Equal Binary Subsequences
Contest: Codeforces Round 825 (Div. 2)
Difficulty rating: 2200
Tags: constructive algorithms, geometry, greedy, implementation, strings

Everool has a binary string $$$s$$$ of length $$$2n$$$. Note that a binary string is a string consisting of only characters $$$0$$$ and $$$1$$$. He wants to partition $$$s$$$ into two disjoint equal subsequences. He needs your help to do it.

You are allowed to do the following operation exactly once.

- You can choose any subsequence (possibly empty) of $$$s$$$ and rotate it right by one position.

In other words, you can select a sequence of indices $$$b_1, b_2, \ldots, b_m$$$, where $$$1 \le b_1 < b_2 < \ldots < b_m \le 2n$$$. After that you simultaneously set $$$$$$s_{b_1} := s_{b_m},$$$$$$ $$$$$$s_{b_2} := s_{b_1},$$$$$$ $$$$$$\ldots,$$$$$$ $$$$$$s_{b_m} := s_{b_{m-1}}.$$$$$$

Can you partition $$$s$$$ into two disjoint equal subsequences after performing the allowed operation exactly once?

A partition of $$$s$$$ into two disjoint equal subsequences $$$s^p$$$ and $$$s^q$$$ is two increasing arrays of indices $$$p_1, p_2, \ldots, p_n$$$ and $$$q_1, q_2, \ldots, q_n$$$, such that each integer from $$$1$$$ to $$$2n$$$ is encountered in either $$$p$$$ or $$$q$$$ exactly once, $$$s^p = s_{p_1} s_{p_2} \ldots s_{p_n}$$$, $$$s^q = s_{q_1} s_{q_2} \ldots s_{q_n}$$$, and $$$s^p = s^q$$$.

If it is not possible to partition after performing any kind of operation, report $$$-1$$$.

If it is possible to do the operation and partition $$$s$$$ into two disjoint subsequences $$$s^p$$$ and $$$s^q$$$, such that $$$s^p = s^q$$$, print elements of $$$b$$$ and indices of $$$s^p$$$, i. e. the values $$$p_1, p_2, \ldots, p_n$$$.

Input format:
Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^5$$$). Description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$), where $$$2n$$$ is the length of the binary string.

The second line of each test case contains the binary string $$$s$$$ of length $$$2n$$$.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$.

Output format:
For each test case, follow the following output format.

If there is no solution, print $$$-1$$$.

Otherwise,

- In the first line, print an integer $$$m$$$ ($$$0 \leq m \leq 2n$$$), followed by $$$m$$$ distinct indices $$$b_1$$$, $$$b_2$$$, ..., $$$b_m$$$(in increasing order).
- In the second line, print $$$n$$$ distinct indices $$$p_1$$$, $$$p_2$$$, ..., $$$p_n$$$ (in increasing order).

If there are multiple solutions, print any.

Constraints:
Time limit: 1.0 s
Memory limit: 256.0 MB

Note:
In the first test case, $$$b$$$ is empty. So string $$$s$$$ is not changed. Now $$$s^p = s_1 s_2 = \mathtt{10}$$$, and $$$s^q = s_3s_4 = \mathtt{10}$$$.

In the second test case, $$$b=[3,5]$$$. Initially $$$s_3=\mathtt{0}$$$, and $$$s_5=\mathtt{1}$$$. On performing the operation, we simultaneously set $$$s_3=\mathtt{1}$$$, and $$$s_5=\mathtt{0}$$$.

So $$$s$$$ is updated to 101000 on performing the operation.

Now if we take characters at indices $$$[1,2,5]$$$ in $$$s^p$$$, we get $$$s_1=\mathtt{100}$$$. Also characters at indices $$$[3,4,6]$$$ are in $$$s^q$$$. Thus $$$s^q=100$$$. We are done as $$$s^p=s^q$$$.

In fourth test case, it can be proved that it is not possible to partition the string after performing any operation.

Example input:
4
2
1010
3
100010
2
1111
2
1110

Example output:
0
1 2
2 3 5
1 2 5
3 2 3 4
1 4
-1

Editorial:
It is easy to see that a necessary condition for a solution to exist is that the number of $$$1$$$ in $$$s$$$ should be even. It turns out that this condition is sufficient too.
Here is one valid construction:
We make $$$n$$$ pairs of the form $$$(s[2i-1],s[2i])$$$ for $$$(1 \leq i \leq n)$$$.
Assume we have $$$x$$$ pairs in which both elements are different and $$$n-x$$$ pairs in which both elements are same.
Claim — $$$x$$$ should be even.
Proof — Assume that among the $$$n-x$$$ pairs in which both elements are same, we have $$$y$$$ pairs in which both elements are $$$1$$$. So number of $$$1$$$ in $$$s$$$ is $$$x+2 \cdot y$$$. We know that number of $$$1$$$ in $$$s$$$ is even, so for $$$x+2 \cdot y$$$ to be even, $$$x$$$ should also be even.
Now we will select $$$x$$$ indices; exactly one index from each of the $$$x$$$ pairs in which both elements are distinct. Take the index of $$$0$$$ from $$$i_{th}$$$ pair if $$$i$$$ is odd, else take the index of $$$1$$$. Thus our selected characters = $$${0,1,0,1, \dots ,0,1}$$$
Now on cyclically shifting the selected characters clockwise once, we can see that elements at selected indices got flipped.
Since, elements in those $$$x$$$ pairs were distinct initially, and we flipped exactly one character from each of those $$$x$$$ pairs, both elements of those $$$x$$$ pairs are same now.
Hence, in updated $$$s$$$, $$$s[2i-1]=s[2i]$$$.
So, for $$$s_1$$$, we can select characters of all odd indices.
Finally we'll have $$$s_1 = s_2$$$.