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$ curl repovive.com/roadmaps/pattern-22/segment-tree/count-of-smaller-numbers-after-self-implementation

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$ curl repovive.com/roadmaps/pattern-22/segment-tree/count-of-smaller-numbers-after-self-implementation

Repovive

LeetCode 315 Count of Smaller Numbers After Self - Implementation - Segment Tree | Pattern 22: LeetCode Interview Patterns | Repovive

Repovive.

Pattern 22: LeetCode Interview Patterns24 sections · 736 units32h total

Open in Course

LeetCode 315 Count of Smaller Numbers After Self - Implementation

The approach

Here's the solution:

function countSmaller(nums):
    def mergeSort(arr):
        if len(arr) <= 1:
            return arr
        mid = len(arr) // 2
        left = mergeSort(arr[:mid])
        right = mergeSort(arr[mid:])
        result = []
        i = j = 0
        while i < len(left):
            while j < len(right) and right[j][0] < left[i][0]:
                j += 1
            counts[left[i][1]] += j
            // merge logic
        return merged
    counts = [0] * len(nums)
    mergeSort(list(enumerate(nums)))
    return counts

$O(n \log n)$ time, $O(n)$ space.