k -= 1
s = list(input())
t = ""
for i in range(k):
t += "()"
for i in range(n//2-k):
t += "("
for i in range(n//2-k):
t += ")"
print(n)
for i in range(n):
j = i
while(s[j] != t[i]):
j += 1
s[i],s[j] = s[j],s[i]
print(i+1,j+1)
``` | instruction | 0 | 7,682 | 21 | 15,364 |
Yes | output | 1 | 7,682 | 21 | 15,365 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are fed up with your messy room, so you decided to clean it up.
Your room is a bracket sequence s=s_{1}s_{2}... s_{n} of length n. Each character of this string is either an opening bracket '(' or a closing bracket ')'.
In one operation you can choose any consecutive substring of s and reverse it. In other words, you can choose any substring s[l ... r]=s_l, s_{l+1}, ..., s_r and change the order of elements in it into s_r, s_{r-1}, ..., s_{l}.
For example, if you will decide to reverse substring s[2 ... 4] of string s="((()))" it will be equal to s="()(())".
A regular (aka balanced) bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example, bracket sequences "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.
A prefix of a string s is a substring that starts at position 1. For example, for s="(())()" there are 6 prefixes: "(", "((", "(()", "(())", "(())(" and "(())()".
In your opinion, a neat and clean room s is a bracket sequence that:
* the whole string s is a regular bracket sequence;
* and there are exactly k prefixes of this sequence which are regular (including whole s itself).
For example, if k = 2, then "(())()" is a neat and clean room.
You want to use at most n operations to make your room neat and clean. Operations are applied one after another sequentially.
It is guaranteed that the answer exists. Note that you do not need to minimize the number of operations: find any way to achieve the desired configuration in n or less operations.
Input
The first line contains integer number t (1 β€ t β€ 100) β the number of test cases in the input. Then t test cases follow.
The first line of a test case contains two integers n and k (1 β€ k β€ n/2, 2 β€ n β€ 2000, n is even) β length of s and required number of regular prefixes.
The second line of a test case contains s of length n β the given bracket sequence. It contains only '(' and ')'.
It is guaranteed that there are exactly n/2 characters '(' and exactly n/2 characters ')' in the given string.
The sum of all values n over all the test cases in the input doesn't exceed 2000.
Output
For each test case print an answer.
In the first line print integer m (0 β€ m β€ n) β the number of operations. You do not need to minimize m, any value is suitable.
In the following m lines print description of the operations, each line should contain two integers l,r (1 β€ l β€ r β€ n), representing single reverse operation of s[l ... r]=s_{l}s_{l+1}... s_{r}. Operations are applied one after another sequentially.
The final s after all operations should be a regular, also it should be exactly k prefixes (including s) which are regular.
It is guaranteed that the answer exists. If there are several possible answers you can print any.
Example
Input
4
8 2
()(())()
10 3
))()()()((
2 1
()
2 1
)(
Output
4
3 4
1 1
5 8
2 2
3
4 10
1 4
6 7
0
1
1 2
Note
In the first example, the final sequence is "()(()())", where two prefixes are regular, "()" and "()(()())". Note, that all the operations except "5 8" in the example output are useless (they do not change s).
Submitted Solution:
```
t = int(input())
for i in range(t):
m = 0
l = []
r = []
result = 0
prefix = 0
oval = -1
n, k = map(int, input().split())
s = input()
# pos = [0] * (n // 2)
for j in range(n):
if s[j] == "(":
result += 1
# if (j != n - 1) and (s[j + 1] == ")") and (oval == -1):
# pos[result - 1] += 1
if (prefix >= k) and (result == 1):
# result = 0
m += 1
l.append(j)
r.append(j + 1)
s = s[:j - 1] + "()" + s[j + 1:]
else:
result -= 1
# if (j != n - 1) and (s[j + 1] == "(") and (oval != -1):
# pos[-result - 1] += 1
if (result < 0) and (oval == -1):
oval = j
if (oval != -1) and (result == 0):
if oval == 0:
if j != n - 1:
s = s[:oval] + s[j::-1] + s[j + 1:]
else:
s = s[:oval] + s[j::-1]
else:
if j != n - 1:
s = s[:oval] + s[j:oval - 1:-1] + s[j + 1:]
else:
s = s[:oval] + s[j:oval - 1:-1]
m += 1
l.append(oval + 1)
r.append(j + 1)
if prefix >= k:
s = s[:oval - 1] + "()" + s[oval + 1:]
m += 1
l.append(oval)
r.append(oval + 1)
oval = -1
if (result == 0) and (j != 0) and (prefix < k):
prefix += 1
"""if prefix < k:
jim = [0]
u = 1
while prefix < k:
print(u, pos, k)
prefix += pos[u] + u - 1 if pos[u] != 0 else 0
u += 1"""
while prefix < k:
new_result = 0
for u in range(n - 1):
if s[u] == "(":
new_result += 1
if (s[u + 1] == ")") and (new_result != 1):
m += 1
l.append(u + 1)
r.append(u + 2)
s = s[:u] + ")(" + s[u + 2:]
if new_result == 2:
prefix += 1
new_result -= 1
if s[u] == ")":
new_result -= 1
if prefix == k:
break
print(m)
for LIL in range(m):
print(l[LIL], r[LIL])
print(s)
``` | instruction | 0 | 7,683 | 21 | 15,366 |
No | output | 1 | 7,683 | 21 | 15,367 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are fed up with your messy room, so you decided to clean it up.
Your room is a bracket sequence s=s_{1}s_{2}... s_{n} of length n. Each character of this string is either an opening bracket '(' or a closing bracket ')'.
In one operation you can choose any consecutive substring of s and reverse it. In other words, you can choose any substring s[l ... r]=s_l, s_{l+1}, ..., s_r and change the order of elements in it into s_r, s_{r-1}, ..., s_{l}.
For example, if you will decide to reverse substring s[2 ... 4] of string s="((()))" it will be equal to s="()(())".
A regular (aka balanced) bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example, bracket sequences "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.
A prefix of a string s is a substring that starts at position 1. For example, for s="(())()" there are 6 prefixes: "(", "((", "(()", "(())", "(())(" and "(())()".
In your opinion, a neat and clean room s is a bracket sequence that:
* the whole string s is a regular bracket sequence;
* and there are exactly k prefixes of this sequence which are regular (including whole s itself).
For example, if k = 2, then "(())()" is a neat and clean room.
You want to use at most n operations to make your room neat and clean. Operations are applied one after another sequentially.
It is guaranteed that the answer exists. Note that you do not need to minimize the number of operations: find any way to achieve the desired configuration in n or less operations.
Input
The first line contains integer number t (1 β€ t β€ 100) β the number of test cases in the input. Then t test cases follow.
The first line of a test case contains two integers n and k (1 β€ k β€ n/2, 2 β€ n β€ 2000, n is even) β length of s and required number of regular prefixes.
The second line of a test case contains s of length n β the given bracket sequence. It contains only '(' and ')'.
It is guaranteed that there are exactly n/2 characters '(' and exactly n/2 characters ')' in the given string.
The sum of all values n over all the test cases in the input doesn't exceed 2000.
Output
For each test case print an answer.
In the first line print integer m (0 β€ m β€ n) β the number of operations. You do not need to minimize m, any value is suitable.
In the following m lines print description of the operations, each line should contain two integers l,r (1 β€ l β€ r β€ n), representing single reverse operation of s[l ... r]=s_{l}s_{l+1}... s_{r}. Operations are applied one after another sequentially.
The final s after all operations should be a regular, also it should be exactly k prefixes (including s) which are regular.
It is guaranteed that the answer exists. If there are several possible answers you can print any.
Example
Input
4
8 2
()(())()
10 3
))()()()((
2 1
()
2 1
)(
Output
4
3 4
1 1
5 8
2 2
3
4 10
1 4
6 7
0
1
1 2
Note
In the first example, the final sequence is "()(()())", where two prefixes are regular, "()" and "()(()())". Note, that all the operations except "5 8" in the example output are useless (they do not change s).
Submitted Solution:
```
import os, sys, math
def solve(seq, k):
seq = [ 1 if a == '(' else -1 for a in seq ]
size = len(seq)
result = []
def rotate(fr, to):
assert fr <= to
result.append((fr, to))
while fr < to:
seq[fr], seq[to] = seq[to], seq[fr]
fr += 1
to -= 1
# print(''.join('(' if q > 0 else ')' for q in seq))
def split(p1, p2):
if p1 + 1 == p2:
return False
assert seq[p1] > 0 and seq[p2] < 0, (seq, p1, p2)
i = p1
while seq[i] > 0:
i += 1
rotate(p1 + 1, i)
return True
def merge(p):
assert seq[p] < 0 and seq[p + 1] > 0, (p, seq)
rotate(p, p + 1)
d = 0
x = 0
while x < size:
d += seq[x]
x += 1
if d < 0:
start = x - 1
while d < 0:
d += seq[x]
x += 1
assert d == 0
rotate(start, x - 1)
zero_points = [ -1 ]
d = 0
for x in range(size):
d += seq[x]
if d == 0:
zero_points.append(x)
start = len(zero_points) - 1
if start < k:
zero_points_index = 0
while start < k:
p1 = zero_points[zero_points_index] + 1
p2 = zero_points[zero_points_index + 1]
if not split(p1, p2):
zero_points_index += 1
else:
zero_points[zero_points_index] = p1 - 1 + 2
start += 1
elif start > k:
zero_points_index = 1
while start > k:
merge(zero_points[zero_points_index])
start -= 1
zero_points_index += 1
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())
for x in range(num):
n, k = map(int, input().strip().split())
n = input().strip()[:n]
res = solve(n, k)
print(len(res))
for q in res:
print(' '.join(map(str, q)))
``` | instruction | 0 | 7,684 | 21 | 15,368 |
No | output | 1 | 7,684 | 21 | 15,369 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are fed up with your messy room, so you decided to clean it up.
Your room is a bracket sequence s=s_{1}s_{2}... s_{n} of length n. Each character of this string is either an opening bracket '(' or a closing bracket ')'.
In one operation you can choose any consecutive substring of s and reverse it. In other words, you can choose any substring s[l ... r]=s_l, s_{l+1}, ..., s_r and change the order of elements in it into s_r, s_{r-1}, ..., s_{l}.
For example, if you will decide to reverse substring s[2 ... 4] of string s="((()))" it will be equal to s="()(())".
A regular (aka balanced) bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example, bracket sequences "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.
A prefix of a string s is a substring that starts at position 1. For example, for s="(())()" there are 6 prefixes: "(", "((", "(()", "(())", "(())(" and "(())()".
In your opinion, a neat and clean room s is a bracket sequence that:
* the whole string s is a regular bracket sequence;
* and there are exactly k prefixes of this sequence which are regular (including whole s itself).
For example, if k = 2, then "(())()" is a neat and clean room.
You want to use at most n operations to make your room neat and clean. Operations are applied one after another sequentially.
It is guaranteed that the answer exists. Note that you do not need to minimize the number of operations: find any way to achieve the desired configuration in n or less operations.
Input
The first line contains integer number t (1 β€ t β€ 100) β the number of test cases in the input. Then t test cases follow.
The first line of a test case contains two integers n and k (1 β€ k β€ n/2, 2 β€ n β€ 2000, n is even) β length of s and required number of regular prefixes.
The second line of a test case contains s of length n β the given bracket sequence. It contains only '(' and ')'.
It is guaranteed that there are exactly n/2 characters '(' and exactly n/2 characters ')' in the given string.
The sum of all values n over all the test cases in the input doesn't exceed 2000.
Output
For each test case print an answer.
In the first line print integer m (0 β€ m β€ n) β the number of operations. You do not need to minimize m, any value is suitable.
In the following m lines print description of the operations, each line should contain two integers l,r (1 β€ l β€ r β€ n), representing single reverse operation of s[l ... r]=s_{l}s_{l+1}... s_{r}. Operations are applied one after another sequentially.
The final s after all operations should be a regular, also it should be exactly k prefixes (including s) which are regular.
It is guaranteed that the answer exists. If there are several possible answers you can print any.
Example
Input
4
8 2
()(())()
10 3
))()()()((
2 1
()
2 1
)(
Output
4
3 4
1 1
5 8
2 2
3
4 10
1 4
6 7
0
1
1 2
Note
In the first example, the final sequence is "()(()())", where two prefixes are regular, "()" and "()(()())". Note, that all the operations except "5 8" in the example output are useless (they do not change s).
