First, consider the following problem:

(Problem X) at most how many buildings can be chosen so that:

• the chosen buildings have the same height, and
• the chosen buildings are contiguous?

This can be solved in \(O(N)\) time by scanning the buildings from the front.

Pseudocode

ans=0
height=0
crrent=0
for i in 1..N:
  if height!=H[i]:
    crrent=0
    height=H[i]
  crrent+=1
  ans=max(ans,crrent)
print(ans)

Next, consider the following problem.

(Problem X) at most how many buildings can be chosen so that:

• the chosen buildings have the same height, and
• the chosen buildings are chosen at the interval of \(K\)?

The buildings that can be chosen at the same time can be divided into the following \(K\) disjoint groups:

• \(1,\ K+1,\ 2K+1,\ 3K+1\ldots\)
• \(2,\ K+2,\ 2K+2,\ 3K+2\ldots\)
• \(3,\ K+3,\ 2K+3,\ 3K+3\ldots\)
• \(\dots\)

So it is sufficient to solve (Problem X) for these \(K\) independent groups.
Finding the answer for each group costs \(O(N/K)\) time, which is repeated \(K\) times, so the overall complexity is \(O(N)\).

Finally, we consider the original problem. By solving (Problem Y) for all \(K=1,2,\ldots,\) the answer can be found in a total of \(O(N^2)\) time.