Count smaller elements right
Given an integer array `nums`, return a new array `counts` where counts[i] is the number of elements to the RIGHT of nums[i] that are strictly smaller than nums[i]. Values may be negative. Length up to 10^5, so an O(n^2) double loop is too slow — use a modified merge sort (count cross-pair inversions) or a Binary Indexed Tree over compressed values for O(n log n).
Implement
count_smaller(nums: list[int]) → list[int]Examples
in
[[5,2,6,1]]out[2,1,1,0]What a strong answer looks like
State your approach and its time/space complexity out loud before you optimize. Handle the edge cases (empty input, duplicates, overflow), and say why you chose this over the brute force. Green tests are the floor, not the grade.
0:00 of about 35 min
solution.py
InputExpectedGot
[[5,2,6,1]][2,1,1,0]not run yetsample