Submitted Solution:
```
t = int(input())
for i in range(t):
n, k = map(int, input().split())
s = input()
j = 0
ans = []
while j < n:
if s[j] == '(':
q = j + 1
while s[q] != ')':
q += 1
first, second = j + 2, q + 1
if first > second:
first, second = second, first
if q + 1 >= n:
p = ''
else:
p = s[q + 1:]
b = s[j + 1:q + 1]
s = s[:j + 1] + b[::-1] + p
j += 2
else:
q = j + 1
while s[q] != '(':
q += 1
first, second = j + 1, q + 1
if first > second:
first, second = second, first
if q + 1 >= n:
p = ''
else:
p = s[q + 1:]
a = s[:j]
b = s[j:q + 1]
s = s[:j] + b[::-1] + p
if first != second:
ans.append([first, second])
print(len(ans) + 1)
for i in ans:
print(i[0], i[1])
print(1, n - 2 * k + 2)
``` | instruction | 0 | 7,685 | 21 | 15,370 |
No | output | 1 | 7,685 | 21 | 15,371 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are fed up with your messy room, so you decided to clean it up.
Your room is a bracket sequence s=s_{1}s_{2}... s_{n} of length n. Each character of this string is either an opening bracket '(' or a closing bracket ')'.
In one operation you can choose any consecutive substring of s and reverse it. In other words, you can choose any substring s[l ... r]=s_l, s_{l+1}, ..., s_r and change the order of elements in it into s_r, s_{r-1}, ..., s_{l}.
For example, if you will decide to reverse substring s[2 ... 4] of string s="((()))" it will be equal to s="()(())".
A regular (aka balanced) bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example, bracket sequences "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.
A prefix of a string s is a substring that starts at position 1. For example, for s="(())()" there are 6 prefixes: "(", "((", "(()", "(())", "(())(" and "(())()".
In your opinion, a neat and clean room s is a bracket sequence that:
* the whole string s is a regular bracket sequence;
* and there are exactly k prefixes of this sequence which are regular (including whole s itself).
For example, if k = 2, then "(())()" is a neat and clean room.
You want to use at most n operations to make your room neat and clean. Operations are applied one after another sequentially.
It is guaranteed that the answer exists. Note that you do not need to minimize the number of operations: find any way to achieve the desired configuration in n or less operations.
Input
The first line contains integer number t (1 β€ t β€ 100) β the number of test cases in the input. Then t test cases follow.
The first line of a test case contains two integers n and k (1 β€ k β€ n/2, 2 β€ n β€ 2000, n is even) β length of s and required number of regular prefixes.
The second line of a test case contains s of length n β the given bracket sequence. It contains only '(' and ')'.
It is guaranteed that there are exactly n/2 characters '(' and exactly n/2 characters ')' in the given string.
The sum of all values n over all the test cases in the input doesn't exceed 2000.
Output
For each test case print an answer.
In the first line print integer m (0 β€ m β€ n) β the number of operations. You do not need to minimize m, any value is suitable.
In the following m lines print description of the operations, each line should contain two integers l,r (1 β€ l β€ r β€ n), representing single reverse operation of s[l ... r]=s_{l}s_{l+1}... s_{r}. Operations are applied one after another sequentially.
The final s after all operations should be a regular, also it should be exactly k prefixes (including s) which are regular.
It is guaranteed that the answer exists. If there are several possible answers you can print any.
Example
Input
4
8 2
()(())()
10 3
))()()()((
2 1
()
2 1
)(
Output
4
3 4
1 1
5 8
2 2
3
4 10
1 4
6 7
0
1
1 2
Note
In the first example, the final sequence is "()(()())", where two prefixes are regular, "()" and "()(()())". Note, that all the operations except "5 8" in the example output are useless (they do not change s).
Submitted Solution:
```
t = int(input())
def conv1(v) :
global z
index, q = 0, 0
for i in range(len(v)) :
if v[i] == '(' : q += 1
else : q -= 1
if q == 0 and v[i] == '(' :
if i != len(v) : v = v[:index] + list(reversed(v[index:i+1])) + v[i+1:]
else : v = v[:index] + list(reversed(v[index:i+1]))
z.append([index+1, i+1])
index = i+1
elif q == 0 : index = i+1
return v
def count(v) :
q, k = 0, 0
for i in v :
if i == '(' : q += 1
else : q -= 1
if q == 0 : k += 1
return k
def conv_min(v, k, n) :
global z
q = 0
for i in range(0, len(v)) :
if k == n : return v
if v[i] == '(' : q += 1
else : q -= 1
if q == 0 :
z.append([i+1, i+2])
n -= 1
def conv_max(v, k, n) :
global z
q = 0
for i in range(0, len(v)) :
if k == n : return v
if v[i] == '(' : q += 1
else :
if q == 2 :
v[i-1], v[i] = v[i], v[i-1]
q = 1
z.append([i, i+1])
n += 1
elif q > 2 :
v[i-q+1], v[i] = v[i], v[i-q+1]
z.append([i-q+1, i+1])
z.append([i-q+1, i-q+2])
q -= 1
n += 1
for _ in range(t) :
_, k = map(lambda x : int(x), input().split())
s = list(input())
z = []
s = conv1(s)
ct = count(s)
if ct >= k : conv_min(s, k, ct)
else : conv_max(s, k, ct)
print(len(z))
print('\n'.join(list(map(lambda x : str(x[0])+' '+str(x[1]), z))))
``` | instruction | 0 | 7,686 | 21 | 15,372 |
No | output | 1 | 7,686 | 21 | 15,373 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A balanced bracket sequence is defined as an integer sequence that can be built with the following rules:
* The empty sequence is balanced.
* If [a_1,β¦,a_n] and [b_1,β¦, b_m] are balanced, then their concatenation [a_1,β¦,a_n,b_1,β¦,b_m] is balanced.
* If x is a positive integer and [a_1,β¦,a_n] is balanced, then [x,a_1,β¦,a_n,-x] is balanced.
The positive numbers can be imagined as opening brackets and the negative numbers as closing brackets, where matching brackets must have the same type (absolute value). For example, [1, 2, -2, -1] and [1, 3, -3, 2, -2, -1] are balanced, but [1, 2, -1, -2] and [-1, 1] are not balanced.
There are 2n cards. Each card has a number on the front and a number on the back. Each integer 1,-1,2,-2,β¦,n,-n appears exactly once on the front of some card and exactly once on the back of some (not necessarily the same) card.
You can reorder the cards however you like. You are not allowed to flip cards, so numbers cannot move between the front and back. Your task is to order the cards so that the sequences given by the front numbers and the back numbers are both balanced, or report that it is impossible.
Input
The first line contains a single integer n (1β€ nβ€ 2β
10^5) β the number of bracket types, and half the number of cards.
The next 2n lines describe the cards. The i-th of these lines contains two integers a_i, b_i (-nβ€ a_i,b_iβ€ n, a_iβ 0, b_iβ 0) β the numbers on the front and back of the i-th card, respectively. Every integer 1,-1,2,-2,β¦,n,-n appears exactly once as a_i and exactly once as b_i.
Output
On the first line, output "YES" if it's possible to reorder these cards to satisfy the condition. Otherwise, output "NO". You can print each letter in any case (upper or lower).
If it is possible, on the next 2n lines output the cards in an order so that the front and back are both balanced. If there are multiple solutions, you may output any.
Examples
Input
5
1 3
-3 -5
4 -3
2 2
-1 -4
-2 5
3 -1
5 1
-4 4
-5 -2
Output
YES
1 3
4 -3
-4 4
-1 -4
5 1
3 -1
2 2
-2 5
-3 -5
-5 -2
Input
2
1 1
-1 2
2 -1
-2 -2
Output
NO
Note
In the first test case, the front numbers create the balanced sequence [1,4,-4,-1,5,3,2,-2,-3,-5] and the back numbers create the balanced sequence [3,-3,4,-4,1,-1,2,5,-5,-2].
In the second test case, the cards are given in an order so that the front numbers are balanced, but the back numbers create the unbalanced sequence [1,2,-1,-2]. If we swapped the second and third cards, we would balance the back numbers and unbalance the front numbers. But there is no order that balances both.
Submitted Solution:
```
import itertools
from typing import List
n = int(input())
d = dict(map(int, input().split()) for _ in range(2 * n))
def isValid(nums: List[int]) -> bool:
"""Removes same consecutive brackets one at a time until length of nums becomes zero and return True, otherwise false.
Args:
nums (List[int]): array of front or back values.
Returns:
bool: Returns True if a valid sequence, else False.
"""
while len(nums) > 0:
l = len(nums)
for i, v in enumerate(nums):
if i > 0:
if abs(v) == abs(nums[i - 1]):
if v < 0:
nums.pop(i)
nums.pop(i - 1)
if len(nums) == l:
return False
return True
def solve(dic) -> str:
"""Creates permuntations of card order, unzips front and back values in x, y and passes them in isValid() function as a parameter.
Args:
arr (List[List[int]]): 2d array of front and back values of the card. [[front,back],..].
Returns:
str: returns 'Yes' if both front values and back values are valid bracket sequence. Otherwise returns 'No'
"""
for i in itertools.permutations(list(d.keys()), len(dic)):
front = list(i)
back = [dic[x] for x in front]
if isValid(front[:]) == True and isValid(back[:]) == True:
return ['YES', front, back]
return ['N0']
ans = solve(d)
if len(ans) > 1:
print(ans[0])
res = "\n".join("{} {}".format(x, y) for x, y in zip(ans[1], ans[2]))
print(res)
else:
print(ans[0])
``` | instruction | 0 | 7,796 | 21 | 15,592 |
No | output | 1 | 7,796 | 21 | 15,593 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A balanced bracket sequence is defined as an integer sequence that can be built with the following rules:
* The empty sequence is balanced.
* If [a_1,β¦,a_n] and [b_1,β¦, b_m] are balanced, then their concatenation [a_1,β¦,a_n,b_1,β¦,b_m] is balanced.
* If x is a positive integer and [a_1,β¦,a_n] is balanced, then [x,a_1,β¦,a_n,-x] is balanced.
The positive numbers can be imagined as opening brackets and the negative numbers as closing brackets, where matching brackets must have the same type (absolute value). For example, [1, 2, -2, -1] and [1, 3, -3, 2, -2, -1] are balanced, but [1, 2, -1, -2] and [-1, 1] are not balanced.
There are 2n cards. Each card has a number on the front and a number on the back. Each integer 1,-1,2,-2,β¦,n,-n appears exactly once on the front of some card and exactly once on the back of some (not necessarily the same) card.
You can reorder the cards however you like. You are not allowed to flip cards, so numbers cannot move between the front and back. Your task is to order the cards so that the sequences given by the front numbers and the back numbers are both balanced, or report that it is impossible.
Input
The first line contains a single integer n (1β€ nβ€ 2β
10^5) β the number of bracket types, and half the number of cards.
The next 2n lines describe the cards. The i-th of these lines contains two integers a_i, b_i (-nβ€ a_i,b_iβ€ n, a_iβ 0, b_iβ 0) β the numbers on the front and back of the i-th card, respectively. Every integer 1,-1,2,-2,β¦,n,-n appears exactly once as a_i and exactly once as b_i.
Output
On the first line, output "YES" if it's possible to reorder these cards to satisfy the condition. Otherwise, output "NO". You can print each letter in any case (upper or lower).
If it is possible, on the next 2n lines output the cards in an order so that the front and back are both balanced. If there are multiple solutions, you may output any.
Examples
Input
5
1 3
-3 -5
4 -3
2 2
-1 -4
-2 5
3 -1
5 1
-4 4
-5 -2
Output
YES
1 3
4 -3
-4 4
-1 -4
5 1
3 -1
2 2
-2 5
-3 -5
-5 -2
Input
2
1 1
-1 2
2 -1
-2 -2
Output
NO
Note
In the first test case, the front numbers create the balanced sequence [1,4,-4,-1,5,3,2,-2,-3,-5] and the back numbers create the balanced sequence [3,-3,4,-4,1,-1,2,5,-5,-2].
In the second test case, the cards are given in an order so that the front numbers are balanced, but the back numbers create the unbalanced sequence [1,2,-1,-2]. If we swapped the second and third cards, we would balance the back numbers and unbalance the front numbers. But there is no order that balances both.
Submitted Solution:
```
import itertools
from typing import List
n = int(input())
d = dict(map(int, input().split()) for _ in range(2 * n))
def isValid(nums: List[int]) -> bool:
"""Removes same consecutive brackets one at a time until length of nums becomes zero and return True, otherwise false.
Args:
nums (List[int]): array of front or back values.
Returns:
bool: Returns True if a valid sequence, else False.
"""
while len(nums) > 0:
l = len(nums)
for i, v in enumerate(nums):
if i > 0:
if abs(v) == abs(nums[i - 1]):
if v < 0:
nums.pop(i)
nums.pop(i - 1)
if len(nums) == l:
return False
return True
def solve(dic) -> str:
"""Creates permuntations of card order, unzips front and back values in x, y and passes them in isValid() function as a parameter.
Args:
arr (List[List[int]]): 2d array of front and back values of the card. [[front,back],..].
