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How Many Numbers Are Smaller Than the Current Number

Problem statement​

Given the array nums, for each nums[i] find out how many numbers in the array are smaller than it. That is, for each nums[i] you have to count the number of valid j's such that j != i and nums[j] < nums[i].

Return the answer in an array.

Example 1:

Input: nums = [8,1,2,2,3]Output: [4,0,1,1,3]Explanation: For nums[0]=8 there exist four smaller numbers than it (1, 2, 2 and 3). For nums[1]=1 does not exist any smaller number than it.For nums[2]=2 there exist one smaller number than it (1). For nums[3]=2 there exist one smaller number than it (1). For nums[4]=3 there exist three smaller numbers than it (1, 2 and 2).

Example 2:

Input: nums = [6,5,4,8]Output: [2,1,0,3]

Example 3:

Input: nums = [7,7,7,7]Output: [0,0,0,0]

Constraints:

  • 2 <= nums.length <= 500
  • 0 <= nums[i] <= 100

My solution​

/**
 * @param {number[]} nums
 * @return {number[]}
 */
var smallerNumbersThanCurrent = function(nums) {
    const freq = Array.from({
        length: 101
    }, () => 0)
    let smaller = Array.from({
        length: 101
    }, () => 0)
    
    nums.forEach((num) => freq[num]++)
    
    for (let i = 1; i < 101; i++)  {
        smaller[i] = smaller[i - 1] + freq[i - 1]
    }
    
    return nums.map(num => smaller[num])
};