What is the time complexity of Sum of Distances?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Sum of Distances?

This problem uses Array, Hash Table, Prefix Sum. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Sum of Distances?

Insight 1: Using a hash map (or equivalent) to store the indices for each distinct number in the array allows for efficient lookup. Insight 2: Calculating the sum directly for each element would result in O(n^2) time complexity. A better approach is to use prefix sums to optimize the distance calculation.

Which companies ask LeetCode 2615. Sum of Distances in interviews?

This problem is frequently asked at BNY Mellon.

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Sum of Distances - LeetCode 2615 Solution

Sum of Distances - Complete Solution Guide

Sum of Distances is LeetCode problem 2615, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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

Detailed Explanation

The problem requires calculating the sum of absolute distances between each index `i` in an array `nums` and all other indices `j` that hold the same value as `nums[i]`. The result is an array `arr` where `arr[i]` is this calculated sum. If no other index has the same value as `nums[i]`, then `arr[i]` is 0. Essentially, for each number in the input array, we need to find all other positions where the same number appears and compute the sum of the absolute differences between their indices.

Solution Approach

The provided solution utilizes a two-pass approach combined with hash maps to achieve an O(n) time complexity. The algorithm first iterates through the array from left to right, and then from right to left. In each pass, it maintains a count and a prefix sum for each distinct number encountered. The prefix sum stores the sum of indices where the number has appeared so far. The count stores how many times the number has appeared so far. These values allow for calculating the sum of distances to the 'left' and to the 'right' of an index efficiently, and then these two partial sums are added together to compute the final result.

Step-by-Step Algorithm

Step 1: Initialize an array `ans` of the same size as `nums` to store the results, filled with zeros.
Step 2: Create two hash maps (or equivalent): `counts` to store the number of occurrences of each number encountered so far, and `prefix_sums` to store the sum of the indices where each number has appeared so far.
Step 3: Iterate through the `nums` array from left to right (first pass). For each element `nums[i]`, calculate the sum of distances to the elements on the left with same value using the formula: `counts[nums[i]] * i - prefix_sums[nums[i]]`. Add this value to `ans[i]`.
Step 4: After computing the left distances, update the `counts` and `prefix_sums` hash maps for the current element `nums[i]` by incrementing the count and adding the index `i` to the prefix sum.
Step 5: Clear the `counts` and `prefix_sums` hash maps to prepare for the second pass.
Step 6: Iterate through the `nums` array from right to left (second pass). For each element `nums[i]`, calculate the sum of distances to the elements on the right with the same value using the formula: `prefix_sums[nums[i]] - counts[nums[i]] * i`. Add this value to `ans[i]`.
Step 7: After computing the right distances, update the `counts` and `prefix_sums` hash maps for the current element `nums[i]` by incrementing the count and adding the index `i` to the prefix sum.
Step 8: Return the `ans` array.

Key Insights

Insight 1: Using a hash map (or equivalent) to store the indices for each distinct number in the array allows for efficient lookup.
Insight 2: Calculating the sum directly for each element would result in O(n^2) time complexity. A better approach is to use prefix sums to optimize the distance calculation.
Insight 3: Two passes are required: one from left to right and another from right to left, accumulating the distances incrementally to avoid redundant calculations.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, Prefix Sum.

Companies

Asked at: BNY Mellon.

Sum of Distances | LeetCode 2615

Medium

Array Hash Table Prefix Sum

Time: O(n)Space: O(n)

Solution Code

Available in:

import java.util.HashMap;
import java.util.Map;

class Solution {
    public long[] distance(int[] nums) {
        int n = nums.length;
        long[] ans = new long[n];

        // Left-to-right pass
        Map<Integer, Integer> counts = new HashMap<>();
        Map<Integer, Long> prefixSums = new HashMap<>();
        for (int i = 0; i < n; i++) {
            int num = nums[i];
            long currentAns = ans[i];
            currentAns += (long) counts.getOrDefault(num, 0) * i - prefixSums.getOrDefault(num, 0L);
            ans[i] = currentAns;
            counts.put(num, counts.getOrDefault(num, 0) + 1);
            prefixSums.put(num, prefixSums.getOrDefault(num, 0L) + i);
        }

        // Right-to-left pass
        counts.clear();
        prefixSums.clear();
        for (int i = n - 1; i >= 0; i--) {
            int num = nums[i];
            long currentAns = ans[i];
            currentAns += prefixSums.getOrDefault(num, 0L) - (long) counts.getOrDefault(num, 0) * i;
            ans[i] = currentAns;
            counts.put(num, counts.getOrDefault(num, 0) + 1);
            prefixSums.put(num, prefixSums.getOrDefault(num, 0L) + i);
        }

        return ans;
    }
}

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Problem Info

Description

Problem 2615: Sum of Distances - Detailed explanation and solution approaches.

