What is the time complexity of Collect Coins in a Tree?

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

What data structures are used to solve Collect Coins in a Tree?

This problem uses Array, Tree, Graph, Topological Sort. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Collect Coins in a Tree?

Insight 1: The problem can be simplified by focusing on the 'core' subtree that directly or indirectly leads to collecting all coins. Nodes and edges not contributing to coin collection can be removed. Insight 2: Pruning leaves without coins simplifies the tree without affecting the overall solution since starting at these leaves is not optimal.

Which companies ask LeetCode 2603. Collect Coins in a Tree in interviews?

This problem is frequently asked at Graviton, Lucid.

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Collect Coins in a Tree - LeetCode 2603 Solution

Collect Coins in a Tree - Complete Solution Guide

Collect Coins in a Tree is LeetCode problem 2603, 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 exists an undirected and unrooted tree with n nodes indexed from 0 to n - 1 . You are given an integer n and a 2D integer array edges of length n - 1 , where edges[i] = [a i , b i ] indicates that there is an edge between nodes a i and b i in the tree. You are also given an array coins of size n where coins[i] can be either 0 or 1 , where 1 indicates the presence of a coin in the vertex i . Initially, you choose to start at any vertex in the tree. Then, you can perform the following operat

Detailed Explanation

The problem presents a tree structure where each node may or may not have a coin. The objective is to find the minimum number of edges that need to be traversed to collect all coins and return to the starting node. You can collect all coins within a distance of 2 from your current position, and move to any adjacent node. The problem emphasizes that edges can be traversed multiple times, and each traversal must be counted.

Solution Approach

The solution employs a pruning strategy to reduce the tree to its essential components. First, it identifies and removes all leaf nodes that do not contain coins. This is done iteratively using a queue. Next, it removes the next two layers of leaves. Each of these prunes is equivalent to being able to collect all coins within a distance of two. After pruning, the solution calculates the number of edges in the remaining tree and returns twice this value. The algorithm leverages the concept of graph traversal using queues and utilizes degree information for efficient leaf node identification.

Step-by-Step Algorithm

Step 1: Build the adjacency list representation of the tree and calculate the degree (number of neighbors) for each node.
Step 2: Identify and remove leaf nodes that do not contain coins. Add nodes with degree 1 and no coins to a queue, then iteratively remove these nodes from the graph and update the degrees of their neighbors. Repeat as long as there are nodes to remove.
Step 3: Remove the first layer of leaves from the remaining tree. Add all nodes with degree 1 to a queue, then remove them.
Step 4: Remove the second layer of leaves from the remaining tree. Remove leaves in the same way as in Step 3, as we can assume we collected the coins in a radius of 2.
Step 5: Calculate the number of edges in the remaining tree, which is equal to the number of remaining nodes minus 1. Multiply this value by 2 to get the total number of edges needed to traverse the tree and return to the starting node.
Step 6: Handle the edge case where there are 1 or fewer remaining nodes by returning 0.

Key Insights

Insight 1: The problem can be simplified by focusing on the 'core' subtree that directly or indirectly leads to collecting all coins. Nodes and edges not contributing to coin collection can be removed.
Insight 2: Pruning leaves without coins simplifies the tree without affecting the overall solution since starting at these leaves is not optimal.
Insight 3: After pruning non-coin leaves, pruning the next two layers of leaves will produce the minimum subtree. The distance constraint of '2' allows for this pruning to occur in two iterations.
Insight 4: The minimum number of edges needed to traverse a tree and return to the starting node is twice the number of edges in the core subtree (2 * (number of nodes - 1)).

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Tree, Graph, Topological Sort.

Companies

Asked at: Graviton, Lucid.

Collect Coins in a Tree | LeetCode 2603

Hard

Array Tree Graph Topological Sort

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

Solution Code

Available in:

import java.util.*;

class Solution {
    public int collectTheCoins(int[] coins, int[][] edges) {
        int n = coins.length;
        if (n <= 2) {
            return 0;
        }

        List<List<Integer>> adj = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            adj.add(new ArrayList<>());
        }

        int[] degree = new int[n];
        for (int[] edge : edges) {
            int u = edge[0];
            int v = edge[1];
            adj.get(u).add(v);
            adj.get(v).add(u);
            degree[u]++;
            degree[v]++;
        }

        int remaining_nodes = n;

        // Step 1: Prune leaves with no coins.
        // These nodes are irrelevant as they have no coins and are at the periphery.
        Queue<Integer> q = new LinkedList<>();
        for (int i = 0; i < n; i++) {
            if (degree[i] == 1 && coins[i] == 0) {
                q.offer(i);
            }
        }

        while (!q.isEmpty()) {
            int u = q.poll();
            remaining_nodes--;
            for (int v : adj.get(u)) {
                degree[v]--;
                if (degree[v] == 1 && coins[v] == 0) {
                    q.offer(v);
                }
            }
        }

        // After this, we have the minimal subtree connecting all coin nodes.
        // All leaves of this subtree must have coins.

