LeetCode #1685 — MEDIUM

Sum of Absolute Differences in a Sorted Array

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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).

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math

Example 1

[2,3,5]

Example 2

[1,4,6,8,10]

Step 02

Core Insight

What unlocks the optimal approach

Absolute difference is the same as max(a, b) - min(a, b). How can you use this fact with the fact that the array is sorted?
For nums[i], the answer is (nums[i] - nums[0]) + (nums[i] - nums[1]) + ... + (nums[i] - nums[i-1]) + (nums[i+1] - nums[i]) + (nums[i+2] - nums[i]) + ... + (nums[n-1] - nums[i]).
It can be simplified to (nums[i] * i - (nums[0] + nums[1] + ... + nums[i-1])) + ((nums[i+1] + nums[i+2] + ... + nums[n-1]) - nums[i] * (n-i-1)). One can build prefix and suffix sums to compute this quickly.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1685: Sum of Absolute Differences in a Sorted Array
class Solution {
    public int[] getSumAbsoluteDifferences(int[] nums) {
        // int s = Arrays.stream(nums).sum();
        int s = 0, t = 0;
        for (int x : nums) {
            s += x;
        }
        int n = nums.length;
        int[] ans = new int[n];
        for (int i = 0; i < n; ++i) {
            int v = nums[i] * i - t + s - t - nums[i] * (n - i);
            ans[i] = v;
            t += nums[i];
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1685: Sum of Absolute Differences in a Sorted Array
func getSumAbsoluteDifferences(nums []int) (ans []int) {
	var s, t int
	for _, x := range nums {
		s += x
	}
	for i, x := range nums {
		v := x*i - t + s - t - x*(len(nums)-i)
		ans = append(ans, v)
		t += x
	}
	return
}

# Accepted solution for LeetCode #1685: Sum of Absolute Differences in a Sorted Array
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

// Accepted solution for LeetCode #1685: Sum of Absolute Differences in a Sorted Array
struct Solution;

impl Solution {
    fn get_sum_absolute_differences(nums: Vec<i32>) -> Vec<i32> {
        let n = nums.len();
        let mut prev = nums[0];
        let mut prev_sum = 0;
        let mut res = vec![0; n];
        for i in 0..n {
            let diff = nums[i] - prev;
            prev_sum += diff * i as i32;
            res[i] += prev_sum;
            prev = nums[i];
        }
        prev = nums[n - 1];
        prev_sum = 0;
        for i in (0..n).rev() {
            let diff = (nums[i] - prev).abs();
            prev_sum += diff * (n - 1 - i) as i32;
            res[i] += prev_sum;
            prev = nums[i];
        }
        res
    }
}

#[test]
fn test() {
    let nums = vec![2, 3, 5];
    let res = vec![4, 3, 5];
    assert_eq!(Solution::get_sum_absolute_differences(nums), res);
    let nums = vec![1, 4, 6, 8, 10];
    let res = vec![24, 15, 13, 15, 21];
    assert_eq!(Solution::get_sum_absolute_differences(nums), res);
}

// Accepted solution for LeetCode #1685: Sum of Absolute Differences in a Sorted Array
function getSumAbsoluteDifferences(nums: number[]): number[] {
    const s = nums.reduce((a, b) => a + b);
    let t = 0;
    const n = nums.length;
    const ans = new Array(n);
    for (let i = 0; i < n; ++i) {
        const v = nums[i] * i - t + s - t - nums[i] * (n - i);
        ans[i] = v;
        t += nums[i];
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.