(Analysis by Timothy Qian)

We first show that you can turn all the lights off in $O(N)$ moves. First, turn off all the switches. Next, for every light that is on after you have turned off that switch, turn on the corresponding switch, then turn off the rotated switch on the next move. This in total takes at most $3N$ moves.

Now we solve the following sub-problem using bitmask dynamic programming. Say we start with all the switches off. Let $dp[i][mask]$ be a boolean denoting whether after $i$ moves, you can turn on the lights corresponding to the bitmask $mask$ in exactly $i$ moves. The base case of $dp[0][mask]$ is true if and only if $mask = 0$. Now let's say we've computed $dp[i - 1][mask]$ for some $i \geq 1$. Then to compute $dp[i][mask]$, we can do casework on the first move. The effects of the first move after exactly $i$ moves will be effectively turning on $i$ consecutive lights. Thus, there are $N$ possible first moves, and we can use $dp[i - 1][mask]$ to compute transitions for using exactly $i$ moves. This is done in $O(N^2 \cdot 2^N)$ time. We can remove a factor of $O(N)$ by computing an array $rep$ (standing for representative), such that we have $rep[mask1] = rep[mask2]$ if and only if $mask1, mask2$ that are cyclic rotations of each other. Then, we only need to check the first moves up to cyclic rotation, which means there is only one possible first move with $i$ consecutive lights on. See my code for further details. This will take $O(N\cdot 2^N)$ time.

Now we go back to the full problem, say that we have two bitmasks $lights$, $switches$ corresponding to the lights and switches respectively. Then let us check whether we can turn all lights off in exactly $k$ moves. We can pretend all the switches are off by computing the effect of the switches after $k$ moves if we didn't toggle any lights. Let the bitmask representing the lights after $k$ moves be $mask$. Then $dp[k][mask]$ will denote whether we can do this in $k$ moves. Thus, we simply need to check this for $k$ from $0$ to $3N$. For each test case, this takes $O(N)$ time. This overall will take $O(N\cdot 2^N + TN)$ time. We also note that an algorithm running in $O(N^2 \cdot 2^N + TN)$ time was sufficient for full credit.

$O(N \cdot 2^N + TN)$ time C++ code:

#include <bits/stdc++.h>

using namespace std;

int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int t, n;
cin >> t >> n;
auto cyc_right = [&n](int x) -> int {
x <<= 1;
if (x & (1 << n)) {
x ^= (1 << n) ^ 1;
}
return x;
};
auto str_to_bin = [&n](string s) -> int {
int res = 0;
for (int i = 0; i < n; ++i) {
if (s[i] == '1') {
res ^= (1 << i);
}
}
return res;
};
vector<int> rep(1 << n, -1);
for (int i = 0; i < 1 << n; ++i) {
if (rep[i] == -1) {
int j = i;
while (rep[j] == -1) {
rep[j] = i;
j = cyc_right(j);
}
}
}
vector<vector<bool>> can(3*n + 1, vector<bool>(1 << n));
can[0][rep[0]] = true;
for (int i = 1; i <= 3*n; ++i) {
}
}
}
while (t--) {
string x, y;
cin >> x >> y;
int fin = str_to_bin(x);
int cur = 0;
for (int i = 0; i <= 3*n; ++i) {
if (can[i][rep[cur ^ fin]]) {
cout << i << '\n';
break;
}
}
}
return 0;
}


Danny Mittal's $O(N^2 \cdot 2^N + TN)$ time Java code:

import java.io.BufferedReader;
import java.io.IOException;
import java.util.StringTokenizer;

public class LightsOffOptimized {

public static void main(String[] args) throws IOException {
int t = Integer.parseInt(tokenizer.nextToken());
int n = Integer.parseInt(tokenizer.nextToken());

int[][] pads = new int[3*n + 1][n];
for (int l = 1; l <= 3*n; l++) {
for (int d = 0; d < n; d++) {
pads[l][d] = (1 << d) ^ pads[l - 1][(d + 1) % n];
}
}

boolean[][] dp = new boolean[3*n + 1][1 << n];
dp[0][0] = true;
for (int turns = 1; turns <= 3*n; turns++) {
for (int d = 0; d < n; d++) {
break;
}
}
}
}
StringBuilder out = new StringBuilder();
for (; t > 0; t--) {

for (int turns = 0; turns <= 3*n; turns++) {
break;
}
if (pad >= 1 << n) {
}
}