Calculates how many numbers are smaller in array

4th exercise - #1365
https://leetcode.com/problems/how-many-numbers-are-smaller-than-the-current-number/submissions/

smallestNumberArray.js

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+/*
+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
+};