Count Later Smaller Ranks Problem

Count Later Smaller Ranks Problem — ExecCode Hard DSA Practice

Solve the Count Later Smaller Ranks problem on ExecCode. Free online hard DSA practice in Two Pointers. Write and run code in Java, C++, Python — no signup required to run.

Problem description

For every index i you must count how many later indices hold a strictly smaller value. This is inversion counting per starting index and is a common OA upgrade of 'count inversions'. Nested loops fail. Return an array ans where ans[i] is the number of indices j > i with nums[j] < nums[i]. You are given: nums — integer array Return integer array of the same length. The last index always contributes 0.

Examples

Input {'nums': [5, 2, 6, 1]}; Output [2, 1, 1, 0]. Input {'nums': [1, 2, 3]}; Output [0, 0, 0]. Input {'nums': [3, 3, 3]}; Output [0, 0, 0]

Constraints

Equal values are not counted

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