To solve the problem efficiently, observe that, if square A is a killer, then every square inside it (sharing the same center point) is also a killer. Conversely, if a square is not a killer then no square surrounding it can be a killer.

This forms the basis of an ~\mathcal O(N^4)~ solution: try and extend a square from each memory cell. The worst-case complexity is ~\mathcal O(N^4)~ because there are ~N^2~ cells, ~\mathcal O(N)~ dimensions to be checked and checking the frame (when trying to extend a square killer) takes ~\mathcal O(N)~.

Naively implementing the above approach will not be efficient enough. One way of speeding it up is to, for each cell and each of the 4 directions, precalculate 32 or 64 binary digits starting at the cell ~(y,x)~ in the chosen direction. This allows us to check frames in blocks of 32 or 64 bits (instead of one by one) and speed up the solution by a factor.

Other heuristics that reduce the number of checks (eliminating those that surely can't contribute to the solution) could be used to speed up the solution further.