1685. Sum of Absolute Differences in a Sorted Array

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 <= 10^5

• 1 <= nums[i] <= nums[i + 1] <= 10^4

Solution (Java)

class Solution {
    public int[] getSumAbsoluteDifferences(int[] nums) {
        int len = nums.length;
        int[] preSums = new int[len];
        for (int i = 1; i < len; i++) {
            preSums[i] = preSums[i - 1] + nums[i - 1];
        }
        int[] postSums = new int[len];
        for (int i = len - 2; i >= 0; i--) {
            postSums[i] = postSums[i + 1] + nums[i + 1];
        }
        int[] result = new int[len];
        for (int i = 0; i < len; i++) {
            result[i] = nums[i] * i - preSums[i] + postSums[i] - nums[i] * (len - i - 1);
        }
        return result;
    }
}

Explain:

nope.

Complexity:

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