1685. Sum of Absolute Differences in a Sorted Array

Description

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

Solution: Prefix Sum

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

 

JavaScript

/**
 * @param {number[]} nums
 * @return {number[]}
 */
const getSumAbsoluteDifferences = function (nums) {
  const prefixSum = [nums[0]];
  const size = nums.length;

  for (let index = 1; index < size; index++) {
    prefixSum[index] = prefixSum[index - 1] + nums[index];
  }
  const sum = prefixSum.at(-1);

  return nums.map((num, index) => {
    const currentSum = prefixSum[index];
    const left = num * (index + 1) - currentSum;
    const right = sum - currentSum - num * (size - index - 1);

    return left + right;
  });
};