I'm learning divide-and-conquer now, and here below is the Sort-and-Count algorithm for counting teh inversions which means given a sequence of n numbers a1, . . . , an, which we assume are all distinct, and we define an inversion to be a pair i j such that ai > aj. . Now if I want to count teh "significant" inversion (ij and ai > 2aj ), what changes I need to do base on Sort-and-Count algorithm ? why the running time complexity will still be same ?

Sort-and-Count (L) t if list L has one element. return o and the list L Divide the list into two halves A and B AI A) Sort-and-Count (A) (ra, B) Sort-and-Count (B) (r L) Merge-and-count. (A B) return r r, r, r and the sorted list L TOn)