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clarification...
- ywliao July 12, 20151. sorry for the duplicated replies...
2. this is basically a modified version of zortlord's answer above, with an additional list of "what elements to search" (flagList). Basically, when counting the surpassers for an element x, if any compared element y is not smaller than x, you don't need to count the surpassers for y.
3. further, if x has with k surpassers, then there's no need to calculate the number of surpassers for the last k elements in the array.
So the worst case time complexity will be O(n^2), with memory O(n) I think...