Question

A local maximum in a 2D array can be defined as a value such that all it’s 4 neighbours are less than or equal to it, ie, for a[i][j] to be a local maximum,

a[i+1][j] <= a[i][j] 
&& a[i-1][j] <= a[i][j]
&& a[i][j+1] <= a[i][j]
&& a[i+1][j-1] <= a[i][j]

I was asked to find all the local maxima in a given 2D array.

The naive way to do this would be to just go through all the elements, and checking whether or not each element is a local maximum. This would be O( n^2 ). I feel that you cannot do better than this, although my friend insists that an asymptotically better algorithm should exist. Any hints ?

I was thinking along the lines of Divide and Conquer, yet I feel that it would be impossible to detect all the local maxima without going through all the numbers, which would necessarily be O( n^2 ). Am I right or am I missing out on something ?

Answer 1

I am pretty sure this cannot be solved in less than O(n^2) comparisons. Assume a chess board 2d matrix where all the white squares are 1 and blacks are 0. It wiil will have O(n^2) solutions and each solution requires at least one comparison.