Returns:
str: returns 'Yes' if both front values and back values are valid bracket sequence. Otherwise returns 'No'
"""
for i in itertools.permutations(list(d.keys()), len(dic)):
front = list(i)
back = [dic[x] for x in front]
if isValid(front[:]) == True and isValid(back[:]) == True:
print(front)
print(back)
return ['YES', front, back]
return ['N0']
ans = solve(d)
if len(ans) > 1:
print(ans[0])
res = "\n".join("{} {}".format(x, y) for x, y in zip(ans[1], ans[2]))
print(res)
else:
print(ans[0])
``` | instruction | 0 | 7,797 | 21 | 15,594 |
No | output | 1 | 7,797 | 21 | 15,595 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A balanced bracket sequence is defined as an integer sequence that can be built with the following rules:
* The empty sequence is balanced.
* If [a_1,β¦,a_n] and [b_1,β¦, b_m] are balanced, then their concatenation [a_1,β¦,a_n,b_1,β¦,b_m] is balanced.
* If x is a positive integer and [a_1,β¦,a_n] is balanced, then [x,a_1,β¦,a_n,-x] is balanced.
The positive numbers can be imagined as opening brackets and the negative numbers as closing brackets, where matching brackets must have the same type (absolute value). For example, [1, 2, -2, -1] and [1, 3, -3, 2, -2, -1] are balanced, but [1, 2, -1, -2] and [-1, 1] are not balanced.
There are 2n cards. Each card has a number on the front and a number on the back. Each integer 1,-1,2,-2,β¦,n,-n appears exactly once on the front of some card and exactly once on the back of some (not necessarily the same) card.
You can reorder the cards however you like. You are not allowed to flip cards, so numbers cannot move between the front and back. Your task is to order the cards so that the sequences given by the front numbers and the back numbers are both balanced, or report that it is impossible.
Input
The first line contains a single integer n (1β€ nβ€ 2β
10^5) β the number of bracket types, and half the number of cards.
The next 2n lines describe the cards. The i-th of these lines contains two integers a_i, b_i (-nβ€ a_i,b_iβ€ n, a_iβ 0, b_iβ 0) β the numbers on the front and back of the i-th card, respectively. Every integer 1,-1,2,-2,β¦,n,-n appears exactly once as a_i and exactly once as b_i.
Output
On the first line, output "YES" if it's possible to reorder these cards to satisfy the condition. Otherwise, output "NO". You can print each letter in any case (upper or lower).
If it is possible, on the next 2n lines output the cards in an order so that the front and back are both balanced. If there are multiple solutions, you may output any.
Examples
Input
5
1 3
-3 -5
4 -3
2 2
-1 -4
-2 5
3 -1
5 1
-4 4
-5 -2
Output
YES
1 3
4 -3
-4 4
-1 -4
5 1
3 -1
2 2
-2 5
-3 -5
-5 -2
Input
2
1 1
-1 2
2 -1
-2 -2
Output
NO
Note
In the first test case, the front numbers create the balanced sequence [1,4,-4,-1,5,3,2,-2,-3,-5] and the back numbers create the balanced sequence [3,-3,4,-4,1,-1,2,5,-5,-2].
In the second test case, the cards are given in an order so that the front numbers are balanced, but the back numbers create the unbalanced sequence [1,2,-1,-2]. If we swapped the second and third cards, we would balance the back numbers and unbalance the front numbers. But there is no order that balances both.
Submitted Solution:
```
import itertools
from typing import List
n = int(input())
d = dict(map(int, input().split()) for _ in range(2 * n))
def isValid(nums: List[int]) -> bool:
"""Removes same consecutive brackets one at a time until length of nums becomes zero and return True, otherwise false.
Args:
nums (List[int]): array of front or back values.
Returns:
bool: Returns True if a valid sequence, else False.
"""
while len(nums) > 0:
l = len(nums)
for i, v in enumerate(nums):
if i > 0:
if abs(v) == abs(nums[i - 1]):
if v < 0:
nums.pop(i)
nums.pop(i - 1)
if len(nums) == l:
return False
return True
def solve(dic) -> str:
"""Creates permuntations of card order, unzips front and back values in x, y and passes them in isValid() function as a parameter.
Args:
arr (List[List[int]]): 2d array of front and back values of the card. [[front,back],..].
Returns:
str: returns 'Yes' if both front values and back values are valid bracket sequence. Otherwise returns 'No'
"""
for i in itertools.permutations(list(d.keys()), len(dic)):
front = list(i)
back = [dic[x] for x in front]
if isValid(front) == True and isValid(back) == True:
return ['YES', front, back]
return ['NO']
ans = solve(d)
if len(ans) > 1:
print(ans[0])
res = "\n".join("{} {}".format(x, y) for x, y in zip(ans[1], ans[2]))
print(res)
else:
print(ans[0])
``` | instruction | 0 | 7,798 | 21 | 15,596 |
No | output | 1 | 7,798 | 21 | 15,597 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A balanced bracket sequence is defined as an integer sequence that can be built with the following rules:
* The empty sequence is balanced.
* If [a_1,β¦,a_n] and [b_1,β¦, b_m] are balanced, then their concatenation [a_1,β¦,a_n,b_1,β¦,b_m] is balanced.
* If x is a positive integer and [a_1,β¦,a_n] is balanced, then [x,a_1,β¦,a_n,-x] is balanced.
The positive numbers can be imagined as opening brackets and the negative numbers as closing brackets, where matching brackets must have the same type (absolute value). For example, [1, 2, -2, -1] and [1, 3, -3, 2, -2, -1] are balanced, but [1, 2, -1, -2] and [-1, 1] are not balanced.
There are 2n cards. Each card has a number on the front and a number on the back. Each integer 1,-1,2,-2,β¦,n,-n appears exactly once on the front of some card and exactly once on the back of some (not necessarily the same) card.
You can reorder the cards however you like. You are not allowed to flip cards, so numbers cannot move between the front and back. Your task is to order the cards so that the sequences given by the front numbers and the back numbers are both balanced, or report that it is impossible.
Input
The first line contains a single integer n (1β€ nβ€ 2β
10^5) β the number of bracket types, and half the number of cards.
The next 2n lines describe the cards. The i-th of these lines contains two integers a_i, b_i (-nβ€ a_i,b_iβ€ n, a_iβ 0, b_iβ 0) β the numbers on the front and back of the i-th card, respectively. Every integer 1,-1,2,-2,β¦,n,-n appears exactly once as a_i and exactly once as b_i.
Output
On the first line, output "YES" if it's possible to reorder these cards to satisfy the condition. Otherwise, output "NO". You can print each letter in any case (upper or lower).
If it is possible, on the next 2n lines output the cards in an order so that the front and back are both balanced. If there are multiple solutions, you may output any.
Examples
Input
5
1 3
-3 -5
4 -3
2 2
-1 -4
-2 5
3 -1
5 1
-4 4
-5 -2
Output
YES
1 3
4 -3
-4 4
-1 -4
5 1
3 -1
2 2
-2 5
-3 -5
-5 -2
Input
2
1 1
-1 2
2 -1
-2 -2
Output
NO
Note
In the first test case, the front numbers create the balanced sequence [1,4,-4,-1,5,3,2,-2,-3,-5] and the back numbers create the balanced sequence [3,-3,4,-4,1,-1,2,5,-5,-2].
In the second test case, the cards are given in an order so that the front numbers are balanced, but the back numbers create the unbalanced sequence [1,2,-1,-2]. If we swapped the second and third cards, we would balance the back numbers and unbalance the front numbers. But there is no order that balances both.
Submitted Solution:
```
import itertools
from typing import List
n = int(input())
d = dict(map(int, input().split()) for _ in range(2 * n))
def isValid(nums: List[int]) -> bool:
"""Removes same consecutive brackets one at a time until length of nums becomes zero and return True, otherwise false.
Args:
nums (List[int]): array of front or back values.
Returns:
bool: Returns True if a valid sequence, else False.
"""
while len(nums) > 0:
l = len(nums)
for i, v in enumerate(nums):
if i > 0:
if -1 * v == nums[i - 1]:
nums.pop(i)
nums.pop(i - 1)
if len(nums) == l:
return False
return True
def solve(dic) -> str:
"""Creates permuntations of card order, unzips front and back values in x, y and passes them in isValid() function as a parameter.
Args:
arr (List[List[int]]): 2d array of front and back values of the card. [[front,back],..].
Returns:
str: returns 'Yes' if both front values and back values are valid bracket sequence. Otherwise returns 'No'
"""
for i in itertools.permutations(dic, len(dic)):
i = list(i)
if isValid(i) == True and isValid([dic[x] for x in i]) == True:
return 'YES'
return 'NO'
print(solve(d))
``` | instruction | 0 | 7,799 | 21 | 15,598 |
No | output | 1 | 7,799 | 21 | 15,599 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is a harder version of the problem. In this version, n β€ 300 000.
Vasya is an experienced developer of programming competitions' problems. As all great minds at some time, Vasya faced a creative crisis. To improve the situation, Petya gifted him a string consisting of opening and closing brackets only. Petya believes, that the beauty of the bracket string is a number of its cyclical shifts, which form a correct bracket sequence.
To digress from his problems, Vasya decided to select two positions of the string (not necessarily distinct) and swap characters located at this positions with each other. Vasya will apply this operation exactly once. He is curious what is the maximum possible beauty he can achieve this way. Please help him.
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.
The cyclical shift of the string s of length n by k (0 β€ k < n) is a string formed by a concatenation of the last k symbols of the string s with the first n - k symbols of string s. For example, the cyclical shift of string "(())()" by 2 equals "()(())".
Cyclical shifts i and j are considered different, if i β j.
Input
The first line contains an integer n (1 β€ n β€ 300 000), the length of the string.
The second line contains a string, consisting of exactly n characters, where each of the characters is either "(" or ")".
Output
The first line should contain a single integer β the largest beauty of the string, which can be achieved by swapping some two characters.
The second line should contain integers l and r (1 β€ l, r β€ n) β the indices of two characters, which should be swapped in order to maximize the string's beauty.
In case there are several possible swaps, print any of them.
Examples
Input
10
()()())(()
Output
5
8 7
Input
12
)(()(()())()
Output
4
5 10
Input
6
)))(()
Output
0
1 1
Note
In the first example, we can swap 7-th and 8-th character, obtaining a string "()()()()()". The cyclical shifts by 0, 2, 4, 6, 8 of this string form a correct bracket sequence.
In the second example, after swapping 5-th and 10-th character, we obtain a string ")(())()()(()". The cyclical shifts by 11, 7, 5, 3 of this string form a correct bracket sequence.
In the third example, swap of any two brackets results in 0 cyclical shifts being correct bracket sequences. | instruction | 0 | 8,456 | 21 | 16,912 |
Tags: implementation
Correct Solution:
```
n = int(input())
s = input()
r = 0
l = n-1
root = []
buf = []
to_the_right = True
for count in range(n):
if to_the_right:
i = r
r += 1
else:
i = l
l -= 1
b = s[i]
if b == '(':
if len(buf) == 0 or buf[-1][0] != -1:
buf.append([-1,-1,[]])
buf[-1][0] = i
else:
if len(buf) == 0 or buf[-1][1] != -1:
buf.append([-1,-1,root])
root = []
to_the_right = False
buf[-1][1] = i
if buf[-1][0] != -1 and buf[-1][1] != -1:
tmp = buf.pop()
if len(buf):
buf[-1][2].append(tmp)
else:
root.append(tmp)
to_the_right = True
sol = [[0,1,1]]
if len(buf) == 0:
sol.append([len(root), 1, 1])
for child in root:
sol.append([len(child[2])+1, child[0]+1, child[1]+1])
for gr_child in child[2]:
sol.append([len(root)+len(gr_child[2])+1, gr_child[0]+1, gr_child[1]+1])
print('%d\n%d %d'%tuple(max(sol)))
``` | output | 1 | 8,456 | 21 | 16,913 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is a harder version of the problem. In this version, n β€ 300 000.
Vasya is an experienced developer of programming competitions' problems. As all great minds at some time, Vasya faced a creative crisis. To improve the situation, Petya gifted him a string consisting of opening and closing brackets only. Petya believes, that the beauty of the bracket string is a number of its cyclical shifts, which form a correct bracket sequence.
To digress from his problems, Vasya decided to select two positions of the string (not necessarily distinct) and swap characters located at this positions with each other. Vasya will apply this operation exactly once. He is curious what is the maximum possible beauty he can achieve this way. Please help him.
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.
The cyclical shift of the string s of length n by k (0 β€ k < n) is a string formed by a concatenation of the last k symbols of the string s with the first n - k symbols of string s. For example, the cyclical shift of string "(())()" by 2 equals "()(())".
Cyclical shifts i and j are considered different, if i β j.
Input
The first line contains an integer n (1 β€ n β€ 300 000), the length of the string.
The second line contains a string, consisting of exactly n characters, where each of the characters is either "(" or ")".
Output
The first line should contain a single integer β the largest beauty of the string, which can be achieved by swapping some two characters.
The second line should contain integers l and r (1 β€ l, r β€ n) β the indices of two characters, which should be swapped in order to maximize the string's beauty.
In case there are several possible swaps, print any of them.
