Prefix Flip (Hard Version)
Contest: Codeforces Round 658 (Div. 1)
Difficulty rating: 1700
Tags: constructive algorithms, data structures, implementation, strings, two pointers

This is the hard version of the problem. The difference between the versions is the constraint on $$$n$$$ and the required number of operations. You can make hacks only if all versions of the problem are solved.

There are two binary strings $$$a$$$ and $$$b$$$ of length $$$n$$$ (a binary string is a string consisting of symbols $$$0$$$ and $$$1$$$). In an operation, you select a prefix of $$$a$$$, and simultaneously invert the bits in the prefix ($$$0$$$ changes to $$$1$$$ and $$$1$$$ changes to $$$0$$$) and reverse the order of the bits in the prefix.

For example, if $$$a=001011$$$ and you select the prefix of length $$$3$$$, it becomes $$$011011$$$. Then if you select the entire string, it becomes $$$001001$$$.

Your task is to transform the string $$$a$$$ into $$$b$$$ in at most $$$2n$$$ operations. It can be proved that it is always possible.

Input format:
The first line contains a single integer $$$t$$$ ($$$1\le t\le 1000$$$)  — the number of test cases. Next $$$3t$$$ lines contain descriptions of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$1\le n\le 10^5$$$)  — the length of the binary strings.

The next two lines contain two binary strings $$$a$$$ and $$$b$$$ of length $$$n$$$.

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

Output format:
For each test case, output an integer $$$k$$$ ($$$0\le k\le 2n$$$), followed by $$$k$$$ integers $$$p_1,\ldots,p_k$$$ ($$$1\le p_i\le n$$$). Here $$$k$$$ is the number of operations you use and $$$p_i$$$ is the length of the prefix you flip in the $$$i$$$-th operation.

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

Note:
In the first test case, we have $$$01\to 11\to 00\to 10$$$.

In the second test case, we have $$$01011\to 00101\to 11101\to 01000\to 10100\to 00100\to 11100$$$.

In the third test case, the strings are already the same. Another solution is to flip the prefix of length $$$2$$$, which will leave $$$a$$$ unchanged.

Example input:
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1

Example output:
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1

Editorial:
There are several ways to solve the hard version as well.
Solution 1
Given an arbitrary binary string $$$s$$$, we can make all bits $$$0$$$ in at most $$$n$$$ operations. Simply scan the string from left to right. If bits $$$i$$$ and $$$i+1$$$ disagree, apply the operation to the prefix of length $$$i$$$. This is also easy to simulate in $$$O(n)$$$ time.
We can make $$$s$$$ all zeros in at most $$$n$$$ operations, and we can make $$$t$$$ all zeros in at most $$$n$$$ operations. By reversing the order of the operations on $$$t$$$, we have transformed $$$s$$$ into $$$t$$$ in at most $$$2n$$$ operations, as desired.
Solution 2
Another approach is to optimize the simulation for solution 2 from the easy version. You can do this with a data structure such as a balanced binary search tree in $$$O(n \log n)$$$ time, but there is no need.
Instead, we can observe that after making the last $$$k$$$ bits correct with our procedure, the prefix of $$$s$$$ of length $$$n-k$$$ will correspond to some segment of the original string $$$s$$$, except it will be possibly flipped (inverted and reversed). So, we need only keep track of the starting index of this segment, and a flag for whether it is flipped. Complexity is $$$O(n)$$$.
Solution 3
A third solution uses randomization to improve the number of operations from $$$3n$$$ in solution 1 of the easy version. We can observe that in a random test case, approximately half the bits of $$$s$$$ will be mismatches with $$$t$$$. Solution 1 in the easy version uses $$$3$$$ operations per mismatch, which is $$$3n/2$$$ operations in expectation. Obviously, you don't get to decide that all test cases are random. But you can spend a small number of operations initially, flipping random prefixes to make the string more random. If it doesn't work, you can try again repeatedly.
Flipping random prefixes is a complicated process that might be hard to compute the exact probability. But if the probability is $$$p$$$, and we try to flip $$$k$$$ prefixes randomly, the time complexity is $$$O((k+1)n)\log p$$$.
If you find a deterministic solution with a strictly lower ratio than $$$2$$$ operations per bit, we would love to hear about it!