Qingshan Loves Strings 2 (Codeforces 1890C)
Tags: Implement, two pointers

Qingshan has a string $$$s$$$ which only contains $$$\texttt{0}$$$ and $$$\texttt{1}$$$.A string $$$a$$$ of length $$$k$$$ is good if and only if  $$$a_i \ne a_{k-i+1}$$$ for all $$$i=1,2,\ldots,k$$$. For Div. 2 contestants, note that this condition is different from the condition in problem B.For example, $$$\texttt{10}$$$, $$$\texttt{1010}$$$, $$$\texttt{111000}$$$ are good, while $$$\texttt{11}$$$, $$$\texttt{101}$$$, $$$\texttt{001}$$$, $$$\texttt{001100}$$$ are not good.Qingshan wants to make $$$s$$$ good. To do this, she can do the following operation at most $$$300$$$ times (possibly, zero):   insert $$$\texttt{01}$$$ to any position of $$$s$$$ (getting a new $$$s$$$). Please tell Qingshan if it is possible to make $$$s$$$ good. If it is possible, print a sequence of operations that makes $$$s$$$ good.

Input: The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows.The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n\le 100$$$) — the length of string $$$s$$$, respectively.The second line of each test case contains a string $$$s$$$ with length $$$n$$$.It is guaranteed that $$$s$$$ only consists of $$$\texttt{0}$$$ and $$$\texttt{1}$$$.

Output: For each test case, if it impossible to make $$$s$$$ good, output $$$-1$$$.Otherwise, output $$$p$$$ ($$$0 \le p \le 300$$$) — the number of operations, in the first line.Then, output $$$p$$$ integers in the second line. The $$$i$$$-th integer should be an index $$$x_i$$$ ($$$0 \le x_i \le n+2i-2$$$) — the position where you want to insert $$$\texttt{01}$$$ in the current $$$s$$$. If $$$x_i=0$$$, you insert $$$\texttt{01}$$$ at the beginning of $$$s$$$. Otherwise, you insert $$$\texttt{01}$$$ immediately after the $$$x_i$$$-th character of $$$s$$$.We can show that under the constraints in this problem, if an answer exists, there is always an answer that requires at most $$$300$$$ operations.

Note: In the first test case, you can do zero operations and get $$$s=\texttt{01}$$$, which is good.Another valid solution is to do one operation: (the inserted $$$\texttt{01}$$$ is underlined)  $$$\texttt{0}\underline{\texttt{01}}\texttt{1}$$$ and get $$$s = \texttt{0011}$$$, which is good.In the second and the third test case, it is impossible to make $$$s$$$ good.In the fourth test case, you can do two operations:  $$$\texttt{001110}\underline{\texttt{01}}$$$  $$$\texttt{0011100}\underline{\texttt{01}}\texttt{1}$$$ and get $$$s = \texttt{0011100011}$$$, which is good.

Sample Inputs:
6
2
01
3
000
4
1111
6
001110
10
0111001100
3
001

Sample Outputs:
0

-1
-1
2
6 7
1
10
-1

Solution (Python):
for _ in range(int(input())):
    h = int(input())
    j = input()
    m = []
    l = 0
    k = 0
    if j.count('0') == j.count('1'):
        while k <= h // 2 + l - 1:
            if j[k] == j[h + l * 2 - k - 1] and j[k] == '1':
                j = j[0:k] + '01' + j[k:h + l * 2]
                l += 1
                m.append(k)
            elif j[k] == j[h + l * 2 - k - 1] and j[k] == '0':
                j = j[0:h + l * 2 - k] + '01' + j[h + l * 2 - k:h + l * 2]
                m.append(h + l * 2 - k)
                l += 1
            k += 1
        print(l)
        print(*m)
    else:
        print('-1')

Solution (Python):
for _ in range(int(input())):
    n = int(input())
    s = input()
    if s.count('0') != s.count('1'):
        print(-1)
    else:
        ans = []
        left = 0
        right = n - 1
        cnt = 0
        while left < right:
            if s[left] == s[right]:
                if s[left] == '0':
                    s = s[:right + 1] + '01' + s[right + 1:]
                    ans.append(right + 1)
                    left += 1
                    right += 1
                else:
                    s = s[:left] + '01' + s[left:]
                    ans.append(left)
                    left += 1
                    right += 1
            else:
                left += 1
                right -= 1
        print(len(ans))
        print(*ans)

Solution (Python):
def is_good(s: str) -> None:
    left = 0
    right = len(s) - 1
    res = []
    if s.count('0') != s.count('1'):
        print(-1)
    else:
        while left < right:
            if s[left] == s[right]:
                if s[left] == '0':
                    s = s[:right + 1] + '01' + s[right + 1:]
                    res.append(right + 1)
                    left += 1
                    right += 1
                else:
                    s = s[:left] + '01' + s[left:]
                    res.append(left)
                    left += 1
                    right += 1
            else:
                left += 1
                right -= 1
        print(len(res))
        print(*res)
t = int(input())
for _ in range(t):
    n = int(input())
    s = input()
    is_good(s)