Examples
Input
10
()()())(()
Output
5
8 7
Input
12
)(()(()())()
Output
4
5 10
Input
6
)))(()
Output
0
1 1
Note
In the first example, we can swap 7-th and 8-th character, obtaining a string "()()()()()". The cyclical shifts by 0, 2, 4, 6, 8 of this string form a correct bracket sequence.
In the second example, after swapping 5-th and 10-th character, we obtain a string ")(())()()(()". The cyclical shifts by 11, 7, 5, 3 of this string form a correct bracket sequence.
In the third example, swap of any two brackets results in 0 cyclical shifts being correct bracket sequences. | instruction | 0 | 8,457 | 21 | 16,914 |
Tags: implementation
Correct Solution:
```
n = int(input())
ddd = input()
d = [0]
for dd in ddd:
if dd == '(':
d.append(d[-1] + 1)
else:
d.append(d[-1] - 1)
if d[-1] != 0:
print("0\n1 1")
exit(0)
d.pop()
mn = min(d)
ind = d.index(mn)
d = d[ind:] + d[:ind]
d = [i - mn for i in d]
fi = -1
crfi = -1
li = -1
mx = 0
cr = 0
cnt0 = 0
for i in range(n):
dd = d[i]
if dd == 0:
cnt0 += 1
if dd == 2:
if cr == 0:
crfi = i
cr += 1
if cr > mx:
fi = crfi
li = i
mx = cr
elif dd < 2:
cr = 0
# print('=========')
# print(d)
# print(cnt0)
# print(fi, li)
# print(mx)
# print("=========")
# if fi == -1:
# print(cnt0)
# print(1, 1)
# else:
# print(cnt0 + mx)
# print(fi, li + 2)
if fi == -1:
ans1 = [cnt0, 0, 0]
else:
ans1 = [cnt0 + mx, fi-1, li]
fi = -1
crfi = -1
li = -1
mx = 0
cr = 0
for i in range(n):
dd = d[i]
if dd == 1:
if cr == 0:
crfi = i
cr += 1
if cr > mx:
fi = crfi
li = i
mx = cr
elif dd < 1:
cr = 0
ans2 = [mx, fi-1, li]
if ans1[0] > ans2[0]:
print(ans1[0])
print(((ans1[1] + ind)%n) + 1, ((ans1[2] + ind)%n) + 1)
else:
print(ans2[0])
print(((ans2[1] + ind)%n) + 1, ((ans2[2] + ind)%n) + 1)
``` | output | 1 | 8,457 | 21 | 16,915 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is a harder version of the problem. In this version, n β€ 300 000.
Vasya is an experienced developer of programming competitions' problems. As all great minds at some time, Vasya faced a creative crisis. To improve the situation, Petya gifted him a string consisting of opening and closing brackets only. Petya believes, that the beauty of the bracket string is a number of its cyclical shifts, which form a correct bracket sequence.
To digress from his problems, Vasya decided to select two positions of the string (not necessarily distinct) and swap characters located at this positions with each other. Vasya will apply this operation exactly once. He is curious what is the maximum possible beauty he can achieve this way. Please help him.
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.
The cyclical shift of the string s of length n by k (0 β€ k < n) is a string formed by a concatenation of the last k symbols of the string s with the first n - k symbols of string s. For example, the cyclical shift of string "(())()" by 2 equals "()(())".
Cyclical shifts i and j are considered different, if i β j.
Input
The first line contains an integer n (1 β€ n β€ 300 000), the length of the string.
The second line contains a string, consisting of exactly n characters, where each of the characters is either "(" or ")".
Output
The first line should contain a single integer β the largest beauty of the string, which can be achieved by swapping some two characters.
The second line should contain integers l and r (1 β€ l, r β€ n) β the indices of two characters, which should be swapped in order to maximize the string's beauty.
In case there are several possible swaps, print any of them.
Examples
Input
10
()()())(()
Output
5
8 7
Input
12
)(()(()())()
Output
4
5 10
Input
6
)))(()
Output
0
1 1
Note
In the first example, we can swap 7-th and 8-th character, obtaining a string "()()()()()". The cyclical shifts by 0, 2, 4, 6, 8 of this string form a correct bracket sequence.
In the second example, after swapping 5-th and 10-th character, we obtain a string ")(())()()(()". The cyclical shifts by 11, 7, 5, 3 of this string form a correct bracket sequence.
In the third example, swap of any two brackets results in 0 cyclical shifts being correct bracket sequences. | instruction | 0 | 8,458 | 21 | 16,916 |
Tags: implementation
Correct Solution:
```
n = int(input().strip())
s= input().strip()
ss= 0
mina = 0
ti = 0
for k in range(len(s)):
if(s[k] == "("):
ss+=1
else:
ss-=1
if(ss<0):
ti = k+1
ss = 0
s=s[ti:]+s[:ti]
#print(s)
ss= 0
for k in range(len(s)):
if(s[k] == "("):
ss+=1
else:
ss-=1
if(ss<0):
print(0)
print(1,1)
break
else:
if(ss == 0):
pre=[0 for k in range(len(s))]
ss=0
for k in range(len(s)):
if (s[k] == "("):
ss += 1
else:
ss -= 1
pre[k] = ss
tt = 0
a =(1,1)
for k in range(0,len(s)):
if(pre[k] == 0):
tt+=1
maxi= tt
#print(pre)
g =0
gg =0
while(gg<len(s)):
if(pre[gg] == 0):
#print(gg,g,"g")
if(gg != g+1):
yy = g+1
y = g+1
q = 0
while(yy<gg):
if(pre[yy] == 1):
# print(yy,y,"y")
if(yy !=y+1):
rr = y+1
r = y+1
h = 0
while(rr<yy):
if(pre[rr] == 2):
h+=1
rr+=1
if(tt+h+1>maxi):
maxi = tt + h + 1
a=(y,yy)
else:
if(tt+1>maxi):
maxi =tt+1
a=(y,yy)
#print(a, a)
q+=1
y = yy+1
yy = y
else:
yy+=1
if (q + 1 > maxi):
maxi = q+1
a = (g, gg)
g= gg+1
gg= g
else:
gg+=1
print(maxi)
# print(a)
print((a[0]+ti)%len(s)+1,(a[1]+ti)%len(s)+1)
else:
print(0)
print(1,1)
``` | output | 1 | 8,458 | 21 | 16,917 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is a harder version of the problem. In this version, n β€ 300 000.
Vasya is an experienced developer of programming competitions' problems. As all great minds at some time, Vasya faced a creative crisis. To improve the situation, Petya gifted him a string consisting of opening and closing brackets only. Petya believes, that the beauty of the bracket string is a number of its cyclical shifts, which form a correct bracket sequence.
To digress from his problems, Vasya decided to select two positions of the string (not necessarily distinct) and swap characters located at this positions with each other. Vasya will apply this operation exactly once. He is curious what is the maximum possible beauty he can achieve this way. Please help him.
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.
The cyclical shift of the string s of length n by k (0 β€ k < n) is a string formed by a concatenation of the last k symbols of the string s with the first n - k symbols of string s. For example, the cyclical shift of string "(())()" by 2 equals "()(())".
Cyclical shifts i and j are considered different, if i β j.
Input
The first line contains an integer n (1 β€ n β€ 300 000), the length of the string.
The second line contains a string, consisting of exactly n characters, where each of the characters is either "(" or ")".
Output
The first line should contain a single integer β the largest beauty of the string, which can be achieved by swapping some two characters.
The second line should contain integers l and r (1 β€ l, r β€ n) β the indices of two characters, which should be swapped in order to maximize the string's beauty.
In case there are several possible swaps, print any of them.
Examples
Input
10
()()())(()
Output
5
8 7
Input
12
)(()(()())()
Output
4
5 10
Input
6
)))(()
Output
0
1 1
Note
In the first example, we can swap 7-th and 8-th character, obtaining a string "()()()()()". The cyclical shifts by 0, 2, 4, 6, 8 of this string form a correct bracket sequence.
In the second example, after swapping 5-th and 10-th character, we obtain a string ")(())()()(()". The cyclical shifts by 11, 7, 5, 3 of this string form a correct bracket sequence.
In the third example, swap of any two brackets results in 0 cyclical shifts being correct bracket sequences.
Submitted Solution:
```
n = int(input().strip())
s= input().strip()
ss= 0
mina = 0
ti = 0
for k in range(len(s)):
if(s[k] == "("):
ss+=1
else:
ss-=1
if(ss<0):
ti = k+1
ss = 0
s=s[ti:]+s[:ti]
ss= 0
for k in range(len(s)):
if(s[k] == "("):
ss+=1
else:
ss-=1
if(ss<0):
print(0)
print(1,1)
break
else:
if(ss == 0):
pre=[0 for k in range(len(s))]
for k in range(len(s)):
if (s[k] == "("):
ss += 1
else:
ss -= 1
pre[k] = ss
tt = 0
a =(1,1)
for k in range(0,len(s)):
if(pre[k] == 0):
tt+=1
maxi= tt
g =0
gg =0
while(gg<len(s)):
if(pre[gg] == 0):
if(gg != g+1):
yy = g+1
y = g+1
q = 0
while(yy<gg):
if(pre[yy] == 1):
if(yy !=y+1):
rr = y+1
r = y+1
h = 0
while(rr<yy):
if(pre[rr] == 2):
h+=1
rr+=1
if(tt+h+1>maxi):
maxi = tt + h + 1
a=(y,yy)
q+=1
y = yy+1
yy = y
else:
yy+=1
if (q + 1 > maxi):
maxi = q+1
a = (g, gg)
g= gg+1
gg= g
else:
gg+=1
print(maxi)
print((a[0]+ti)%len(s)+1,(a[1]+ti)%len(s)+1)
else:
print(0)
print(1,1)
``` | instruction | 0 | 8,459 | 21 | 16,918 |
No | output | 1 | 8,459 | 21 | 16,919 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is a harder version of the problem. In this version, n β€ 300 000.
Vasya is an experienced developer of programming competitions' problems. As all great minds at some time, Vasya faced a creative crisis. To improve the situation, Petya gifted him a string consisting of opening and closing brackets only. Petya believes, that the beauty of the bracket string is a number of its cyclical shifts, which form a correct bracket sequence.
To digress from his problems, Vasya decided to select two positions of the string (not necessarily distinct) and swap characters located at this positions with each other. Vasya will apply this operation exactly once. He is curious what is the maximum possible beauty he can achieve this way. Please help him.
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.
The cyclical shift of the string s of length n by k (0 β€ k < n) is a string formed by a concatenation of the last k symbols of the string s with the first n - k symbols of string s. For example, the cyclical shift of string "(())()" by 2 equals "()(())".
Cyclical shifts i and j are considered different, if i β j.
Input
The first line contains an integer n (1 β€ n β€ 300 000), the length of the string.
The second line contains a string, consisting of exactly n characters, where each of the characters is either "(" or ")".
Output
The first line should contain a single integer β the largest beauty of the string, which can be achieved by swapping some two characters.
The second line should contain integers l and r (1 β€ l, r β€ n) β the indices of two characters, which should be swapped in order to maximize the string's beauty.
In case there are several possible swaps, print any of them.
Examples
Input
10
()()())(()
Output
5
8 7
Input
12
)(()(()())()
Output
4
5 10
Input
6
)))(()
Output
0
1 1
Note
In the first example, we can swap 7-th and 8-th character, obtaining a string "()()()()()". The cyclical shifts by 0, 2, 4, 6, 8 of this string form a correct bracket sequence.
In the second example, after swapping 5-th and 10-th character, we obtain a string ")(())()()(()". The cyclical shifts by 11, 7, 5, 3 of this string form a correct bracket sequence.
In the third example, swap of any two brackets results in 0 cyclical shifts being correct bracket sequences.
Submitted Solution:
```
n = int(input().strip())
s= input().strip()
ss= 0
mina = 0
ti = 0
for k in range(len(s)):
if(s[k] == "("):
ss+=1
else:
ss-=1
if(ss<0):
ti = k+1
ss = 0
s=s[ti:]+s[:ti]
print(s)
ss= 0
for k in range(len(s)):
if(s[k] == "("):
ss+=1
else:
ss-=1
if(ss<0):
print(0)
print(1,1)
break
else:
if(ss == 0):
pre=[0 for k in range(len(s))]
for k in range(len(s)):
if (s[k] == "("):
ss += 1
else:
ss -= 1
pre[k] = ss
tt = 0
a =(1,1)
for k in range(0,len(s)):
if(pre[k] == 0):
tt+=1
maxi= tt
g =0
gg =0
while(gg<len(s)):
if(pre[gg] == 0):
if(gg != g+1):
yy = g+1
y = g+1
q = 0
while(yy<gg):
if(pre[yy] == 1):
if(yy !=y+1):
rr = y+1
r = y+1
h = 0
while(rr<yy):
if(pre[rr] == 2):
h+=1
rr+=1
if(tt+h+1>maxi):
maxi = tt + h + 1
a=(y,yy)
q+=1
y = yy+1
yy = y
else:
yy+=1
if (q + 1 > maxi):
maxi = q+1
a = (g, gg)
g= gg+1
gg= g
else:
gg+=1
print(maxi)
print((a[0]+ti)%len(s)+1,(a[1]+ti)%len(s)+1)
else:
print(0)
print(1,1)
``` | instruction | 0 | 8,460 | 21 | 16,920 |
No | output | 1 | 8,460 | 21 | 16,921 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is a harder version of the problem. In this version, n β€ 300 000.