        // Step 2: Prune the first layer of leaves from the remaining tree.
        // To collect a coin at a leaf, we can be at its parent (distance 1).
        // So we don't need to travel to the leaf itself.
        Queue<Integer> leaves = new LinkedList<>();
        for (int i = 0; i < n; i++) {
            if (degree[i] == 1 && remaining_nodes > 2) {
                leaves.offer(i);
            }
        }

        int leavesSize = leaves.size();
        for (int i = 0; i < leavesSize; i++) {
            int u = leaves.poll();
            remaining_nodes--;
            for (int v : adj.get(u)) {
                degree[v]--;
                if (degree[v] == 1 && remaining_nodes > 2) {
                    leaves.offer(v);
                }
            }
        }

        // Step 3: Prune the second layer of leaves.
        // After pruning the first layer, new leaves are formed. To collect coins
        // at these new leaves, we can be at their parents (distance 1), which
        // is distance 2 from the original leaves.
        leavesSize = leaves.size();
        for (int i = 0; i < leavesSize; i++) {
            int u = leaves.poll();
            remaining_nodes--;
            // No need to update degrees further, as we only need the count of remaining nodes.
        }

        if (remaining_nodes <= 1) {
            return 0;
        } else {
            // The remaining nodes form the core subtree we must traverse.
            // A traversal that visits every edge and returns to start takes 2 * |E| steps.
            // For a tree with k nodes, there are k-1 edges.
            return (remaining_nodes - 1) * 2;
        }
    }
}

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Problem Info

Description

Problem 2603: Collect Coins in a Tree - Detailed explanation and solution approaches.

Topics

Array Tree Graph Topological Sort

Companies

GravitonLucid

Stats

Difficulty:Hard

Languages:4

Companies:2

Problem #:2603

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

6 steps

1
Step 1: Build the adjacency list representation of the tree and calculate the degree (number of neighbors) for each node.
2
Step 2: Identify and remove leaf nodes that do not contain coins. Add nodes with degree 1 and no coins to a queue, then iteratively remove these nodes from the graph and update the degrees of their neighbors. Repeat as long as there are nodes to remove.
3
Step 3: Remove the first layer of leaves from the remaining tree. Add all nodes with degree 1 to a queue, then remove them.
4
Step 4: Remove the second layer of leaves from the remaining tree. Remove leaves in the same way as in Step 3, as we can assume we collected the coins in a radius of 2.
5
Step 5: Calculate the number of edges in the remaining tree, which is equal to the number of remaining nodes minus 1. Multiply this value by 2 to get the total number of edges needed to traverse the tree and return to the starting node.
6
Step 6: Handle the edge case where there are 1 or fewer remaining nodes by returning 0.

💡

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 determined by the graph traversal and pruning steps, which depend on the number of nodes and edges. Since the number of edges is n-1, the time complexity is dominated by O(n). The space complexity is determined by the size of the adjacency list and the queues used for graph traversal, all of which scale linearly with the number of nodes.

📖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

Dynamic Programming

Could potentially use dynamic programming to calculate minimum cost from each node, however it's unlikely to improve the time complexity.

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

Trade-offs: Likely not to be as efficient or simple as the current approach

✨

Key Insights

1Insight 1: The problem can be simplified by focusing on the 'core' subtree that directly or indirectly leads to collecting all coins. Nodes and edges not contributing to coin collection can be removed.
2Insight 2: Pruning leaves without coins simplifies the tree without affecting the overall solution since starting at these leaves is not optimal.
3Insight 3: After pruning non-coin leaves, pruning the next two layers of leaves will produce the minimum subtree. The distance constraint of '2' allows for this pruning to occur in two iterations.
4Insight 4: The minimum number of edges needed to traverse a tree and return to the starting node is twice the number of edges in the core subtree (2 * (number of nodes - 1)).

📚

Key Terms Explained

For Students

Array

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

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: The input tree has only one node. The algorithm correctly handles this case because the loops in the pruning steps will not execute and the returned value will be 0.
Edge case 2: The input tree has only coin-less nodes. The initial pruning will remove all nodes, resulting in the algorithm returning 0.
Edge case 3: The input tree has two or fewer nodes. The algorithm explicitly returns 0 in this case.
Edge case 4: A single node with a coin: the pruning steps do not execute, so the remaining number of nodes is 1, and the algorithm returns 0.
Edge case 5: Star graph with only one coin in the center node: Pruning all leaves leads to remaining node = 1, therefore returning 0

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering that edges can be traversed multiple times. This leads to trying to find the shortest path instead of pruning.

✗Common mistake 2: Failing to prune both layers of leaves. Collecting coins at distance 2 requires two steps of leaf pruning.

✗Common mistake 3: Incorrectly calculating the number of edges needed to traverse the final core subtree. The correct calculation is 2 * (number of nodes - 1).

✗Common mistake 4: Not handling the edge case where the initial tree contains no coins, which simplifies the problem.

✅

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 optimization where the goal is to minimize the cost of traversing a network to collect data from specific nodes. This can apply to sensor networks or delivery networks.

🏦Real-world scenario 2: Resource allocation in a hierarchical system where resources are placed at certain locations and the goal is to minimize the travel distance required to access all resources.

💡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

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

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

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: August 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

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

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

@TechLeadJul 6

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

@DevExpert92Jun 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.

What is the difference between DFS and BFS for trees?

DFS (using recursion or stack) explores depth-first, useful for path-finding and backtracking problems. BFS (using queue) explores level-by-level, ideal for finding shortest paths or processing nodes by depth. Choose based on whether level information matters.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

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 "Collect Coins in a Tree"?

For this hard-level Array 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?