What is the time complexity of Amount of Time for Binary Tree to Be Infected?

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

What data structures are used to solve Amount of Time for Binary Tree to Be Infected?

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

What are the key insights for solving Amount of Time for Binary Tree to Be Infected?

Insight 1: The core idea is to model the binary tree as an undirected graph. This allows us to easily traverse between parent and child nodes during the infection spread. Insight 2: Breadth-First Search (BFS) is suitable for simulating the infection spread because it explores the tree level by level, representing the passage of time (minutes).

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Amount of Time for Binary Tree to Be Infected - LeetCode 2385 Solution

Q: Which companies ask LeetCode 2385. Amount of Time for Binary Tree to Be Infected in interviews?

This problem is frequently asked at Flipkart, Goldman Sachs, Nutanix, PhonePe, ServiceNow and 2 more companies.

Amount of Time for Binary Tree to Be Infected - Complete Solution Guide

Amount of Time for Binary Tree to Be Infected is LeetCode problem 2385, 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 the root of a binary tree with unique values, and an integer start . At minute 0 , an infection starts from the node with value start . Each minute, a node becomes infected if: The node is currently uninfected. The node is adjacent to an infected node. Return the number of minutes needed for the entire tree to be infected. Example 1: Input: root = [1,5,3,null,4,10,6,9,2], start = 3 Output: 4 Explanation: The following nodes are infected during: - Minute 0: Node 3 - Minute 1: Nodes

Detailed Explanation

The problem asks us to find the amount of time it takes for a binary tree to be completely infected, starting from a given node. The infection spreads to adjacent nodes (parent, left child, right child) in each minute. We need to return the number of minutes required for the entire tree to be infected. The tree has unique node values, and the 'start' node is guaranteed to exist.

Solution Approach

The solution converts the binary tree into an undirected graph represented by an adjacency list. Then, it performs a Breadth-First Search (BFS) starting from the given 'start' node. The BFS algorithm explores the graph level by level. Each level represents one minute of infection spread. The algorithm keeps track of visited nodes to avoid revisiting them. The number of levels traversed during BFS represents the total time required to infect the entire tree.

Step-by-Step Algorithm

Step 1: **Convert Binary Tree to Graph:** Traverse the binary tree (e.g., using level-order traversal) and create an adjacency list representing an undirected graph. For each node, add edges to its left child, right child, and its parent (if applicable).
Step 2: **Initialize BFS:** Create a queue for BFS and enqueue the 'start' node. Also, create a set to keep track of visited nodes to avoid cycles.
Step 3: **Perform BFS:** While the queue is not empty, increment the 'time' (minutes). Process all nodes at the current level. For each node, find its neighbors in the graph and enqueue unvisited neighbors. Mark the enqueued neighbors as visited.
Step 4: **Return Time:** After the BFS is complete (queue is empty), return the calculated 'time'. Since the time is incremented at the beginning of each loop, it's initialized to -1, so the final result accounts for starting at minute 0.

Key Insights

Insight 1: The core idea is to model the binary tree as an undirected graph. This allows us to easily traverse between parent and child nodes during the infection spread.
Insight 2: Breadth-First Search (BFS) is suitable for simulating the infection spread because it explores the tree level by level, representing the passage of time (minutes).
Insight 3: The longest path from the 'start' node in the converted graph represents the minimum time needed to infect the entire tree.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Hash Table, Tree, Depth-First Search, Breadth-First Search, Binary Tree.

Companies

Asked at: Flipkart, Goldman Sachs, Nutanix, PhonePe, ServiceNow, ShareChat, Snap.

Amount of Time for Binary Tree to Be Infected | LeetCode 2385

Medium

Hash Table Tree Depth-First Search Breadth-First Search Binary Tree

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

Solution Code

Available in:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
import java.util.*;

class Solution {
    public int amountOfTime(TreeNode root, int start) {
        Map<Integer, List<Integer>> graph = new HashMap<>();
        
        Queue<TreeNode> q_build = new LinkedList<>();
        if (root != null) {
            q_build.offer(root);
        }

        while (!q_build.isEmpty()) {
            TreeNode node = q_build.poll();
            if (node.left != null) {
                graph.computeIfAbsent(node.val, k -> new ArrayList<>()).add(node.left.val);
                graph.computeIfAbsent(node.left.val, k -> new ArrayList<>()).add(node.val);
                q_build.offer(node.left);
            }
            if (node.right != null) {
                graph.computeIfAbsent(node.val, k -> new ArrayList<>()).add(node.right.val);
                graph.computeIfAbsent(node.right.val, k -> new ArrayList<>()).add(node.val);
                q_build.offer(node.right);
            }
        }
        
        int minutes = -1;
        Queue<Integer> q_bfs = new LinkedList<>();
        q_bfs.offer(start);
        Set<Integer> visited = new HashSet<>();
        visited.add(start);
        
        while (!q_bfs.isEmpty()) {
            minutes += 1;
            int size = q_bfs.size();
            for (int i = 0; i < size; i++) {
                int curr_node = q_bfs.poll();
                if (graph.containsKey(curr_node)) {
                    for (int neighbor : graph.get(curr_node)) {
                        if (!visited.contains(neighbor)) {
                            visited.add(neighbor);
                            q_bfs.offer(neighbor);
                        }
                    }
                }
            }
        }
        
        return minutes;
    }
}

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Problem Info

Description

Problem 2385: Amount of Time for Binary Tree to Be Infected - Detailed explanation and solution approaches.