Vasya is an experienced developer of programming competitions' problems. As all great minds at some time, Vasya faced a creative crisis. To improve the situation, Petya gifted him a string consisting of opening and closing brackets only. Petya believes, that the beauty of the bracket string is a number of its cyclical shifts, which form a correct bracket sequence.
To digress from his problems, Vasya decided to select two positions of the string (not necessarily distinct) and swap characters located at this positions with each other. Vasya will apply this operation exactly once. He is curious what is the maximum possible beauty he can achieve this way. Please help him.
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.
The cyclical shift of the string s of length n by k (0 β€ k < n) is a string formed by a concatenation of the last k symbols of the string s with the first n - k symbols of string s. For example, the cyclical shift of string "(())()" by 2 equals "()(())".
Cyclical shifts i and j are considered different, if i β j.
Input
The first line contains an integer n (1 β€ n β€ 300 000), the length of the string.
The second line contains a string, consisting of exactly n characters, where each of the characters is either "(" or ")".
Output
The first line should contain a single integer β the largest beauty of the string, which can be achieved by swapping some two characters.
The second line should contain integers l and r (1 β€ l, r β€ n) β the indices of two characters, which should be swapped in order to maximize the string's beauty.
In case there are several possible swaps, print any of them.
Examples
Input
10
()()())(()
Output
5
8 7
Input
12
)(()(()())()
Output
4
5 10
Input
6
)))(()
Output
0
1 1
Note
In the first example, we can swap 7-th and 8-th character, obtaining a string "()()()()()". The cyclical shifts by 0, 2, 4, 6, 8 of this string form a correct bracket sequence.
In the second example, after swapping 5-th and 10-th character, we obtain a string ")(())()()(()". The cyclical shifts by 11, 7, 5, 3 of this string form a correct bracket sequence.
In the third example, swap of any two brackets results in 0 cyclical shifts being correct bracket sequences.
Submitted Solution:
```
n = int(input())
ddd = input()
d = [0]
for dd in ddd:
if dd == '(':
d.append(d[-1] + 1)
else:
d.append(d[-1] - 1)
if d[-1] != 0:
print("0\n1 1")
exit(0)
d.pop()
mn = min(d)
ind = d.index(mn)
d = d[ind:] + d[:ind]
d = [i - mn for i in d]
fi = -1
crfi = -1
li = -1
mx = 0
cr = 0
cnt0 = 0
for i in range(n):
dd = d[i]
if dd == 0:
cnt0 += 1
if dd == 2:
if cr == 0:
crfi = i
cr += 1
if cr > mx:
fi = crfi
li = i
mx = cr
elif dd < 2:
cr = 0
print('=========')
print(d)
print(cnt0)
print(fi, li)
print(mx)
print("=========")
# if fi == -1:
# print(cnt0)
# print(1, 1)
# else:
# print(cnt0 + mx)
# print(fi, li + 2)
if fi == -1:
ans1 = [cnt0, 0, 0]
else:
ans1 = [cnt0 + mx, fi-1, li]
fi = -1
crfi = -1
li = -1
mx = 0
cr = 0
for i in range(n):
dd = d[i]
if dd == 1:
if cr == 0:
crfi = i
cr += 1
if cr > mx:
fi = crfi
li = i
mx = cr
elif dd < 1:
cr = 0
ans2 = [mx, fi-1, li]
if ans1[0] > ans2[0]:
print(ans1[0])
print(((ans1[1] + ind)%n) + 1, ((ans1[2] + ind)%n) + 1)
else:
print(ans2[0])
print(((ans2[1] + ind)%n) + 1, ((ans2[2] + ind)%n) + 1)
``` | instruction | 0 | 8,461 | 21 | 16,922 |
No | output | 1 | 8,461 | 21 | 16,923 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s, consisting of brackets of two types: '(', ')', '[' and ']'.
A string is called a regular bracket sequence (RBS) if it's of one of the following types:
* empty string;
* '(' + RBS + ')';
* '[' + RBS + ']';
* RBS + RBS.
where plus is a concatenation of two strings.
In one move you can choose a non-empty subsequence of the string s (not necessarily consecutive) that is an RBS, remove it from the string and concatenate the remaining parts without changing the order.
What is the maximum number of moves you can perform?
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of testcases.
Each of the next t lines contains a non-empty string, consisting only of characters '(', ')', '[' and ']'. The total length of the strings over all testcases doesn't exceed 2 β
10^5.
Output
For each testcase print a single integer β the maximum number of moves you can perform on a given string s.
Example
Input
5
()
[]()
([)]
)]([
)[(]
Output
1
2
2
0
1
Note
In the first example you can just erase the whole string.
In the second example you can first erase the brackets on positions 1 and 2: "[]()", then "()" is left. After that you can erase it whole. You could erase the whole string from the beginning but you would get one move instead of two.
In the third example you can first erase the brackets on positions 1 and 3: "([)]". They form an RBS "()". Then "[]" is left, so you can erase it whole.
In the fourth example there is no subsequence that is an RBS, so you can't perform a move at all.
In the fifth example you can erase the brackets on positions 2 and 4: ")[(]" and get ")(" as a result. You can erase nothing from it. | instruction | 0 | 16,994 | 21 | 33,988 |
Tags: greedy
Correct Solution:
```
t = int(input())
for _ in range(t):
s = input()
zso = kso = sum = 0
for i in range(len(s)):
if s[i] == "(":
zso += 1
elif s[i] == "[":
kso += 1
elif s[i] == ")" and zso > 0:
zso -= 1
sum += 1
elif s[i] == "]" and kso > 0:
kso -= 1
sum += 1
print(sum)
``` | output | 1 | 16,994 | 21 | 33,989 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s, consisting of brackets of two types: '(', ')', '[' and ']'.
A string is called a regular bracket sequence (RBS) if it's of one of the following types:
* empty string;
* '(' + RBS + ')';
* '[' + RBS + ']';
* RBS + RBS.
where plus is a concatenation of two strings.
In one move you can choose a non-empty subsequence of the string s (not necessarily consecutive) that is an RBS, remove it from the string and concatenate the remaining parts without changing the order.
What is the maximum number of moves you can perform?
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of testcases.
Each of the next t lines contains a non-empty string, consisting only of characters '(', ')', '[' and ']'. The total length of the strings over all testcases doesn't exceed 2 β
10^5.
Output
For each testcase print a single integer β the maximum number of moves you can perform on a given string s.
Example
Input
5
()
[]()
([)]
)]([
)[(]
Output
1
2
2
0
1
Note
In the first example you can just erase the whole string.
In the second example you can first erase the brackets on positions 1 and 2: "[]()", then "()" is left. After that you can erase it whole. You could erase the whole string from the beginning but you would get one move instead of two.
In the third example you can first erase the brackets on positions 1 and 3: "([)]". They form an RBS "()". Then "[]" is left, so you can erase it whole.
In the fourth example there is no subsequence that is an RBS, so you can't perform a move at all.
In the fifth example you can erase the brackets on positions 2 and 4: ")[(]" and get ")(" as a result. You can erase nothing from it. | instruction | 0 | 16,995 | 21 | 33,990 |
Tags: greedy
Correct Solution:
```
t = int(input())
for _ in range(t):
s = input().strip()
curr0 = 0
num0 = 0
curr1 = 0
num1 = 0
for y in s:
if y == '(':
curr0 += 1
elif y == '[':
curr1 += 1
elif y == ')':
if curr0 > 0:
curr0 -= 1
num0 += 1
elif curr1 > 0:
curr1 -= 1
num1 += 1
print(num0+num1)
``` | output | 1 | 16,995 | 21 | 33,991 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s, consisting of brackets of two types: '(', ')', '[' and ']'.
A string is called a regular bracket sequence (RBS) if it's of one of the following types:
* empty string;
* '(' + RBS + ')';
* '[' + RBS + ']';
* RBS + RBS.
where plus is a concatenation of two strings.
In one move you can choose a non-empty subsequence of the string s (not necessarily consecutive) that is an RBS, remove it from the string and concatenate the remaining parts without changing the order.
What is the maximum number of moves you can perform?
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of testcases.
Each of the next t lines contains a non-empty string, consisting only of characters '(', ')', '[' and ']'. The total length of the strings over all testcases doesn't exceed 2 β
10^5.
Output
For each testcase print a single integer β the maximum number of moves you can perform on a given string s.
Example
Input
5
()
[]()
([)]
)]([
)[(]
Output
1
2
2
0
1
Note
In the first example you can just erase the whole string.
In the second example you can first erase the brackets on positions 1 and 2: "[]()", then "()" is left. After that you can erase it whole. You could erase the whole string from the beginning but you would get one move instead of two.
In the third example you can first erase the brackets on positions 1 and 3: "([)]". They form an RBS "()". Then "[]" is left, so you can erase it whole.
In the fourth example there is no subsequence that is an RBS, so you can't perform a move at all.
In the fifth example you can erase the brackets on positions 2 and 4: ")[(]" and get ")(" as a result. You can erase nothing from it. | instruction | 0 | 16,996 | 21 | 33,992 |
Tags: greedy
Correct Solution:
```
import math
for _ in range(int(input())):
#n = int(input())
ar = list(input())
o1,o2,c1,c2 = 0,0,0,0
count = 0
for i in ar:
if i == '[':
o1 +=1
elif i == ']' and o1>0:
count +=1
o1-=1
for i in ar:
if i == '(':
o2 +=1
elif i == ')' and o2>0:
count +=1
o2 -= 1
print(count)
``` | output | 1 | 16,996 | 21 | 33,993 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s, consisting of brackets of two types: '(', ')', '[' and ']'.
A string is called a regular bracket sequence (RBS) if it's of one of the following types:
* empty string;
* '(' + RBS + ')';
* '[' + RBS + ']';
* RBS + RBS.
where plus is a concatenation of two strings.
In one move you can choose a non-empty subsequence of the string s (not necessarily consecutive) that is an RBS, remove it from the string and concatenate the remaining parts without changing the order.
What is the maximum number of moves you can perform?
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of testcases.
Each of the next t lines contains a non-empty string, consisting only of characters '(', ')', '[' and ']'. The total length of the strings over all testcases doesn't exceed 2 β
10^5.
Output
For each testcase print a single integer β the maximum number of moves you can perform on a given string s.
Example
Input
5
()
[]()
([)]
)]([
)[(]
Output
1
2
2
0
1
Note
In the first example you can just erase the whole string.
In the second example you can first erase the brackets on positions 1 and 2: "[]()", then "()" is left. After that you can erase it whole. You could erase the whole string from the beginning but you would get one move instead of two.
In the third example you can first erase the brackets on positions 1 and 3: "([)]". They form an RBS "()". Then "[]" is left, so you can erase it whole.
In the fourth example there is no subsequence that is an RBS, so you can't perform a move at all.
In the fifth example you can erase the brackets on positions 2 and 4: ")[(]" and get ")(" as a result. You can erase nothing from it. | instruction | 0 | 16,997 | 21 | 33,994 |
Tags: greedy
Correct Solution:
```
for _ in range(int(input())):
t = input()
x, y, a = 0, 0, 0
for ch in t:
if ch == '(': x += 1
if ch == '[': y += 1
if ch == ')' and x:
a += 1
x -= 1
if ch == ']' and y:
a += 1
y -= 1
print(a)
``` | output | 1 | 16,997 | 21 | 33,995 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s, consisting of brackets of two types: '(', ')', '[' and ']'.
A string is called a regular bracket sequence (RBS) if it's of one of the following types:
* empty string;
* '(' + RBS + ')';
* '[' + RBS + ']';
* RBS + RBS.
where plus is a concatenation of two strings.
In one move you can choose a non-empty subsequence of the string s (not necessarily consecutive) that is an RBS, remove it from the string and concatenate the remaining parts without changing the order.
What is the maximum number of moves you can perform?
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of testcases.
Each of the next t lines contains a non-empty string, consisting only of characters '(', ')', '[' and ']'. The total length of the strings over all testcases doesn't exceed 2 β
10^5.
Output
For each testcase print a single integer β the maximum number of moves you can perform on a given string s.
Example
Input
5
()
[]()
([)]
)]([
)[(]
Output
1
2
2
0
1
Note
In the first example you can just erase the whole string.
In the second example you can first erase the brackets on positions 1 and 2: "[]()", then "()" is left. After that you can erase it whole. You could erase the whole string from the beginning but you would get one move instead of two.
