smallestNumberArray.js

/*
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
*/

//Answer//

/**
 * @param {number[]} nums
 * @return {number[]}
 */
var smallerNumbersThanCurrent = function(nums) {
let A = [];
    for (let i = 0; i <nums.length ; i++) {
A.push(nums.filter(x=>x<nums[i]).length)
} 
    return A
};