One approach we can take is to iterate over Benji's mask names ~B_i~ (for ~i = 1 \dots M~), and for each one, compare it against each of Ethan's mask names ~A_j~ (for ~j = 1 \dots N~), incrementing our running answer if we find a match (such that ~B_i = A_j~). Over the course of this algorithm, we may compare each of the strings ~A_{1 \dots N}~ against each of the strings ~B_{1 \dots M}~, resulting in a time complexity of ~\mathcal O(NM)~.

Another possible approach is to start by inserting Ethan's mask names ~A_{1 \dots N}~ into a data structure such as a set (a balanced binary search tree) or hash table, and then iterate over each of Benji's mask names ~B_i~ (for ~i = 1 \dots M~), querying the data structure for the existence of ~B_j~ rather than iterating over all of the strings ~A_{1 \dots N}~ each time. The time complexity of this algorithm is either ~\mathcal O((N+M) \log N)~ or ~\mathcal O(N+M)~, depending on the data structure used.