leetcode - 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 <= 10^5
1 <= nums[i] <= nums[i + 1] <= 10^4

Solution

Since it’s absolute difference, the result[i] is:
$$R[i]=\underset{\text{numbers that are at the left of i}}{nums[i]-nums[j_1]+nums[i]-nums[j_2]+\dots }+\underset{\text{numbers that are at the right of i}}{nums[k_1]-nums[i]+nums[k_2]-nums[i]+\dots }$$

So we use 2 variables to store the sum, and go through the list once.

Time complexity: $o(n)$
Space complexity: $o(1)$

Code

class Solution:
    def getSumAbsoluteDifferences(self, nums: List[int]) -> List[int]:
        all_sum = sum(nums)
        pre_sum = 0
        res = []
        n = len(nums)
        for i, each_num in enumerate(nums):
            all_sum -= each_num
            res.append(all_sum - (n - i - 1) * each_num + i * each_num - pre_sum)
            pre_sum += each_num
        return res