Author: Unknown
Subject: Inversion sort
Date: Tuesday, 03 Mar 2020, 21:38:10

Hi, I remember that during class, we learned that the number of steps for the lowest number of swaps for inversion sort is 
(n(n-1)/2)^2 = O(n^4)
where we have n(n-1)/2 possible sum of swaps.
However, I am not sure how we reach (n(n-1)/2)^2