What is the time complexity of Sum of Distances in Tree?

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

What data structures are used to solve Sum of Distances in Tree?

This problem uses Dynamic Programming, Tree, Depth-First Search, Graph. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Sum of Distances in Tree?

Insight 1: Directly calculating distances from each node using BFS or DFS for every node would be O(n^2), which is too slow for the given constraints (n <= 3 * 10^4). We need a more efficient approach. Insight 2: The problem can be solved using dynamic programming and two Depth-First Searches (DFS). The first DFS calculates the sum of distances from an arbitrary root node (e.g., node 0) and the number of nodes in each subtree. The second DFS leverages this information to calculate the sum of distances for all other nodes by propagating information down the tree.

Which companies ask LeetCode 834. Sum of Distances in Tree in interviews?

This problem is frequently asked at ByteDance, MathWorks, Media.net, PhonePe.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Sum of Distances in Tree - LeetCode 834 Solution

Q: How to solve Sum of Distances in Tree?

The solution uses a two-pass Depth-First Search (DFS) algorithm to efficiently calculate the sum of distances for each node. The first DFS calculates the sum of distances from an arbitrary root node (node 0) and the number of nodes in each subtree. The second DFS propagates this information to calculate the sum of distances for all other nodes. This approach avoids calculating distances from scratch for each node.

Q: Which companies ask LeetCode 834. Sum of Distances in Tree in interviews?

This problem is frequently asked at ByteDance, MathWorks, Media.net, PhonePe.

Sum of Distances in Tree - Complete Solution Guide

Sum of Distances in Tree is LeetCode problem 834, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

There is an undirected connected tree with n nodes labeled from 0 to n - 1 and n - 1 edges. You are given the integer n and the array edges where edges[i] = [a i , b i ] indicates that there is an edge between nodes a i and b i in the tree. Return an array answer of length n where answer[i] is the sum of the distances between the i th node in the tree and all other nodes. Example 1: Input: n = 6, edges = [[0,1],[0,2],[2,3],[2,4],[2,5]] Output: [8,12,6,10,10,10] Explanation: The tree is shown abo

Detailed Explanation

The problem asks us to calculate the sum of distances between each node in a given undirected tree and all other nodes. For each node `i`, we need to find the sum of the shortest path lengths from node `i` to every other node in the tree. The input consists of the number of nodes `n` and a list of edges representing the tree's connections. The output should be an array `answer` of length `n`, where `answer[i]` is the sum of distances from node `i` to all other nodes.

Solution Approach

The solution uses a two-pass Depth-First Search (DFS) algorithm to efficiently calculate the sum of distances for each node. The first DFS calculates the sum of distances from an arbitrary root node (node 0) and the number of nodes in each subtree. The second DFS propagates this information to calculate the sum of distances for all other nodes. This approach avoids calculating distances from scratch for each node.

Step-by-Step Algorithm

Step 1: Build the graph representation from the given edges. Use an adjacency list where each node stores a list of its neighbors.
Step 2: Initialize two arrays, `count` and `ans`, of size `n`. `count[i]` will store the number of nodes in the subtree rooted at node `i` (including node `i` itself). `ans[i]` will store the sum of distances from node `i` to all other nodes. Initialize `count[i]` to 1 for all nodes, and `ans[i]` to 0.
Step 3: Perform the first DFS (dfs1) starting from an arbitrary root (node 0). During this DFS, calculate the `count` and `ans` arrays. For each node `u`, recursively calculate the `count` and `ans` values for its children. Update `count[u]` by adding the `count` values of its children. Update `ans[u]` by adding the `ans` values of its children plus the `count` of each child (representing the distance increase for each node in the child's subtree).
Step 4: Perform the second DFS (dfs2) starting from the root node (node 0). During this DFS, propagate the sum of distances to all other nodes. For each node `u` and its child `v`, the sum of distances for `v` can be calculated as `ans[v] = ans[u] + n - 2 * count[v]`. This formula comes from the fact that when moving from `u` to `v`, the distances to nodes in `v`'s subtree decrease by 1 (count[v] nodes), and the distances to the other nodes increase by 1 (n - count[v] nodes).
Step 5: Return the `ans` array, which contains the sum of distances for each node.

Key Insights

Insight 1: Directly calculating distances from each node using BFS or DFS for every node would be O(n^2), which is too slow for the given constraints (n <= 3 * 10^4). We need a more efficient approach.
Insight 2: The problem can be solved using dynamic programming and two Depth-First Searches (DFS). The first DFS calculates the sum of distances from an arbitrary root node (e.g., node 0) and the number of nodes in each subtree. The second DFS leverages this information to calculate the sum of distances for all other nodes by propagating information down the tree.
Insight 3: The key idea is to realize that when moving from a parent node `u` to a child node `v`, the distances to all nodes in the subtree rooted at `v` decrease by 1, and the distances to all other nodes (n - count[v]) increase by 1. This allows us to efficiently update the sum of distances.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Dynamic Programming, Tree, Depth-First Search, Graph.

Companies

Asked at: ByteDance, MathWorks, Media.net, PhonePe.

