LeetCode #2615 — MEDIUM

Sum of Distances

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

Solve on LeetCode
The Problem

Problem Statement

You are given a 0-indexed integer array nums. There exists an array arr of length nums.length, where arr[i] is the sum of |i - j| over all j such that nums[j] == nums[i] and j != i. If there is no such j, set arr[i] to be 0.

Return the array arr.

Example 1:

Input: nums = [1,3,1,1,2]
Output: [5,0,3,4,0]
Explanation: 
When i = 0, nums[0] == nums[2] and nums[0] == nums[3]. Therefore, arr[0] = |0 - 2| + |0 - 3| = 5. 
When i = 1, arr[1] = 0 because there is no other index with value 3.
When i = 2, nums[2] == nums[0] and nums[2] == nums[3]. Therefore, arr[2] = |2 - 0| + |2 - 3| = 3. 
When i = 3, nums[3] == nums[0] and nums[3] == nums[2]. Therefore, arr[3] = |3 - 0| + |3 - 2| = 4. 
When i = 4, arr[4] = 0 because there is no other index with value 2.

Example 2:

Input: nums = [0,5,3]
Output: [0,0,0]
Explanation: Since each element in nums is distinct, arr[i] = 0 for all i.

Constraints:

  • 1 <= nums.length <= 105
  • 0 <= nums[i] <= 109

Note: This question is the same as 2121: Intervals Between Identical Elements.

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed integer array nums. There exists an array arr of length nums.length, where arr[i] is the sum of |i - j| over all j such that nums[j] == nums[i] and j != i. If there is no such j, set arr[i] to be 0. Return the array arr.

Baseline thinking

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

Pattern signal: Array · Hash Map

Example 1

[1,3,1,1,2]

Example 2

[0,5,3]

Related Problems

  • Remove Duplicates from Sorted Array (remove-duplicates-from-sorted-array)
  • Find All Duplicates in an Array (find-all-duplicates-in-an-array)
  • Minimum Operations to Make All Array Elements Equal (minimum-operations-to-make-all-array-elements-equal)
Step 02

Core Insight

What unlocks the optimal approach

  • Can we use the prefix sum here?
  • For each number x, collect all the indices where x occurs, and calculate the prefix sum of the array.
  • For each occurrence of x, the indices to the right will be regular subtraction while the indices to the left will be reversed subtraction.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. 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 #2615: Sum of Distances
class Solution {
    public long[] distance(int[] nums) {
        int n = nums.length;
        long[] ans = new long[n];
        Map<Integer, List<Integer>> d = new HashMap<>();
        for (int i = 0; i < n; ++i) {
            d.computeIfAbsent(nums[i], k -> new ArrayList<>()).add(i);
        }
        for (var idx : d.values()) {
            int m = idx.size();
            long left = 0;
            long right = -1L * m * idx.get(0);
            for (int i : idx) {
                right += i;
            }
            for (int i = 0; i < m; ++i) {
                ans[idx.get(i)] = left + right;
                if (i + 1 < m) {
                    left += (idx.get(i + 1) - idx.get(i)) * (i + 1L);
                    right -= (idx.get(i + 1) - idx.get(i)) * (m - i - 1L);
                }
            }
        }
        return ans;
    }
}
Step 06

Interactive Study Demo

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

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Related Problems
#1Two Sumeasy→#36Valid Sudokumedium→#41First Missing Positivehard→#73Set Matrix Zeroesmedium→#105Construct Binary Tree from Preorder and Inorder Traversalmedium→