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: $\mathcal{O}(NM{2}^{NM})$

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: $\mathcal{O}(N{2}^{3N})$

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: $\mathcal{O}(N^3)$