Bracket Xoring (Codeforces 1896F)
Tags: Greedy, constructive algorithms

You are given a binary string $$$s$$$ of length $$$2n$$$ where each element is $$$\mathtt{0}$$$ or $$$\mathtt{1}$$$. You can do the following operation: Choose a balanced bracket sequence$$$^\dagger$$$ $$$b$$$ of length $$$2n$$$.  For every index $$$i$$$ from $$$1$$$ to $$$2n$$$ in order, where $$$b_i$$$ is an open bracket, let $$$p_i$$$ denote the minimum index such that $$$b[i,p_i]$$$ is a balanced bracket sequence. Then, we perform a range toggle operation$$$^\ddagger$$$ from $$$i$$$ to $$$p_i$$$ on $$$s$$$. Note that since a balanced bracket sequence of length $$$2n$$$ will have $$$n$$$ open brackets, we will do $$$n$$$ range toggle operations on $$$s$$$. Your task is to find a sequence of no more than $$$10$$$ operations that changes all elements of $$$s$$$ to $$$\mathtt{0}$$$, or determine that it is impossible to do so. Note that you do not have to minimize the number of operations.Under the given constraints, it can be proven that if it is possible to change all elements of $$$s$$$ to $$$\mathtt{0}$$$, there exists a way that requires no more than $$$10$$$ operations.$$$^\dagger$$$ A sequence of brackets is called balanced if one can turn it into a valid math expression by adding characters $$$+$$$ and $$$1$$$. For example, sequences "(())()", "()", and "(()(()))" are balanced, while ")(", "(()", and "(()))(" are not.$$$^\ddagger$$$ If we perform a range toggle operation from $$$l$$$ to $$$r$$$ on a binary string $$$s$$$, then we toggle all values of $$$s_i$$$ such that $$$l \leq i \leq r$$$. If $$$s_i$$$ is toggled, we will set $$$s_i := \mathtt{0}$$$ if $$$s_i = \mathtt{1}$$$ or vice versa. For example, if $$$s=\mathtt{1000101}$$$ and we perform a range toggle operation from $$$3$$$ to $$$5$$$, $$$s$$$ will be changed to $$$s=\mathtt{1011001}$$$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 1000$$$). The description of the test cases follows.The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2\cdot 10^5$$$) — where $$$2n$$$ is the length of string $$$s$$$.The second line of each test case contains a binary string $$$s$$$ of length $$$2n$$$ ($$$s_i = \mathtt{0}$$$ or $$$s_i = \mathtt{1}$$$).It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2\cdot 10^5$$$.

Output: For each test case, output $$$-1$$$ in a single line if it is impossible to change all elements of $$$s$$$ to $$$\mathtt{0}$$$.Otherwise, output a single integer $$$k$$$ ($$$0 \le k \le 10$$$) representing the number of operations needed to change all elements of $$$s$$$ to $$$\mathtt{0}$$$. Then, on each of the next $$$k$$$ lines, output a balanced bracket sequence of length $$$2n$$$ representing the operations needed to change all elements of $$$s$$$ to $$$0$$$s.If there are multiple ways to change all elements of $$$s$$$ to $$$\mathtt{0}$$$ that require not more than $$$10$$$ operations, you can output any of them.

Note: In the first test case, it can be proven that it is impossible to change all elements of $$$s$$$ to $$$\mathtt{0}$$$.In the second test case, the first operation using the bracket sequence $$$b = \mathtt{()()}$$$ will convert the binary string $$$s=\mathtt{0000}$$$ to $$$s=\mathtt{1111}$$$. Then, the second operation using the same bracket sequence $$$b = \mathtt{()()}$$$ will convert the binary string $$$s=\mathtt{1111}$$$ back to $$$s=\mathtt{0000}$$$. Note that since all elements of $$$s$$$ is already $$$\mathtt{0}$$$ initially, using $$$0$$$ operations is also a valid answer.In the third test case, a single operation using the bracket sequence $$$b = \mathtt{(())()}$$$ will change all elements of $$$s$$$ to $$$\mathtt{0}$$$. The operation will happen as follows.  $$$b_1$$$ is an open bracket and $$$p_1 = 4$$$ since $$$b[1,4]=\mathtt{(())}$$$ is a balanced bracket sequence. Hence, we do a range toggle operation from $$$1$$$ to $$$4$$$ on the binary string $$$s = \mathtt{100111}$$$ to obtain $$$s = \mathtt{011011}$$$.  $$$b_2$$$ is an open bracket and $$$p_2 = 3$$$ since $$$b[2,3]=\mathtt{()}$$$ is a balanced bracket sequence. Hence, we do a range toggle operation from $$$2$$$ to $$$3$$$ on the binary string $$$s = \mathtt{011011}$$$ to obtain $$$s = \mathtt{000011}$$$.  $$$b_3$$$ is not an open bracket, so no range toggle operation is done at this step.  $$$b_4$$$ is not an open bracket, so no range toggle operation is done at this step.  $$$b_5$$$ is an open bracket and $$$p_5 = 6$$$ since $$$b[5,6]=\mathtt{()}$$$ is a balanced bracket sequence. Hence, we do a range toggle operation from $$$5$$$ to $$$6$$$ on the binary string $$$s = \mathtt{000011}$$$ to obtain $$$s = \mathtt{000000}$$$.  $$$b_6$$$ is not an open bracket, so no range toggle operation is done at this step. In the fourth test case, the first operation using the bracket sequence $$$b = \mathtt{(((())))}$$$ will convert the binary string $$$s = \mathtt{01011100}$$$ to $$$s = \mathtt{11111001}$$$. Then, the second operation using the bracket sequence $$$b = \mathtt{()()(())}$$$ will convert the binary string $$$s = \mathtt{11111001}$$$ to $$$s=\mathtt{00000000}$$$.