In the third example you can first erase the brackets on positions 1 and 3: "([)]". They form an RBS "()". Then "[]" is left, so you can erase it whole.
In the fourth example there is no subsequence that is an RBS, so you can't perform a move at all.
In the fifth example you can erase the brackets on positions 2 and 4: ")[(]" and get ")(" as a result. You can erase nothing from it. | instruction | 0 | 16,998 | 21 | 33,996 |
Tags: greedy
Correct Solution:
```
#: Author - Soumya Saurav
import sys,io,os,time
from collections import defaultdict
from collections import Counter
from collections import deque
from itertools import combinations
from itertools import permutations
import bisect,math,heapq
alphabet = "abcdefghijklmnopqrstuvwxyz"
#input = sys.stdin.readline
########################################
for ii in range(int(input())):
s = input()
s1 = []
s2 = []
ans = 0
for i in s:
if i == "(":
s1.append(1)
if i == ")":
if s1:
s1.pop()
ans += 1
if i == "[":
s2.append(1)
if i == "]":
if s2:
s2.pop()
ans += 1
print(ans)
'''
# Wrap solution for more recursion depth
for dx,dy in [[1,0],[0,1],[-1,0],[0,-1]]:
#submit as python3
import collections,sys,threading
sys.setrecursionlimit(10**9)
threading.stack_size(10**8)
threading.Thread(target=solve).start()
'''
``` | output | 1 | 16,998 | 21 | 33,997 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s, consisting of brackets of two types: '(', ')', '[' and ']'.
A string is called a regular bracket sequence (RBS) if it's of one of the following types:
* empty string;
* '(' + RBS + ')';
* '[' + RBS + ']';
* RBS + RBS.
where plus is a concatenation of two strings.
In one move you can choose a non-empty subsequence of the string s (not necessarily consecutive) that is an RBS, remove it from the string and concatenate the remaining parts without changing the order.
What is the maximum number of moves you can perform?
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of testcases.
Each of the next t lines contains a non-empty string, consisting only of characters '(', ')', '[' and ']'. The total length of the strings over all testcases doesn't exceed 2 β
10^5.
Output
For each testcase print a single integer β the maximum number of moves you can perform on a given string s.
Example
Input
5
()
[]()
([)]
)]([
)[(]
Output
1
2
2
0
1
Note
In the first example you can just erase the whole string.
In the second example you can first erase the brackets on positions 1 and 2: "[]()", then "()" is left. After that you can erase it whole. You could erase the whole string from the beginning but you would get one move instead of two.
In the third example you can first erase the brackets on positions 1 and 3: "([)]". They form an RBS "()". Then "[]" is left, so you can erase it whole.
In the fourth example there is no subsequence that is an RBS, so you can't perform a move at all.
In the fifth example you can erase the brackets on positions 2 and 4: ")[(]" and get ")(" as a result. You can erase nothing from it. | instruction | 0 | 16,999 | 21 | 33,998 |
Tags: greedy
Correct Solution:
```
for i in range(int(input())):
x,y,c=0,0,0
for i in input():
if i=='(': x+=1
elif i==')' and x: x-=1;c+=1
elif i=='[': y+=1
elif i==']' and y: y-=1;c+=1
print(c)
``` | output | 1 | 16,999 | 21 | 33,999 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s, consisting of brackets of two types: '(', ')', '[' and ']'.
A string is called a regular bracket sequence (RBS) if it's of one of the following types:
* empty string;
* '(' + RBS + ')';
* '[' + RBS + ']';
* RBS + RBS.
where plus is a concatenation of two strings.
In one move you can choose a non-empty subsequence of the string s (not necessarily consecutive) that is an RBS, remove it from the string and concatenate the remaining parts without changing the order.
What is the maximum number of moves you can perform?
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of testcases.
Each of the next t lines contains a non-empty string, consisting only of characters '(', ')', '[' and ']'. The total length of the strings over all testcases doesn't exceed 2 β
10^5.
Output
For each testcase print a single integer β the maximum number of moves you can perform on a given string s.
Example
Input
5
()
[]()
([)]
)]([
)[(]
Output
1
2
2
0
1
Note
In the first example you can just erase the whole string.
In the second example you can first erase the brackets on positions 1 and 2: "[]()", then "()" is left. After that you can erase it whole. You could erase the whole string from the beginning but you would get one move instead of two.
In the third example you can first erase the brackets on positions 1 and 3: "([)]". They form an RBS "()". Then "[]" is left, so you can erase it whole.
In the fourth example there is no subsequence that is an RBS, so you can't perform a move at all.
In the fifth example you can erase the brackets on positions 2 and 4: ")[(]" and get ")(" as a result. You can erase nothing from it. | instruction | 0 | 17,000 | 21 | 34,000 |
Tags: greedy
Correct Solution:
```
def solve(string):
p_stack = []
s_stack = []
res = 0
for ch in string:
if ch == "(":
p_stack.append(1)
elif len(p_stack) != 0 and ch == ")":
p_stack.pop()
res += 1
if ch == "[":
s_stack.append(1)
elif len(s_stack) != 0 and ch == "]":
s_stack.pop()
res += 1
print(res)
if __name__ == '__main__':
# Test case
for _ in range(int(input())):
string = input()
# Calling function
solve(string)
``` | output | 1 | 17,000 | 21 | 34,001 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s, consisting of brackets of two types: '(', ')', '[' and ']'.
A string is called a regular bracket sequence (RBS) if it's of one of the following types:
* empty string;
* '(' + RBS + ')';
* '[' + RBS + ']';
* RBS + RBS.
where plus is a concatenation of two strings.
In one move you can choose a non-empty subsequence of the string s (not necessarily consecutive) that is an RBS, remove it from the string and concatenate the remaining parts without changing the order.
What is the maximum number of moves you can perform?
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of testcases.
Each of the next t lines contains a non-empty string, consisting only of characters '(', ')', '[' and ']'. The total length of the strings over all testcases doesn't exceed 2 β
10^5.
Output
For each testcase print a single integer β the maximum number of moves you can perform on a given string s.
Example
Input
5
()
[]()
([)]
)]([
)[(]
Output
1
2
2
0
1
Note
In the first example you can just erase the whole string.
In the second example you can first erase the brackets on positions 1 and 2: "[]()", then "()" is left. After that you can erase it whole. You could erase the whole string from the beginning but you would get one move instead of two.
In the third example you can first erase the brackets on positions 1 and 3: "([)]". They form an RBS "()". Then "[]" is left, so you can erase it whole.
In the fourth example there is no subsequence that is an RBS, so you can't perform a move at all.
In the fifth example you can erase the brackets on positions 2 and 4: ")[(]" and get ")(" as a result. You can erase nothing from it. | instruction | 0 | 17,001 | 21 | 34,002 |
Tags: greedy
Correct Solution:
```
import math
t=int(input())
for _ in range(t):
s=str(input())
buffer_1=0 # ( or )
buffer_2=0 # [ or ]
answer=0
for i in range(len(s)):
r=s[i]
if r=="(" or r=="[":
if r=="(":
buffer_1+=1
else:
buffer_2+=1
else:
if r==")":
if buffer_1>0:
buffer_1-=1
answer+=1
else:
if buffer_2>0:
buffer_2-=1
answer+=1
print(answer)
``` | output | 1 | 17,001 | 21 | 34,003 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s, consisting of brackets of two types: '(', ')', '[' and ']'.
A string is called a regular bracket sequence (RBS) if it's of one of the following types:
* empty string;
* '(' + RBS + ')';
* '[' + RBS + ']';
* RBS + RBS.
where plus is a concatenation of two strings.
In one move you can choose a non-empty subsequence of the string s (not necessarily consecutive) that is an RBS, remove it from the string and concatenate the remaining parts without changing the order.
What is the maximum number of moves you can perform?
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of testcases.
Each of the next t lines contains a non-empty string, consisting only of characters '(', ')', '[' and ']'. The total length of the strings over all testcases doesn't exceed 2 β
10^5.
Output
For each testcase print a single integer β the maximum number of moves you can perform on a given string s.
Example
Input
5
()
[]()
([)]
)]([
)[(]
Output
1
2
2
0
1
Note
In the first example you can just erase the whole string.
In the second example you can first erase the brackets on positions 1 and 2: "[]()", then "()" is left. After that you can erase it whole. You could erase the whole string from the beginning but you would get one move instead of two.
In the third example you can first erase the brackets on positions 1 and 3: "([)]". They form an RBS "()". Then "[]" is left, so you can erase it whole.
In the fourth example there is no subsequence that is an RBS, so you can't perform a move at all.
In the fifth example you can erase the brackets on positions 2 and 4: ")[(]" and get ")(" as a result. You can erase nothing from it.
Submitted Solution:
```
for _ in range(int(input())):
s = input()
res = [0,0]
c1,c2=0,0
for i in s:
if i=='(':
res[0]+=1
elif i=='[':
res[1]+=1
elif i==')':
if res[0]!=0:
res[0]-=1
c1+=1
else:
if res[1]!=0:
res[1]-=1
c2+=1
print(c1+c2)
``` | instruction | 0 | 17,002 | 21 | 34,004 |
Yes | output | 1 | 17,002 | 21 | 34,005 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s, consisting of brackets of two types: '(', ')', '[' and ']'.
A string is called a regular bracket sequence (RBS) if it's of one of the following types:
* empty string;
* '(' + RBS + ')';
* '[' + RBS + ']';
* RBS + RBS.
where plus is a concatenation of two strings.
In one move you can choose a non-empty subsequence of the string s (not necessarily consecutive) that is an RBS, remove it from the string and concatenate the remaining parts without changing the order.
What is the maximum number of moves you can perform?
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of testcases.
Each of the next t lines contains a non-empty string, consisting only of characters '(', ')', '[' and ']'. The total length of the strings over all testcases doesn't exceed 2 β
10^5.
Output
For each testcase print a single integer β the maximum number of moves you can perform on a given string s.
Example
Input
5
()
[]()
([)]
)]([
)[(]
Output
1
2
2
0
1
Note
In the first example you can just erase the whole string.
In the second example you can first erase the brackets on positions 1 and 2: "[]()", then "()" is left. After that you can erase it whole. You could erase the whole string from the beginning but you would get one move instead of two.
In the third example you can first erase the brackets on positions 1 and 3: "([)]". They form an RBS "()". Then "[]" is left, so you can erase it whole.
In the fourth example there is no subsequence that is an RBS, so you can't perform a move at all.
In the fifth example you can erase the brackets on positions 2 and 4: ")[(]" and get ")(" as a result. You can erase nothing from it.
Submitted Solution:
```
#list(map(int, input().rstrip().split()))
def f(A, B):
i, j = 0, 0
c = 0
while i < len(A) and j < len(B):
if A[i] < B[j]:
c += 1
i += 1
j += 1
else:
j += 1
return c
t = int(input())
for test in range(t):
s = input()
ta,tc,qa,qc = [], [], [], []
for i in range(len(s)):
if s[i] == "(":
ta.append(i)
elif s[i] == ")":
tc.append(i)
elif s[i] == "[":
qa.append(i)
else:
qc.append(i)
print( f(ta,tc) + f(qa,qc))
``` | instruction | 0 | 17,003 | 21 | 34,006 |
Yes | output | 1 | 17,003 | 21 | 34,007 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s, consisting of brackets of two types: '(', ')', '[' and ']'.
A string is called a regular bracket sequence (RBS) if it's of one of the following types:
* empty string;
* '(' + RBS + ')';
* '[' + RBS + ']';
* RBS + RBS.
where plus is a concatenation of two strings.
In one move you can choose a non-empty subsequence of the string s (not necessarily consecutive) that is an RBS, remove it from the string and concatenate the remaining parts without changing the order.
What is the maximum number of moves you can perform?
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of testcases.
Each of the next t lines contains a non-empty string, consisting only of characters '(', ')', '[' and ']'. The total length of the strings over all testcases doesn't exceed 2 β
10^5.
Output
For each testcase print a single integer β the maximum number of moves you can perform on a given string s.
Example
Input
5
()
[]()
([)]
)]([
)[(]
Output
1
2
2
0
1
Note
In the first example you can just erase the whole string.
In the second example you can first erase the brackets on positions 1 and 2: "[]()", then "()" is left. After that you can erase it whole. You could erase the whole string from the beginning but you would get one move instead of two.
In the third example you can first erase the brackets on positions 1 and 3: "([)]". They form an RBS "()". Then "[]" is left, so you can erase it whole.
In the fourth example there is no subsequence that is an RBS, so you can't perform a move at all.
In the fifth example you can erase the brackets on positions 2 and 4: ")[(]" and get ")(" as a result. You can erase nothing from it.
