What is the time complexity of Most Profitable Path in a Tree?

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

What data structures are used to solve Most Profitable Path in a Tree?

This problem uses Array, Tree, Depth-First Search, Breadth-First Search, Graph. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Most Profitable Path in a Tree?

Insight 1: Represent the tree as an adjacency list to efficiently traverse its structure. Insight 2: Calculate the distance of each node from the root (node 0) using Breadth-First Search (BFS). This represents Alice's arrival time at each node.

Which companies ask LeetCode 2467. Most Profitable Path in a Tree in interviews?

This problem is frequently asked at Intuit, Snowflake.

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Most Profitable Path in a Tree - LeetCode 2467 Solution

Most Profitable Path in a Tree - Complete Solution Guide

Most Profitable Path in a Tree is LeetCode problem 2467, 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

There is an undirected tree with n nodes labeled from 0 to n - 1 , rooted at node 0 . You are given 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. At every node i , there is a gate. You are also given an array of even integers amount , where amount[i] represents: the price needed to open the gate at node i , if amount[i] is negative, or, the cash reward obtained on opening the gate at node i , otherwis

Detailed Explanation

The problem presents an undirected tree where nodes represent locations with associated prices (negative) or rewards (positive). Alice starts at node 0 and wants to maximize her income by moving to a leaf node. Bob starts at a specified node `bob` and moves towards node 0. Both Alice and Bob move simultaneously each second. If they arrive at a node at the same time, they split the price/reward. If Bob arrives before Alice, Alice gets nothing at that node (Bob has already opened the gate). If Alice arrives before Bob, she gets the full price/reward. The goal is to determine the maximum possible net income Alice can achieve by choosing the optimal path to a leaf node.

Solution Approach

The solution involves three main steps: building the adjacency list representation of the tree, using BFS to calculate the distance of each node from the root (Alice's time), and then determining Bob's path from `bob` to 0 and adjusting the `amount` array based on who reaches the node first. Finally, DFS is used to traverse the tree from the root to all leaf nodes, calculating Alice's profit along each path, and identifying the maximum profit she can make.

Step-by-Step Algorithm

Step 1: **Build Adjacency List:** Construct an adjacency list from the input `edges` to represent the tree structure. This allows easy access to neighboring nodes.
Step 2: **BFS for Alice's Time and Parent Pointers:** Perform a BFS from node 0. Calculate `dist_from_0[i]` for each node `i`, which stores the distance from node 0. This represents the time Alice takes to reach node `i`. Also, construct the `parent` array which stores the parent node of each node. This is crucial for reconstructing Bob's path.
Step 3: **Determine Bob's Path and Adjust Amounts:** Trace Bob's path back to node 0 using the `parent` array. For each node on Bob's path, compare his arrival time (`bob_time`) with Alice's arrival time (`alice_time`). If `bob_time < alice_time`, Alice receives nothing (amount[curr] = 0). If `bob_time == alice_time`, the reward/price is divided (amount[curr] /= 2).
Step 4: **DFS for Maximum Profit:** Perform a DFS starting from node 0 to explore all paths to leaf nodes. At each node, add the adjusted `amount[u]` to the current profit. When a leaf node is reached, update `max_profit` if the current profit is greater.

Key Insights

Insight 1: Represent the tree as an adjacency list to efficiently traverse its structure.
Insight 2: Calculate the distance of each node from the root (node 0) using Breadth-First Search (BFS). This represents Alice's arrival time at each node.
Insight 3: Determine the path Bob takes to reach the root and the time he takes to reach each node on his path. Compare Bob's arrival time with Alice's arrival time to determine how the price/reward at each node should be divided or assigned.
Insight 4: Use Depth-First Search (DFS) to explore all possible paths from the root to leaf nodes and compute Alice's net income for each path. Keep track of the maximum income found so far.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Tree, Depth-First Search, Breadth-First Search, Graph.

Companies

Asked at: Intuit, Snowflake.

Most Profitable Path in a Tree | LeetCode 2467

Medium

Array Tree Depth-First Search Breadth-First Search Graph

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

Solution Code

Available in:

import java.util.*;

class Solution {
    private int max_profit;

    public int mostProfitablePath(int[][] edges, int bob, int[] amount) {
        int n = amount.length;
        List<List<Integer>> adj = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            adj.add(new ArrayList<>());
        }

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

        int[] dist_from_0 = new int[n];
        Arrays.fill(dist_from_0, -1);
        int[] parent = new int[n];
        Arrays.fill(parent, -1);

        Queue<Integer> q = new LinkedList<>();
        q.offer(0);
        dist_from_0[0] = 0;

        // BFS from root 0 to find distances (Alice's time) and parent pointers
        while (!q.isEmpty()) {
            int u = q.poll();
            for (int v : adj.get(u)) {
                if (v != parent[u]) {
                    parent[v] = u;
                    dist_from_0[v] = dist_from_0[u] + 1;
                    q.offer(v);
                }
            }
        }

        // Modify amounts based on Bob's path from `bob` to `0`
        int curr = bob;
        int bob_time = 0;
        while (curr != 0) {
            int alice_time = dist_from_0[curr];
            if (bob_time < alice_time) {
                amount[curr] = 0;
            } else if (bob_time == alice_time) {
                amount[curr] /= 2;
            }

            curr = parent[curr];
            bob_time += 1;
        }

        this.max_profit = Integer.MIN_VALUE;