Sample Inputs:
4
1
01
2
0000
3
100111
4
01011100

Sample Outputs:
-1
2
()()
()()
1
(())()
2
(((())))
()()(())

Solution (Python):
import sys

def inp():
    return sys.stdin.readline()
for _ in range(int(inp())):
    n = int(inp())
    s = inp()
    if s[-1:] == '\n':
        s = s[:-1]
    assert len(s) == 2 * n
    c = s.count('1')
    if c % 2 == 1 or s[0] != s[-1]:
        print(-1)
        continue
    if c == 0:
        print(0)
        continue
    if n == 1:
        print(1)
        print('()')
        continue
    v = False
    t = (2 * n - 3) * [False]
    bbs = []
    for i in range(0, 2 * n - 3):
        v = t[i] = v ^ (s[i + 1] == '1')
    for j in range(4):
        start = ['', '(', '()', '(()'][j]
        l = len(start)
        seq = [start]
        for i in range(j, 2 * n - 3, 4):
            seq.append('(())' if t[i] else '()()')
            l += 4
        while l < 2 * n - 1:
            seq.append('()')
            l += 2
        if l < 2 * n:
            seq.append(')')
        u = ''.join(seq)
        assert len(u) == 2 * n
        bbs.append(u)
    if s[0] == '1':
        bbs.append('(' + (n - 1) * '()' + ')')
    print(len(bbs))
    for ss in bbs:
        print(ss)

Solution (Python):
from sys import *
input = stdin.readline
inputlist = lambda : map(int, stdin.readline().split())

def getopposite(sequence):
    return ''.join(('1' if x == '0' else '0' for x in sequence))

def solvetwo(sequence):
    string1 = [0] * (2 * n)
    string2 = [0] * (2 * n)
    ones = []
    zeros = []
    for i in range(2 * n):
        if sequence[i] == '0':
            zeros.append(i)
        else:
            ones.append(i)
    for index in range(len(zeros) // 2):
        string1[zeros[index]] = '('
        string2[zeros[index]] = '('
        string1[zeros[len(zeros) - 1 - index]] = ')'
        string2[zeros[len(zeros) - 1 - index]] = ')'
    for index in range(len(ones)):
        if index % 2:
            string1[ones[index]] = ')'
            string2[ones[index]] = '('
        else:
            string1[ones[index]] = '('
            string2[ones[index]] = ')'
    stdout.write(''.join(string1) + '\n')
    stdout.write(''.join(string2) + '\n')

def solve(sequence):
    if sequence[0] != sequence[-1] or sequence.count('0') % 2:
        stdout.write('-1' + '\n')
    elif sequence[0] == '0':
        stdout.write('2' + '\n')
        solvetwo(sequence)
    else:
        stdout.write('3' + '\n')
        stdout.write('()' * n + '\n')
        solvetwo(getopposite(sequence))
for i in range(int(input())):
    n = int(input())
    sequence = input()[:-1]
    solve(sequence)

Solution (Python):
import sys

def inp():
    return sys.stdin.readline()
for _ in range(int(inp())):
    n = int(inp())
    s = inp()
    if s[-1:] == '\n':
        s = s[:-1]
    assert len(s) == 2 * n
    c = s.count('1')
    if c % 2 == 1 or s[0] != s[-1]:
        print(-1)
        continue
    if c == 0:
        print(0)
        continue
    if n == 1:
        print(1)
        print('()')
        continue
    v = False
    t = (2 * n - 3) * [False]
    bbs = []
    for i in range(0, 2 * n - 3):
        v = t[i] = v ^ (s[i + 1] == '1')
    for j in range(4):
        start = ['', '(', '()', '(()'][j]
        l = len(start)
        seq = [start]
        for i in range(j, 2 * n - 3, 4):
            seq.append('(())' if t[i] else '()()')
            l += 4
        while l < 2 * n - 1:
            seq.append('()')
            l += 2
        if l < 2 * n:
            seq.append(')')
        u = ''.join(seq)
        assert len(u) == 2 * n
        bbs.append(u)
    if s[0] == '1':
        bbs.append('(' + (n - 1) * '()' + ')')
    print(len(bbs))
    for ss in bbs:
        print(ss)