1685

1685. Sum of Absolute Differences in a Sorted Array (Medium)

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

Related Topics:
Array, Math, Prefix Sum

Solution 1.

For each A[i], the corresponding ans[i] is

SUM( A[j] | j >= i ) - COUNT( j >= i )
+ SUM( A[j] | j < i ) - COUNT( j < i )

// OJ: https://leetcode.com/problems/sum-of-absolute-differences-in-a-sorted-array/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1) extra space
class Solution {
public:
    vector<int> getSumAbsoluteDifferences(vector<int>& A) {
        long total = accumulate(begin(A), end(A), 0L), right = total, N = A.size();
        vector<int> ans(N);
        for (int i = 0; i < N; ++i) {
            ans[i] = right - (N - i) * A[i] + i * A[i] - (total - right);
            right -= A[i];
        }
        return ans;
    }
};