Sum of Absolute Differences in a Sorted Array

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:


Solution:

class Solution {
    public int[] getSumAbsoluteDifferences(int[] nums) {
        int n = nums.length;
        int[] prefixSum = new int[n + 1];
        for (int i = 1; i <= n; i ++) {
            prefixSum[i] = prefixSum[i - 1] + nums[i - 1];
        }
        int[] res = new int[n];
        for (int i = 0; i < n; i ++) {
            int left = prefixSum[i];
            int right = prefixSum[n] - prefixSum[i + 1];
            // System.out.println(left + ", " + right);
            res[i] = (nums[i] * i - left) + (right - nums[i] * (n - 1 - i));
        }
        return res;
    }
}