Let's take fat fish, say it is k-th in sorted order. Let's give each fish its mark: it will be 1 for (k - 1)-th fish, 2 for each fish smaller than (k - 1)-th, and 0 for all bigger (including our fat fish). The result of the fight will be fish with mark min(x, y) where x and y are marks of fishes in the fight. In the end we will get fish with mark 0, so it is easy to see that there will be a fight between fishes with marks 0 and 1 at some moment. I claim that this fight is non-dangerous (any fish with mark 0 is more than twice bigger than any possible fish with marks 1 or 2). Also it is not hard to check that these fights are different for all fat fishes.

graph : https://imgur.com/a/Iac4Lm2

Here we see,that s(1) = 1 ==> a(1) = 1.We can say,that a(3) = s(3) — a(1) and a(4) = s(4) — a(1). But can we reduce the answer?Yes.We can put on vertex 2 min(s(3) — a(1), s(4) — a(1)).And this value will embrace all children of vertex(2).So a(1) = 1, a(2) = 2, a(3) = 0, a(4) = 1. Ans = 4.

If a row contains three characters, two neighbor rows of each row will be same. If a column contains three characters, two neighbor columns of each column will be same. Thus, in a good matrix, either each row contain at most two different characters, or each column contain at most two different characters.