        // DFS from root 0 to find the maximum profit path to a leaf
        dfs(0, -1, 0, adj, amount);

        return this.max_profit;
    }

    private void dfs(int u, int p, int current_profit, List<List<Integer>> adj, int[] amount) {
        current_profit += amount[u];

        boolean is_leaf = true;
        for (int v : adj.get(u)) {
            if (v != p) {
                is_leaf = false;
                dfs(v, u, current_profit, adj, amount);
            }
        }

        if (is_leaf) {
            this.max_profit = Math.max(this.max_profit, current_profit);
        }
    }
}

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Problem Info

Description

Problem 2467: Most Profitable Path in a Tree - Detailed explanation and solution approaches.

Topics

Array Tree Depth-First Search Breadth-First Search Graph

Companies

IntuitSnowflake

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:2467

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

4 steps

1
Step 1: **Build Adjacency List:** Construct an adjacency list from the input `edges` to represent the tree structure. This allows easy access to neighboring nodes.
2
Step 2: **BFS for Alice's Time and Parent Pointers:** Perform a BFS from node 0. Calculate `dist_from_0[i]` for each node `i`, which stores the distance from node 0. This represents the time Alice takes to reach node `i`. Also, construct the `parent` array which stores the parent node of each node. This is crucial for reconstructing Bob's path.
3
Step 3: **Determine Bob's Path and Adjust Amounts:** Trace Bob's path back to node 0 using the `parent` array. For each node on Bob's path, compare his arrival time (`bob_time`) with Alice's arrival time (`alice_time`). If `bob_time < alice_time`, Alice receives nothing (amount[curr] = 0). If `bob_time == alice_time`, the reward/price is divided (amount[curr] /= 2).
4
Step 4: **DFS for Maximum Profit:** Perform a DFS starting from node 0 to explore all paths to leaf nodes. At each node, add the adjusted `amount[u]` to the current profit. When a leaf node is reached, update `max_profit` if the current profit is greater.

💡

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?

Each node and edge is visited a constant number of times during BFS and DFS. The adjacency list and the arrays to store distances and parents take O(n) space because their sizes are proportional to 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

Dijkstra's Algorithm

Dijkstra's algorithm could be used to find the shortest path from the root to each leaf node, and then calculate profits accordingly. However, since this is a tree and not a general graph, BFS provides a simpler and more efficient way to calculate distances.

Time: O(n log n) in the general case, O(n) if using a priority queue optimized for tree structures. Since the edge weights are uniform (each edge has a weight of 1 for traversal), it's not necessarily more efficient than BFS.Space: O(n)

Trade-offs: Dijkstra's algorithm is more general and could be applied to weighted graphs, but it's overkill for a tree with unweighted edges. BFS is simpler and faster in this specific problem.

✨

Key Insights

1Insight 1: Represent the tree as an adjacency list to efficiently traverse its structure.
2Insight 2: Calculate the distance of each node from the root (node 0) using Breadth-First Search (BFS). This represents Alice's arrival time at each node.
3Insight 3: Determine the path Bob takes to reach the root and the time he takes to reach each node on his path. Compare Bob's arrival time with Alice's arrival time to determine how the price/reward at each node should be divided or assigned.
4Insight 4: Use Depth-First Search (DFS) to explore all possible paths from the root to leaf nodes and compute Alice's net income for each path. Keep track of the maximum income found so far.

📚

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: Bob starts at a node adjacent to the root. The algorithm correctly handles this as Bob's path will be short.
Edge case 2: The tree consists of only two nodes. The algorithm correctly calculates the profits.
Edge case 3: All amounts are negative. The algorithm will find the path with the least negative cost.
Edge case 4: The amounts are large and positive and negative. Using integer division when splitting the values between Alice and Bob can cause incorrect results, if it is not done correctly. The problem statement guarantees amounts are even, so integer division by 2 is safe and expected.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating Bob's path back to the root. The `parent` array must be populated correctly during the BFS to trace back Bob's path.

✗Common mistake 2: Not considering the case when Alice and Bob arrive at the same node simultaneously and failing to divide the reward/price.

✗Common mistake 3: Forgetting to handle the case when Bob arrives at a node before Alice and incorrectly assigning the reward to Alice in that case. The reward needs to be zeroed out in this scenario.

✗Common mistake 4: Not correctly initializing the `max_profit` to negative infinity, leading to incorrect results when all paths have negative profits.

✗Common mistake 5: Not correctly identifying leaf nodes during the DFS traversal. A leaf node is only a leaf if no other node is visited except its parent

✅

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: Optimizing delivery routes with time-sensitive discounts/penalties at each location. Alice is a delivery truck trying to maximize profit, while Bob is a competing truck.

🏦Real-world scenario 2: Network routing where nodes have costs associated with them. Alice represents a data packet finding the path with the maximum reward and Bob is a denial of service attack.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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: October 6, 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

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

@InterviewProJul 13

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

@CodeNinjaJul 6

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

@TechLeadJun 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.

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 "Most Profitable Path in a Tree"?

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 tree techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?