Topics

Hash Table Tree Depth-First Search Breadth-First Search Binary Tree

Companies

FlipkartGoldman SachsNutanixPhonePeServiceNowShareChatSnap

Stats

Difficulty:Medium

Languages:4

Companies:7

Problem #:2385

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: **Convert Binary Tree to Graph:** Traverse the binary tree (e.g., using level-order traversal) and create an adjacency list representing an undirected graph. For each node, add edges to its left child, right child, and its parent (if applicable).
2
Step 2: **Initialize BFS:** Create a queue for BFS and enqueue the 'start' node. Also, create a set to keep track of visited nodes to avoid cycles.
3
Step 3: **Perform BFS:** While the queue is not empty, increment the 'time' (minutes). Process all nodes at the current level. For each node, find its neighbors in the graph and enqueue unvisited neighbors. Mark the enqueued neighbors as visited.
4
Step 4: **Return Time:** After the BFS is complete (queue is empty), return the calculated 'time'. Since the time is incremented at the beginning of each loop, it's initialized to -1, so the final result accounts for starting at minute 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 tree traversal to create the graph and the BFS traversal to find the maximum infection time. Both of these operations visit each node once, so the time complexity is O(N). The space complexity is governed by the size of the graph (adjacency list), the BFS queue, and the visited set, all of which can potentially store up to N 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

Depth-First Search (DFS)

Instead of BFS, DFS can be used to find the distance of each node from the 'start' node. The maximum distance found will be the answer. First, convert the tree into a graph. Then, perform a DFS from the start node to find the maximum depth (time).

Time: O(N)Space: O(N)

Trade-offs: DFS has a similar time and space complexity to BFS in this problem. However, BFS is often preferred because it naturally represents the level-by-level infection spread more intuitively.

✨

Key Insights

1Insight 1: The core idea is to model the binary tree as an undirected graph. This allows us to easily traverse between parent and child nodes during the infection spread.
2Insight 2: Breadth-First Search (BFS) is suitable for simulating the infection spread because it explores the tree level by level, representing the passage of time (minutes).
3Insight 3: The longest path from the 'start' node in the converted graph represents the minimum time needed to infect the entire tree.

📚

Key Terms Explained

For Students

Hash Table

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

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: **Single Node Tree:** If the tree contains only the 'start' node, the infection is already complete at minute 0. The BFS will not enter the while loop, and the initialized time (-1) + 1 = 0 will be returned.
Edge case 2: **'start' node is the root:** The algorithm correctly handles this because the graph is constructed regardless of whether the start node is the root or not. BFS will proceed as usual.
Edge case 3: **Skewed Tree:** The skewed tree is handled correctly because the graph construction and BFS implementation don't rely on the tree being balanced.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: **Not converting the binary tree to a graph:** Directly using the binary tree structure for BFS makes it difficult to move from child to parent. Converting to a graph allows for easy traversal in both directions.

✗Common mistake 2: **Not keeping track of visited nodes:** Without a visited set, BFS can get stuck in cycles, leading to infinite loops and incorrect results. The undirected graph introduces the possibility of cycles which a visited set prevents.

✗Common mistake 3: **Incorrectly calculating the time:** Forgetting to initialize 'time' to -1 to correctly handle the base case (infection starting at minute 0), or incrementing the time at the wrong point in the BFS loop.

✅

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 Contagion:** Modeling the spread of a computer virus or malware through a network. The nodes represent computers, and the edges represent network connections. The goal is to estimate the time it takes for the entire network to be infected.

🏦Real-world scenario 2: **Social Network Influence:** Analyzing the spread of information or trends through a social network. The nodes represent individuals, and the edges represent social connections. The algorithm can estimate how quickly a message will reach the entire network, given a starting point.

💡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

Hash tables trade space for time. Before using one, consider if the problem constraints allow the extra memory. In interviews, always mention this trade-off.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Don't forget to handle key collisions and null/undefined keys. Consider what happens with duplicate values.

👨‍💻 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: July 16, 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

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.

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 "Amount of Time for Binary Tree to Be Infected"?

For this medium-level Hash Table 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?