Submitted Solution:
```
from sys import stdin, stdout
import math,sys,heapq
from itertools import permutations, combinations
from collections import defaultdict,deque,OrderedDict
from os import path
import random
import bisect as bi
def yes():print('YES')
def no():print('NO')
if (path.exists('input.txt')):
#------------------Sublime--------------------------------------#
sys.stdin=open('input.txt','r');sys.stdout=open('output.txt','w');
def I():return (int(input()))
def In():return(map(int,input().split()))
else:
#------------------PYPY FAst I/o--------------------------------#
def I():return (int(stdin.readline()))
def In():return(map(int,stdin.readline().split()))
#sys.setrecursionlimit(1500)
def dict(a):
d={}
for x in a:
if d.get(x,-1)!=-1:
d[x]+=1
else:
d[x]=1
return d
def find_gt(a, x):
'Find leftmost value greater than x'
i = bi.bisect_left(a, x)
if i != len(a):
return i
else:
return -1
def main():
try:
l=input()
n=len(l)
nosq,noo=0,0
ans=0
for i in range(n):
if l[i]=='[':
nosq+=1
elif l[i]=='(':
noo+=1
elif l[i]==')':
if noo>0:
noo-=1
ans+=1
elif l[i]==']':
if nosq>0:
nosq-=1
ans+=1
print(ans)
except:
pass
M = 998244353
P = 1000000007
if __name__ == '__main__':
for _ in range(I()):main()
#for _ in range(1):main()
``` | instruction | 0 | 17,004 | 21 | 34,008 |
Yes | output | 1 | 17,004 | 21 | 34,009 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s, consisting of brackets of two types: '(', ')', '[' and ']'.
A string is called a regular bracket sequence (RBS) if it's of one of the following types:
* empty string;
* '(' + RBS + ')';
* '[' + RBS + ']';
* RBS + RBS.
where plus is a concatenation of two strings.
In one move you can choose a non-empty subsequence of the string s (not necessarily consecutive) that is an RBS, remove it from the string and concatenate the remaining parts without changing the order.
What is the maximum number of moves you can perform?
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of testcases.
Each of the next t lines contains a non-empty string, consisting only of characters '(', ')', '[' and ']'. The total length of the strings over all testcases doesn't exceed 2 β
10^5.
Output
For each testcase print a single integer β the maximum number of moves you can perform on a given string s.
Example
Input
5
()
[]()
([)]
)]([
)[(]
Output
1
2
2
0
1
Note
In the first example you can just erase the whole string.
In the second example you can first erase the brackets on positions 1 and 2: "[]()", then "()" is left. After that you can erase it whole. You could erase the whole string from the beginning but you would get one move instead of two.
In the third example you can first erase the brackets on positions 1 and 3: "([)]". They form an RBS "()". Then "[]" is left, so you can erase it whole.
In the fourth example there is no subsequence that is an RBS, so you can't perform a move at all.
In the fifth example you can erase the brackets on positions 2 and 4: ")[(]" and get ")(" as a result. You can erase nothing from it.
Submitted Solution:
```
t = int(input())
for j in range(t):
s = input()
count = 0
q1 = 0
q2 = 0
for i in range(len(s)):
if s[i] == "(":
q1 += 1
if s[i] == "[":
q2 += 1
if s[i] == ")":
if q1 > 0:
count += 1
q1 -= 1
if s[i] == "]":
if q2 > 0:
count += 1
q2 -= 1
print(count)
``` | instruction | 0 | 17,005 | 21 | 34,010 |
Yes | output | 1 | 17,005 | 21 | 34,011 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s, consisting of brackets of two types: '(', ')', '[' and ']'.
A string is called a regular bracket sequence (RBS) if it's of one of the following types:
* empty string;
* '(' + RBS + ')';
* '[' + RBS + ']';
* RBS + RBS.
where plus is a concatenation of two strings.
In one move you can choose a non-empty subsequence of the string s (not necessarily consecutive) that is an RBS, remove it from the string and concatenate the remaining parts without changing the order.
What is the maximum number of moves you can perform?
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of testcases.
Each of the next t lines contains a non-empty string, consisting only of characters '(', ')', '[' and ']'. The total length of the strings over all testcases doesn't exceed 2 β
10^5.
Output
For each testcase print a single integer β the maximum number of moves you can perform on a given string s.
Example
Input
5
()
[]()
([)]
)]([
)[(]
Output
1
2
2
0
1
Note
In the first example you can just erase the whole string.
In the second example you can first erase the brackets on positions 1 and 2: "[]()", then "()" is left. After that you can erase it whole. You could erase the whole string from the beginning but you would get one move instead of two.
In the third example you can first erase the brackets on positions 1 and 3: "([)]". They form an RBS "()". Then "[]" is left, so you can erase it whole.
In the fourth example there is no subsequence that is an RBS, so you can't perform a move at all.
In the fifth example you can erase the brackets on positions 2 and 4: ")[(]" and get ")(" as a result. You can erase nothing from it.
Submitted Solution:
```
t=int(input())
for ttttttt in range(t):
s=input()
a=0
b=0
ans=0
for i in s:
if i=='[':
a+=1
elif i==']' and a>0:
a-=1
ans+=1
elif i==')' and b>0:
b-=1
ans+=1
else:
b+=1
print(ans)
``` | instruction | 0 | 17,006 | 21 | 34,012 |
No | output | 1 | 17,006 | 21 | 34,013 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s, consisting of brackets of two types: '(', ')', '[' and ']'.
A string is called a regular bracket sequence (RBS) if it's of one of the following types:
* empty string;
* '(' + RBS + ')';
* '[' + RBS + ']';
* RBS + RBS.
where plus is a concatenation of two strings.
In one move you can choose a non-empty subsequence of the string s (not necessarily consecutive) that is an RBS, remove it from the string and concatenate the remaining parts without changing the order.
What is the maximum number of moves you can perform?
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of testcases.
Each of the next t lines contains a non-empty string, consisting only of characters '(', ')', '[' and ']'. The total length of the strings over all testcases doesn't exceed 2 β
10^5.
Output
For each testcase print a single integer β the maximum number of moves you can perform on a given string s.
Example
Input
5
()
[]()
([)]
)]([
)[(]
Output
1
2
2
0
1
Note
In the first example you can just erase the whole string.
In the second example you can first erase the brackets on positions 1 and 2: "[]()", then "()" is left. After that you can erase it whole. You could erase the whole string from the beginning but you would get one move instead of two.
In the third example you can first erase the brackets on positions 1 and 3: "([)]". They form an RBS "()". Then "[]" is left, so you can erase it whole.
In the fourth example there is no subsequence that is an RBS, so you can't perform a move at all.
In the fifth example you can erase the brackets on positions 2 and 4: ")[(]" and get ")(" as a result. You can erase nothing from it.
Submitted Solution:
```
t = int(input())
for i in range(t):
s = input()
a, b, c, d = 0, 0, 0, 0
for ch in s:
if ch == ']':
d += 1;
elif ch==')':
b += 1
ans1, ans2 = 0, 0
for ch in s:
if ch==']':
d -= 1
elif ch=='[':
c += 1
ans1 = max(ans1, min(c, d))
for ch in s:
if ch ==')':
b -= 1
elif ch=='(':
a += 1
ans2 = max(ans2, min(a, b))
print(ans1 + ans2)
``` | instruction | 0 | 17,007 | 21 | 34,014 |
No | output | 1 | 17,007 | 21 | 34,015 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s, consisting of brackets of two types: '(', ')', '[' and ']'.
A string is called a regular bracket sequence (RBS) if it's of one of the following types:
* empty string;
* '(' + RBS + ')';
* '[' + RBS + ']';
* RBS + RBS.
where plus is a concatenation of two strings.
In one move you can choose a non-empty subsequence of the string s (not necessarily consecutive) that is an RBS, remove it from the string and concatenate the remaining parts without changing the order.
What is the maximum number of moves you can perform?
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of testcases.
Each of the next t lines contains a non-empty string, consisting only of characters '(', ')', '[' and ']'. The total length of the strings over all testcases doesn't exceed 2 β
10^5.
Output
For each testcase print a single integer β the maximum number of moves you can perform on a given string s.
Example
Input
5
()
[]()
([)]
)]([
)[(]
Output
1
2
2
0
1
Note
In the first example you can just erase the whole string.
In the second example you can first erase the brackets on positions 1 and 2: "[]()", then "()" is left. After that you can erase it whole. You could erase the whole string from the beginning but you would get one move instead of two.
In the third example you can first erase the brackets on positions 1 and 3: "([)]". They form an RBS "()". Then "[]" is left, so you can erase it whole.
In the fourth example there is no subsequence that is an RBS, so you can't perform a move at all.
In the fifth example you can erase the brackets on positions 2 and 4: ")[(]" and get ")(" as a result. You can erase nothing from it.
Submitted Solution:
```
tests = int(input())
for t in range(tests):
brackets = input()
arr = []
n = 0
k = 0
haveLeft = 0
haveLeftBox = 0
for i in brackets:
if i == "(":
haveLeft += 1
elif i == "[":
haveLeftBox += 1
elif i == ")" and haveLeft > 0:
n += 1
elif i == "]" and haveLeftBox > 0:
k += 1
print(n + k)
``` | instruction | 0 | 17,008 | 21 | 34,016 |
No | output | 1 | 17,008 | 21 | 34,017 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s, consisting of brackets of two types: '(', ')', '[' and ']'.
A string is called a regular bracket sequence (RBS) if it's of one of the following types:
* empty string;
* '(' + RBS + ')';
* '[' + RBS + ']';
* RBS + RBS.
where plus is a concatenation of two strings.
In one move you can choose a non-empty subsequence of the string s (not necessarily consecutive) that is an RBS, remove it from the string and concatenate the remaining parts without changing the order.
What is the maximum number of moves you can perform?
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of testcases.
Each of the next t lines contains a non-empty string, consisting only of characters '(', ')', '[' and ']'. The total length of the strings over all testcases doesn't exceed 2 β
10^5.
Output
For each testcase print a single integer β the maximum number of moves you can perform on a given string s.
Example
Input
5
()
[]()
([)]
)]([
)[(]
Output
1
2
2
0
1
Note
In the first example you can just erase the whole string.
In the second example you can first erase the brackets on positions 1 and 2: "[]()", then "()" is left. After that you can erase it whole. You could erase the whole string from the beginning but you would get one move instead of two.
In the third example you can first erase the brackets on positions 1 and 3: "([)]". They form an RBS "()". Then "[]" is left, so you can erase it whole.
In the fourth example there is no subsequence that is an RBS, so you can't perform a move at all.
In the fifth example you can erase the brackets on positions 2 and 4: ")[(]" and get ")(" as a result. You can erase nothing from it.
Submitted Solution:
```
for _ in range(int(input())):
s = input()
graph = {"r":[0,0], "s":[0,0]}
l=[]
for i in range(len(s)):
if(s[i]=='('):
graph["r"][0]+=1
elif(s[i]=='['):
graph["s"][0]+=1
else:
if(i!=0):
if(s[i]==')'):
if(graph["r"][0]>graph["r"][1]):
graph["r"][1]+=1
elif(s[i]==']'):
if(graph["s"][0]>graph["s"][1]):
graph["s"][1]+=1
l.append({"r":[graph["r"][0],graph["r"][1]], "s":[graph["s"][0],graph["s"][1]]})
count = 0
for i in range(len(l)):
if(i!=0):
if(l[i]['r'][0]>=l[i]['r'][1] and l[i]['r'][0]!=0 and l[i]['r'][1]!=0 and not(l[i-1]['r'][0]==l[i]['r'][0] and l[i-1]['r'][1]==l[i]['r'][1])):
count+=1
if(l[i]['s'][0]>=l[i]['s'][1] and l[i]['s'][0]!=0 and l[i]['s'][1]!=0 and not(l[i-1]['s'][0]==l[i]['s'][0] and l[i-1]['s'][1]==l[i]['s'][1])):
count+=1
print(count)
``` | instruction | 0 | 17,009 | 21 | 34,018 |
No | output | 1 | 17,009 | 21 | 34,019 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recall that the bracket sequence is considered regular if it is possible to insert symbols '+' and '1' into it so that the result is a correct arithmetic expression. For example, a sequence "(()())" is regular, because we can get correct arithmetic expression insering symbols '+' and '1': "((1+1)+(1+1))". Also the following sequences are regular: "()()()", "(())" and "()". The following sequences are not regular bracket sequences: ")(", "(()" and "())(()".
In this problem you are given two integers n and k. Your task is to construct a regular bracket sequence consisting of round brackets with length 2Β·n with total sum of nesting of all opening brackets equals to exactly k. The nesting of a single opening bracket equals to the number of pairs of brackets in which current opening bracket is embedded.
For example, in the sequence "()(())" the nesting of first opening bracket equals to 0, the nesting of the second opening bracket equals to 0 and the nesting of the third opening bracket equal to 1. So the total sum of nestings equals to 1.
Input
The first line contains two integers n and k (1 β€ n β€ 3Β·105, 0 β€ k β€ 1018) β the number of opening brackets and needed total nesting.
Output
Print the required regular bracket sequence consisting of round brackets.
If there is no solution print "Impossible" (without quotes).
