Sum of Absolute Differences in a Sorted Array - LeetCode javascript solutions

Problem

You are given an integer array nums sorted in non-decreasing order.

Build and return **an integer array *result* with the same length as nums such that result[i] is equal to the summation of absolute differences between nums[i] and all the other elements in the array.**

In other words, result[i] is equal to sum(|nums[i]-nums[j]|) where 0 <= j < nums.length and j != i (0-indexed).

Example 1:

Input: nums = [2,3,5]
Output: [4,3,5]
Explanation: Assuming the arrays are 0-indexed, then
result[0] = |2-2| + |2-3| + |2-5| = 0 + 1 + 3 = 4,
result[1] = |3-2| + |3-3| + |3-5| = 1 + 0 + 2 = 3,
result[2] = |5-2| + |5-3| + |5-5| = 3 + 2 + 0 = 5.

Example 2:

Input: nums = [1,4,6,8,10]
Output: [24,15,13,15,21]

Constraints:

2 <= nums.length <= 105
1 <= nums[i] <= nums[i + 1] <= 104

Solution

/**
 * @param {number[]} nums
 * @return {number[]}
 */
var getSumAbsoluteDifferences = function(nums) {
    var diffLeft = Array(nums.length);
    for (var i = 0; i < nums.length; i++) {
        if (i === 0) {
            diffLeft[i] = 0;
        } else {
            diffLeft[i] = diffLeft[i - 1] + (nums[i] - nums[i - 1]) * i;
        }
    }
    var diffRight = Array(nums.length);
    for (var j = nums.length - 1; j >= 0; j--) {
        if (j === nums.length - 1) {
            diffRight[j] = 0;
        } else {
            diffRight[j] = diffRight[j + 1] + (nums[j + 1] - nums[j]) * (nums.length - 1 - j);
        }
    }
    var diff = Array(nums.length);
    for (var k = 0; k < nums.length; k++) {
        diff[k] = diffLeft[k] + diffRight[k];
    }
    return diff;
};

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).