"To optimise it to O(n^3) just note that the selection of a and b is independent."

See my second submission to this problem for a solution following this format. We first get r[] = {1,0,1,0,0} and initialize x[i] = {-1,0,-1,0,0}. We loop through a[] and keep track of max and max2, indices for the 1st and 2nd largest term so far. We know that if a[max]>a[i]>a[max2], that's another "contribution" for max so x[max]++.

The only x[max]++ here happens at i = 1, x[0]++. After the loop, you get x[i] = {0, 0, -1, 0, 0}, so we find the smallest a[i] with x[i] = 0 -> 1.

Sadly, bad time zone for this one. Moo.