LeetCode 1685. Sum of Absolute Differences in a Sorted Array Solution in Java, C++, Python & More | Explanation + Code

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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 1: Summation + Enumeration

First, we calculate the sum of all elements in the array nums, denoted as s. We use a variable t to record the sum of the elements that have been enumerated so far.

Next, we enumerate nums[i]. Then ans[i] = nums[i] × i - t + s - t - nums[i] × (n - i). After that, we update t, i.e., t = t + nums[i]. We continue to enumerate the next element until all elements are enumerated.

The time complexity is O(n), where n is the length of the array nums. The space complexity is O(1).

PythonJavaC++GoTypeScriptJavaScriptC#
class Solution: def getSumAbsoluteDifferences(self, nums: List[int]) -> List[int]: ans = [] s, t = sum(nums), 0 for i, x in enumerate(nums): v = x * i - t + s - t - x * (len(nums) - i) ans.append(v) t += x return ans(code-box)

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