1365. How Many Numbers Are Smaller Than the Current Number

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]

var smallerNumbersThanCurrent = function(nums) {
    let result = new Array(nums.length)
    for(let i = 0; i < nums.length; i++) {
        let count = 0
        const value = nums[i]
        nums.forEach( j => { if(j < value) count++ })
        result[i] = count
    }
    return result
}