How do I find the Pareto-optimal points in O(nh) and O(nlog(h)) complexity?

Question

Can anybody suggest an algorithm to find Pareto-optimal points (to form a staircase) just as given in diagram in O(n*h) and O(n*log(h)) time complexity, where h is the number of Pareto-optimal points?

I used gift wrapping algorithm to solve this (in O(n*h)), but it only finds convex hulls type of staircase and misses those points who form concave angles.

Answer 1

To put suggestion in one place.

Idea is to use divide and conquer strategy. If we can find pareto point P which splits rest of pareto points in half's in O(n) than it can be done in O(n log(h)). That can be checked by splitting points in three parts right of P, left and up of P, left and down of P. Third set doesn't have pareto points. And recursively do same procedure on other two set of points. Set that three parts split points in ratio: a, b, 1 - a - b. With that complexity is:

T(n, h) = T(a*n, h/2) + T(b*n, h/2) + O(n)
       <= T(n, h/2) + O(n)
       <= T(n, h/4) + 2 * O(n)
       <= T(n, h/8) + 3 * O(n)
       <= ...
       <= O(n log(h))

Problem is finding pareto point with that characteristic in <= O(n). Some simple heuristic like P where x >= (min(x) + max(x)) / 2, probably makes very good average.

I am not sure, but I think it is possible to use k-th order statistic to find point with that characteristic. Similar to finding median as pivot in quicksort.