Topics

Array Hash Table Prefix Sum

Companies

BNY Mellon

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2615

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Initialize an array `ans` of the same size as `nums` to store the results, filled with zeros.
2
Step 2: Create two hash maps (or equivalent): `counts` to store the number of occurrences of each number encountered so far, and `prefix_sums` to store the sum of the indices where each number has appeared so far.
3
Step 3: Iterate through the `nums` array from left to right (first pass). For each element `nums[i]`, calculate the sum of distances to the elements on the left with same value using the formula: `counts[nums[i]] * i - prefix_sums[nums[i]]`. Add this value to `ans[i]`.
4
Step 4: After computing the left distances, update the `counts` and `prefix_sums` hash maps for the current element `nums[i]` by incrementing the count and adding the index `i` to the prefix sum.
5
Step 5: Clear the `counts` and `prefix_sums` hash maps to prepare for the second pass.
6
Step 6: Iterate through the `nums` array from right to left (second pass). For each element `nums[i]`, calculate the sum of distances to the elements on the right with the same value using the formula: `prefix_sums[nums[i]] - counts[nums[i]] * i`. Add this value to `ans[i]`.
7
Step 7: After computing the right distances, update the `counts` and `prefix_sums` hash maps for the current element `nums[i]` by incrementing the count and adding the index `i` to the prefix sum.
8
Step 8: Return the `ans` array.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The dominant operations are the iterations through the input array (two passes), and the operations inside the loops, which are constant time hash map operations (insertions, lookups, and updates). Therefore, the time complexity is O(n). The space complexity arises from storing the counts and prefix sums in hash maps, which in the worst case can store all the distinct numbers in the array, leading to O(n) space complexity. The output array also contributes O(n) to the space complexity.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute-Force Approach

For each index `i`, iterate through the entire `nums` array to find all other indices `j` where `nums[i] == nums[j]`. Calculate the sum of `|i - j|` for all such `j`. This approach doesn't use prefix sums.

Time: O(n^2), where n is the length of the input array `nums`.Space: O(1), excluding the output array.

Trade-offs: The brute-force approach is simpler to understand but has a quadratic time complexity, making it inefficient for large input arrays. The main solution is more efficient with a linear time complexity, but it requires more space for the hash maps. Use brute-force for small arrays where simplicity is preferred; use the optimized solution for larger arrays where performance is crucial.

✨

Key Insights

1Insight 1: Using a hash map (or equivalent) to store the indices for each distinct number in the array allows for efficient lookup.
2Insight 2: Calculating the sum directly for each element would result in O(n^2) time complexity. A better approach is to use prefix sums to optimize the distance calculation.
3Insight 3: Two passes are required: one from left to right and another from right to left, accumulating the distances incrementally to avoid redundant calculations.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Edge Cases

Edge case 1: When all elements in `nums` are distinct, the prefix sum and count will always be zero when calculating the sum of distances for each index. The solution correctly handles this because the initial value of `ans[i]` is 0 and the accumulated distance will be 0 + 0 = 0. The hashmaps return 0 if the key is not present.
Edge case 2: When all elements in `nums` are the same, the prefix sum and count are correctly maintained across all iterations. The solution will calculate the correct sum of distances in this scenario as well.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to clear the hash maps between the left-to-right and right-to-left passes. If the hash maps are not cleared, the counts and prefix sums from the first pass will incorrectly influence the calculations in the second pass, leading to wrong results.

✗Common mistake 2: Not using long long data type for the `prefix_sums` hash map. The sum of indices can become large, potentially exceeding the maximum value of an integer. Using long long prevents integer overflow and ensures accurate calculations. The Java and C++ solutions correctly used `long` and `long long` respectively.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Analyzing customer behavior in e-commerce. Suppose `nums` represents the purchase amounts of different customers, and `i` represents the customer ID. The algorithm can be used to find how similar each customer's purchase amount is to other customers. This can be useful in targeted marketing and personalized recommendations.

🏦Real-world scenario 2: Network analysis. Consider `nums` as representing the communities that different users belong to, and `i` represents user ID. This algorithm can compute the aggregate distance between a user and all other users in the same community, to highlight central members or detect outliers.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 4, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Sum of Distances"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?