1365. How Many Numbers Are Smaller Than the Current Number.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
*/

var smallerNumbersThanCurrent = function(nums) {
    var sortedArray = [...nums].sort((a, b) => { return a - b });
    return nums.map((num) => {
        return sortedArray.indexOf(num);
    });
}

var smallerNumbersThanCurrent = function(nums) {
    var sortedArray = [...nums].sort((a, b) => { return a - b });
    let arr = []
    for (let val of nums) {
        arr.push(sortedArray.indexOf(val))
    }
    return arr
}


var smallerNumbersThanCurrent = function(nums) {
    var arr = [];
    for (var i = 0; i < nums.length; i++) {
        arr[i] = 0;
        for (var j = 0; j < nums.length; j++) {
            if (nums[j] < nums[i]) {
                arr[i] += 1;
            }
        }
    }
    return arr;
};