Approach: Counting sort

def smallerNumbersThanCurrent(nums):
    count = [0]*101
    for x in nums: count[x] += 1
    prefix = [0]*101
    for i in range(1,101): prefix[i] = prefix[i-1] + count[i-1]
    return [prefix[x] for x in nums]

Complexity: O(n) time, O(1) space.

Full Solution — all languages

class Solution:
    def smallerNumbersThanCurrent(self, nums):
        res = []
        for x in nums:
            res.append(sum(1 for y in nums if y < x))
        return res

import java.util.*;
class Solution {
    public int[] smallerNumbersThanCurrent(int[] nums) {
        int[] res = new int[nums.length];
        for (int i = 0; i < nums.length; i++) {
            int c = 0;
            for (int y : nums) if (y < nums[i]) c++;
            res[i] = c;
        }
        return res;
    }
}

#include <bits/stdc++.h> using namespace std;
class Solution {
    public: vector<int> smallerNumbersThanCurrent(vector<int>& nums) {
        vector<int> res(nums.size());
        for (size_t i = 0; i < nums.size(); i++) {
            int c = 0;
            for (int y : nums) if (y < nums[i]) c++;
            res[i] = c;
        }
        return res;
    }
};

/**
 * @param {number[]} nums
 * @return {number[]}
 */ var smallerNumbersThanCurrent = function(nums) {
    return nums.map(x => nums.reduce((c, y) => c + (y < x ? 1 : 0), 0));
};

How Many Numbers Are Smaller Than the Current Number

Example 1

Example 2

Constraints