1685. 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:

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

Solutions

1. prefix and suffix O(n)

class Solution {
public:
    vector<int> getSumAbsoluteDifferences(vector<int>& nums) {
        if (nums.size() < 2) return vector<int>(nums.size());
        vector<int> res(nums.size());

        int num = 1, sumd = 0;
        for (int i = 1; i < nums.size(); i++) {
            sumd += num++ * (nums[i] - nums[i - 1]);
            res[i] += sumd;
        }
        num = 1; sumd = 0;
        for (int i = (int)nums.size() - 2; i >= 0; i--) {
            sumd += num++ * (nums[i + 1] - nums[i]);
            res[i] += sumd;
        }

        return res;
    }
};