For the first subtask, it suffices to brute force all ways of replacing each ? with either an X or a . and checking if the resulting plan is efficient.

Time complexity: ~\mathcal{O}(NM2^{NM})~

For the second subtask, let's assume ~N=2~ since the ~N=1~ case is trivial. Let's suppose that there are no ? in either string. Then clearly there is an efficient plan if the strings are different and no efficient plan if the strings are equal. If there is a single ? in either string, there is always an efficient plan.

Time complexity: ~\mathcal{O}(NM)~

For the third subtask, observe that rows with more than ~\lceil \log_2 N \rceil~ ? can be ignored due to Hall's marriage theorem. It essentially states that, each row that can yield at least ~N~ distinct rows after replacing the ? with . or X can be ignored because they can always be matched to one of these rows. Therefore, if we ignore rows that don't satisfy the limit, each row contains at most ~3~ ?. Therefore, it suffices to brute force all of these assignments as well.

Time complexity: ~\mathcal{O}(N2^{3N})~

Using the same observation as the previous subtask, we can remove all rows with more than ~9~ ?. Now, each row can yield at most ~N~ distinct rows after performing all possible replacements. We can build a bipartite graph with each initial row on the left and every possible resulting row on the right. There are at most ~N^2~ nodes on the right and at most ~N^2~ edges in the graph. We can then run any bipartite matching algorithm to find a perfect matching.

Time complexity: ~\mathcal{O}(N^3)~