Examples
Input
3 1
Output
()(())
Input
4 6
Output
(((())))
Input
2 5
Output
Impossible
Note
The first example is examined in the statement.
In the second example the answer is "(((())))". The nesting of the first opening bracket is 0, the nesting of the second is 1, the nesting of the third is 2, the nesting of fourth is 3. So the total sum of nestings equals to 0 + 1 + 2 + 3 = 6.
In the third it is impossible to construct a regular bracket sequence, because the maximum possible total sum of nestings for two opening brackets equals to 1. This total sum of nestings is obtained for the sequence "(())". | instruction | 0 | 20,859 | 21 | 41,718 |
Tags: constructive algorithms
Correct Solution:
```
from sys import exit
from math import sqrt
n, k = map(int, input().split())
if n * (n - 1) // 2 < k:
print('Impossible')
else:
h = int((1 + sqrt(8 * k + 1)) / 2)
c = n
for _ in range(h):
print('(', end='')
c -= h
k -= h * (h - 1) // 2
while h != 0:
m = k // h
for _ in range(m):
print('()', end='')
c -= m
k -= h * m
h -= 1
print(')', end='')
while c > 0:
print('()', end='')
c -= 1
``` | output | 1 | 20,859 | 21 | 41,719 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recall that the bracket sequence is considered regular if it is possible to insert symbols '+' and '1' into it so that the result is a correct arithmetic expression. For example, a sequence "(()())" is regular, because we can get correct arithmetic expression insering symbols '+' and '1': "((1+1)+(1+1))". Also the following sequences are regular: "()()()", "(())" and "()". The following sequences are not regular bracket sequences: ")(", "(()" and "())(()".
In this problem you are given two integers n and k. Your task is to construct a regular bracket sequence consisting of round brackets with length 2Β·n with total sum of nesting of all opening brackets equals to exactly k. The nesting of a single opening bracket equals to the number of pairs of brackets in which current opening bracket is embedded.
For example, in the sequence "()(())" the nesting of first opening bracket equals to 0, the nesting of the second opening bracket equals to 0 and the nesting of the third opening bracket equal to 1. So the total sum of nestings equals to 1.
Input
The first line contains two integers n and k (1 β€ n β€ 3Β·105, 0 β€ k β€ 1018) β the number of opening brackets and needed total nesting.
Output
Print the required regular bracket sequence consisting of round brackets.
If there is no solution print "Impossible" (without quotes).
Examples
Input
3 1
Output
()(())
Input
4 6
Output
(((())))
Input
2 5
Output
Impossible
Note
The first example is examined in the statement.
In the second example the answer is "(((())))". The nesting of the first opening bracket is 0, the nesting of the second is 1, the nesting of the third is 2, the nesting of fourth is 3. So the total sum of nestings equals to 0 + 1 + 2 + 3 = 6.
In the third it is impossible to construct a regular bracket sequence, because the maximum possible total sum of nestings for two opening brackets equals to 1. This total sum of nestings is obtained for the sequence "(())". | instruction | 0 | 20,860 | 21 | 41,720 |
Tags: constructive algorithms
Correct Solution:
```
ch=input()
l=ch.split(' ')
n=int(l[0])
k=int(l[1])
l1='('
l2=')'
l3='()'
s=0
mi=0
ch=''
from math import sqrt
q=int(sqrt(2*k))
while k>=(q*(q-1))/2:
q=q+1
q=q-1
d=int(k-(q*(q-1))/2)
mi=q
i=0
if d!=0:
mi=q+1
i=1
if n<mi:
print('Impossible')
else:
c=n-mi
print(c*l3+d*l1+i*l3+(q-d)*l1+q*l2)
# Made By Mostafa_Khaled
``` | output | 1 | 20,860 | 21 | 41,721 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recall that the bracket sequence is considered regular if it is possible to insert symbols '+' and '1' into it so that the result is a correct arithmetic expression. For example, a sequence "(()())" is regular, because we can get correct arithmetic expression insering symbols '+' and '1': "((1+1)+(1+1))". Also the following sequences are regular: "()()()", "(())" and "()". The following sequences are not regular bracket sequences: ")(", "(()" and "())(()".
In this problem you are given two integers n and k. Your task is to construct a regular bracket sequence consisting of round brackets with length 2Β·n with total sum of nesting of all opening brackets equals to exactly k. The nesting of a single opening bracket equals to the number of pairs of brackets in which current opening bracket is embedded.
For example, in the sequence "()(())" the nesting of first opening bracket equals to 0, the nesting of the second opening bracket equals to 0 and the nesting of the third opening bracket equal to 1. So the total sum of nestings equals to 1.
Input
The first line contains two integers n and k (1 β€ n β€ 3Β·105, 0 β€ k β€ 1018) β the number of opening brackets and needed total nesting.
Output
Print the required regular bracket sequence consisting of round brackets.
If there is no solution print "Impossible" (without quotes).
Examples
Input
3 1
Output
()(())
Input
4 6
Output
(((())))
Input
2 5
Output
Impossible
Note
The first example is examined in the statement.
In the second example the answer is "(((())))". The nesting of the first opening bracket is 0, the nesting of the second is 1, the nesting of the third is 2, the nesting of fourth is 3. So the total sum of nestings equals to 0 + 1 + 2 + 3 = 6.
In the third it is impossible to construct a regular bracket sequence, because the maximum possible total sum of nestings for two opening brackets equals to 1. This total sum of nestings is obtained for the sequence "(())". | instruction | 0 | 20,861 | 21 | 41,722 |
Tags: constructive algorithms
Correct Solution:
```
# Legends Always Come Up with Solution
# Author: Manvir Singh
import os
import sys
from io import BytesIO, IOBase
from collections import Counter
from bisect import *
from math import gcd
from itertools import permutations,combinations
from math import sqrt,ceil,floor
def main():
n,k=map(int,input().split())
z=floor((-1+sqrt(1+8*k))/2)
f=int((z*(z+1))//2!=k)
if z+f>n-1:
print("Impossible")
else:
if f:
y=((z+1)*(z+2))//2-k
print("("*(z+1)+")"*y+"("+")"*(z+2-y)+"()"*(n-(z+2)))
else:
print("("*(z+1)+")"*(z+1)+"()"*(n-1-z))
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
``` | output | 1 | 20,861 | 21 | 41,723 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recall that the bracket sequence is considered regular if it is possible to insert symbols '+' and '1' into it so that the result is a correct arithmetic expression. For example, a sequence "(()())" is regular, because we can get correct arithmetic expression insering symbols '+' and '1': "((1+1)+(1+1))". Also the following sequences are regular: "()()()", "(())" and "()". The following sequences are not regular bracket sequences: ")(", "(()" and "())(()".
In this problem you are given two integers n and k. Your task is to construct a regular bracket sequence consisting of round brackets with length 2Β·n with total sum of nesting of all opening brackets equals to exactly k. The nesting of a single opening bracket equals to the number of pairs of brackets in which current opening bracket is embedded.
For example, in the sequence "()(())" the nesting of first opening bracket equals to 0, the nesting of the second opening bracket equals to 0 and the nesting of the third opening bracket equal to 1. So the total sum of nestings equals to 1.
Input
The first line contains two integers n and k (1 β€ n β€ 3Β·105, 0 β€ k β€ 1018) β the number of opening brackets and needed total nesting.
Output
Print the required regular bracket sequence consisting of round brackets.
If there is no solution print "Impossible" (without quotes).
Examples
Input
3 1
Output
()(())
Input
4 6
Output
(((())))
Input
2 5
Output
Impossible
Note
The first example is examined in the statement.
In the second example the answer is "(((())))". The nesting of the first opening bracket is 0, the nesting of the second is 1, the nesting of the third is 2, the nesting of fourth is 3. So the total sum of nestings equals to 0 + 1 + 2 + 3 = 6.
In the third it is impossible to construct a regular bracket sequence, because the maximum possible total sum of nestings for two opening brackets equals to 1. This total sum of nestings is obtained for the sequence "(())". | instruction | 0 | 20,862 | 21 | 41,724 |
Tags: constructive algorithms
Correct Solution:
```
from math import ceil
n, k = map(int, input().split())
if n * (n - 1) // 2 < k:
print('Impossible')
exit()
x = ceil(((8 * k + 1) ** 0.5 + 1) / 2)
p = x * (x - 1) // 2 - k
print('()' * (n - x) + '(' * (x - 1) + ')' * p + '()' + ')' * (x - p - 1))
``` | output | 1 | 20,862 | 21 | 41,725 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recall that the bracket sequence is considered regular if it is possible to insert symbols '+' and '1' into it so that the result is a correct arithmetic expression. For example, a sequence "(()())" is regular, because we can get correct arithmetic expression insering symbols '+' and '1': "((1+1)+(1+1))". Also the following sequences are regular: "()()()", "(())" and "()". The following sequences are not regular bracket sequences: ")(", "(()" and "())(()".
In this problem you are given two integers n and k. Your task is to construct a regular bracket sequence consisting of round brackets with length 2Β·n with total sum of nesting of all opening brackets equals to exactly k. The nesting of a single opening bracket equals to the number of pairs of brackets in which current opening bracket is embedded.
For example, in the sequence "()(())" the nesting of first opening bracket equals to 0, the nesting of the second opening bracket equals to 0 and the nesting of the third opening bracket equal to 1. So the total sum of nestings equals to 1.
Input
The first line contains two integers n and k (1 β€ n β€ 3Β·105, 0 β€ k β€ 1018) β the number of opening brackets and needed total nesting.
Output
Print the required regular bracket sequence consisting of round brackets.
If there is no solution print "Impossible" (without quotes).
Examples
Input
3 1
Output
()(())
Input
4 6
Output
(((())))
Input
2 5
Output
Impossible
Note
The first example is examined in the statement.
In the second example the answer is "(((())))". The nesting of the first opening bracket is 0, the nesting of the second is 1, the nesting of the third is 2, the nesting of fourth is 3. So the total sum of nestings equals to 0 + 1 + 2 + 3 = 6.
In the third it is impossible to construct a regular bracket sequence, because the maximum possible total sum of nestings for two opening brackets equals to 1. This total sum of nestings is obtained for the sequence "(())". | instruction | 0 | 20,863 | 21 | 41,726 |
Tags: constructive algorithms
Correct Solution:
```
n, k = map(int, input().split())
k_now, level, n_now = 0, 0, 0
ans = str()
if k > n * (n - 1) / 2:
print('Impossible')
else:
while n_now < 2 * n:
if k_now + level <= k:
ans += '('
k_now += level
level += 1
else:
ans += ')'
level -= 1
n_now += 1
print(ans)
``` | output | 1 | 20,863 | 21 | 41,727 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recall that the bracket sequence is considered regular if it is possible to insert symbols '+' and '1' into it so that the result is a correct arithmetic expression. For example, a sequence "(()())" is regular, because we can get correct arithmetic expression insering symbols '+' and '1': "((1+1)+(1+1))". Also the following sequences are regular: "()()()", "(())" and "()". The following sequences are not regular bracket sequences: ")(", "(()" and "())(()".
In this problem you are given two integers n and k. Your task is to construct a regular bracket sequence consisting of round brackets with length 2Β·n with total sum of nesting of all opening brackets equals to exactly k. The nesting of a single opening bracket equals to the number of pairs of brackets in which current opening bracket is embedded.
For example, in the sequence "()(())" the nesting of first opening bracket equals to 0, the nesting of the second opening bracket equals to 0 and the nesting of the third opening bracket equal to 1. So the total sum of nestings equals to 1.
Input
The first line contains two integers n and k (1 β€ n β€ 3Β·105, 0 β€ k β€ 1018) β the number of opening brackets and needed total nesting.
Output
Print the required regular bracket sequence consisting of round brackets.
If there is no solution print "Impossible" (without quotes).
Examples
Input
3 1
Output
()(())
Input
4 6
Output
(((())))
Input
2 5
Output
Impossible
Note
The first example is examined in the statement.
In the second example the answer is "(((())))". The nesting of the first opening bracket is 0, the nesting of the second is 1, the nesting of the third is 2, the nesting of fourth is 3. So the total sum of nestings equals to 0 + 1 + 2 + 3 = 6.
In the third it is impossible to construct a regular bracket sequence, because the maximum possible total sum of nestings for two opening brackets equals to 1. This total sum of nestings is obtained for the sequence "(())". | instruction | 0 | 20,864 | 21 | 41,728 |
Tags: constructive algorithms
Correct Solution:
```
n,m = map(int, input().split())
if m > n*(n - 1)//2:
print('Impossible')
else:
ans = []
cur = 0
acc = 0
for i in range(n):
while cur + acc > m:
ans.append(')')
acc -= 1
ans.append('(')
cur += acc
acc += 1
left_cnt = 0
for i in ans:
if i == '(':
left_cnt += 1
else:
left_cnt -= 1
for i in range(left_cnt):
ans.append(')')
print(''.join(ans))
``` | output | 1 | 20,864 | 21 | 41,729 |
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