Count of Smaller Numbers After Self

Hard · binary search solution for Count of Smaller Numbers After Self. Python, Go, and a step-by-step animation.

Problem

Read the prompt, flip through the examples, then steal the idea.

LeetCode 315. Count of Smaller Numbers After Self. The copyrighted prompt is on LeetCode. This page is a working binary search solution for this problem: editable input, Python, Go, and a step-by-step animation.

Constraints

—Player inputs are capped so the animation stays readable.
—Match LeetCode's own constraints when you submit there.

Examples

Input

nums =: [-1, 0, 3, 5, 9, 12]
target =: 9

Output

9 sits at index 4. Each step throws away half the range.

Solution

Python and Go. The walkthrough below runs this same code.

Approach 1Optimal▶ Animated

Count of Smaller Numbers After Self — binary search

Halve the sorted range until the target is found or the range collapses. Built for LeetCode 315. Count of Smaller Numbers After Self.

Because the slice is sorted, comparing nums[mid] to target tells you which half can be thrown away. That is the whole algorithm.

Algorithm

While lo <= hi, pick mid.
Equal → return mid. Too small → search right. Too big → search left.
Empty range → not found (-1).

Time: O(log n)
Space: O(1)

Solution.py

1class Solution:2    def countOfSmallerNumbersAfterSelf(self, nums: list[int], target: int) -> int:3        lo, hi = 0, len(nums) - 14        while lo <= hi:5            mid = (lo + hi) // 26            if nums[mid] == target:7                return mid8            if nums[mid] < target:9                lo = mid + 110            else:11                hi = mid - 112        return -1

Watch this run

Interactive walkthrough

Playing “Count of Smaller Numbers After Self — binary search”. Edit the input, hit Run, then scrub.

Approach: Count of Smaller Numbers After Self — binary search

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