Sum of Distances in Tree | LeetCode 834

Hard

Dynamic Programming Tree Depth-First Search Graph

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

Solution Code

Available in:

import java.util.ArrayList;
import java.util.List;

class Solution {
    public int[] sumOfDistancesInTree(int n, int[][] edges) {
        List<List<Integer>> graph = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            graph.add(new ArrayList<>());
        }
        for (int[] edge : edges) {
            int u = edge[0];
            int v = edge[1];
            graph.get(u).add(v);
            graph.get(v).add(u);
        }

        int[] count = new int[n];
        for (int i = 0; i < n; i++) {
            count[i] = 1;
        }
        int[] ans = new int[n];

        dfs1(0, -1, graph, count, ans);
        dfs2(0, -1, graph, count, ans, n);

        return ans;
    }

    private void dfs1(int u, int p, List<List<Integer>> graph, int[] count, int[] ans) {
        for (int v : graph.get(u)) {
            if (v == p) {
                continue;
            }
            dfs1(v, u, graph, count, ans);
            count[u] += count[v];
            ans[u] += ans[v] + count[v];
        }
    }

    private void dfs2(int u, int p, List<List<Integer>> graph, int[] count, int[] ans, int n) {
        for (int v : graph.get(u)) {
            if (v == p) {
                continue;
            }
            ans[v] = ans[u] + n - 2 * count[v];
            dfs2(v, u, graph, count, ans, n);
        }
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 834: Sum of Distances in Tree - Detailed explanation and solution approaches.

Topics

Dynamic Programming Tree Depth-First Search Graph

Companies

ByteDanceMathWorksMedia.netPhonePe

Stats

Difficulty:Hard

Languages:4

Companies:4

Problem #:834

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

5 steps

1
Step 1: Build the graph representation from the given edges. Use an adjacency list where each node stores a list of its neighbors.
2
Step 2: Initialize two arrays, `count` and `ans`, of size `n`. `count[i]` will store the number of nodes in the subtree rooted at node `i` (including node `i` itself). `ans[i]` will store the sum of distances from node `i` to all other nodes. Initialize `count[i]` to 1 for all nodes, and `ans[i]` to 0.
3
Step 3: Perform the first DFS (dfs1) starting from an arbitrary root (node 0). During this DFS, calculate the `count` and `ans` arrays. For each node `u`, recursively calculate the `count` and `ans` values for its children. Update `count[u]` by adding the `count` values of its children. Update `ans[u]` by adding the `ans` values of its children plus the `count` of each child (representing the distance increase for each node in the child's subtree).
4
Step 4: Perform the second DFS (dfs2) starting from the root node (node 0). During this DFS, propagate the sum of distances to all other nodes. For each node `u` and its child `v`, the sum of distances for `v` can be calculated as `ans[v] = ans[u] + n - 2 * count[v]`. This formula comes from the fact that when moving from `u` to `v`, the distances to nodes in `v`'s subtree decrease by 1 (count[v] nodes), and the distances to the other nodes increase by 1 (n - count[v] nodes).
5
Step 5: Return the `ans` array, which contains the sum of distances for each node.

💡

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 time complexity is dominated by the two DFS traversals, each visiting all nodes and edges once. The space complexity is dominated by the graph representation and the `count` and `ans` arrays, which store information for each node in the tree. Therefore, both time and space complexity are O(n).

📖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

For each node, perform a Breadth-First Search (BFS) or Depth-First Search (DFS) to calculate the distance to all other nodes. Sum these distances for each starting node.

Time: O(n^2)Space: O(n)

Trade-offs: The brute-force approach is simple to understand but has a time complexity of O(n^2), making it inefficient for large trees. The given solution is much faster (O(n)) but requires understanding of dynamic programming and tree traversals. Brute force might be suitable for small input sizes where the difference in execution time is negligible.

✨

Key Insights

1Insight 1: Directly calculating distances from each node using BFS or DFS for every node would be O(n^2), which is too slow for the given constraints (n <= 3 * 10^4). We need a more efficient approach.
2Insight 2: The problem can be solved using dynamic programming and two Depth-First Searches (DFS). The first DFS calculates the sum of distances from an arbitrary root node (e.g., node 0) and the number of nodes in each subtree. The second DFS leverages this information to calculate the sum of distances for all other nodes by propagating information down the tree.
3Insight 3: The key idea is to realize that when moving from a parent node `u` to a child node `v`, the distances to all nodes in the subtree rooted at `v` decrease by 1, and the distances to all other nodes (n - count[v]) increase by 1. This allows us to efficiently update the sum of distances.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Tree

Hierarchical structure with nodes. Like a family tree!

Graph

Nodes connected by edges. Like a social network or city map!

Edge Cases

Edge case 1: n = 1, edges = []. The solution correctly handles this case by returning [0] because the distance from the only node to itself is 0.
Edge case 2: A skewed tree (e.g., nodes connected in a linear chain). The algorithm works correctly for skewed trees as it performs DFS traversals to calculate the sum of distances.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to include the count of a child's subtree when calculating the distance contribution from a node to its parent in `dfs1`. The distances to all nodes within the child's subtree increase by 1 when moving from the child to the parent.

✗Common mistake 2: Incorrectly applying the formula `ans[v] = ans[u] + n - 2 * count[v]` in `dfs2`. It's crucial to understand that `count[v]` represents the number of nodes in the subtree rooted at `v`, and `n - count[v]` represents the number of nodes *not* in that subtree.

✅

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: Network Routing. Calculating the total cost for a message to be delivered from one server to all other servers in a network, where the network topology is a tree.

🏦Real-world scenario 2: Facility Location. Finding the optimal location for a facility (e.g., a hospital, warehouse) within a tree-structured distribution network to minimize the total distance to all customers.

💡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

Start with recursion, then memoize. The hardest part is identifying the state. Ask yourself: "What information do I need to make the current decision?"

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Forgetting base cases or initializing the DP array incorrectly. Start by solving the smallest subproblems.

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 23, 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

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

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

@DevExpert92Jul 6

❓

Frequently Asked Questions

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

How do I choose between recursive and iterative tree traversal?

Use recursive traversal for cleaner code when stack depth isn't a concern (tree height < ~1000). Use iterative with explicit stack for very deep trees to avoid stack overflow. Iterative is also preferred when you need fine-grained control over the traversal process.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

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

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

🤖 Need help with this problem?