Frosh Week

ID: froshweek

Difficulty: 5.6

CPU Time: 3 seconds

Memory: 1024 MB

Solution

Since every single pair-wise must be counted, this is effectively an inversion count in merge-sort. So, keep a global variable of swaps occured (initialized at zero), and implement a basic merge-sort that, when the left-hand element is larger than the right-hand element, there will be leftArraySize minus leftArrayIndex swaps that occur to place the right-hand element in its place.