What is the time complexity of Detonate the Maximum Bombs?

The optimal solution has a time complexity of O(n^3).

What data structures are used to solve Detonate the Maximum Bombs?

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

What are the key insights for solving Detonate the Maximum Bombs?

Insight 1: The problem can be modeled as a directed graph where bombs are nodes, and an edge from bomb A to bomb B exists if detonating bomb A will detonate bomb B (i.e., bomb B is within bomb A's radius). Insight 2: We need to try detonating each bomb as a starting point and find the connected component size (number of reachable nodes) in the graph. The largest such connected component represents the maximum number of bombs that can be detonated.

Which companies ask LeetCode 2101. Detonate the Maximum Bombs in interviews?

This problem is frequently asked at Chime.

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Detonate the Maximum Bombs - LeetCode 2101 Solution

Q: What is the space complexity of Detonate the Maximum Bombs?

The space complexity is O(n^2).

Detonate the Maximum Bombs - Complete Solution Guide

Detonate the Maximum Bombs is LeetCode problem 2101, 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 list of bombs. The range of a bomb is defined as the area where its effect can be felt. This area is in the shape of a circle with the center as the location of the bomb. The bombs are represented by a 0-indexed 2D integer array bombs where bombs[i] = [x i , y i , r i ] . x i and y i denote the X-coordinate and Y-coordinate of the location of the i th bomb, whereas r i denotes the radius of its range. You may choose to detonate a single bomb. When a bomb is detonated, it will det

Detailed Explanation

The problem presents a scenario where you are given a list of bombs, each with a location (x, y) and a radius r. When a bomb is detonated, it sets off all other bombs within its radius. The goal is to find the maximum number of bombs that can be detonated if you are only allowed to initially detonate one bomb. This involves understanding the concept of cascading detonations.

Solution Approach

The solution uses Breadth-First Search (BFS) to simulate the detonation process starting from each bomb. It constructs an adjacency list representing the graph of bombs. For each bomb, it performs a BFS to find all bombs that would be detonated in a chain reaction. The maximum number of bombs detonated from any starting bomb is the answer.

Step-by-Step Algorithm

Step 1: Build the adjacency list (graph). Iterate through all pairs of bombs (i, j). If bomb j is within the radius of bomb i, add an edge from i to j in the adjacency list.
Step 2: Iterate through each bomb from index 0 to n-1. For each bomb `i`, assume we start by detonating this bomb.
Step 3: Perform BFS starting from bomb `i`. Use a queue to keep track of bombs to detonate and a set to keep track of visited/detonated bombs.
Step 4: In the BFS, while the queue is not empty, dequeue a bomb and iterate through its neighbors (bombs within its range). If a neighbor has not been visited yet, add it to the visited set and enqueue it.
Step 5: After the BFS completes, the size of the visited set represents the number of bombs detonated starting from bomb `i`. Update the maximum number of bombs detonated found so far.
Step 6: Repeat steps 2-5 for each bomb in the list.
Step 7: Return the maximum number of bombs detonated found across all starting bombs.

Key Insights

Insight 1: The problem can be modeled as a directed graph where bombs are nodes, and an edge from bomb A to bomb B exists if detonating bomb A will detonate bomb B (i.e., bomb B is within bomb A's radius).
Insight 2: We need to try detonating each bomb as a starting point and find the connected component size (number of reachable nodes) in the graph. The largest such connected component represents the maximum number of bombs that can be detonated.
Insight 3: Since the constraints are relatively small (n <= 100), an O(n^3) solution is acceptable. This allows for a brute-force approach of trying each bomb as a starting point and using Depth-First Search (DFS) or Breadth-First Search (BFS) to find the connected component size.

Complexity Analysis

Time Complexity: O(n^3)

Space Complexity: O(n^2)

Topics

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

Companies

Asked at: Chime.

Detonate the Maximum Bombs | LeetCode 2101

Medium

Array Math Depth-First Search Breadth-First Search Graph Geometry

Time: O(n^3)Space: O(n^2)

Solution Code

Available in:

import java.util.*;

class Solution {
    public int maximumDetonation(int[][] bombs) {
        int n = bombs.length;
        Map<Integer, List<Integer>> adj = new HashMap<>();

        for (int i = 0; i < n; i++) {
            adj.put(i, new ArrayList<>());
        }

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (i == j) {
                    continue;
                }

                int xi = bombs[i][0];
                int yi = bombs[i][1];
                int ri = bombs[i][2];
                int xj = bombs[j][0];
                int yj = bombs[j][1];

                if ((long)(xi - xj) * (xi - xj) + (long)(yi - yj) * (yi - yj) <= (long)ri * ri) {
                    adj.get(i).add(j);
                }
            }
        }

        int max_bombs_detonated = 0;

        for (int i = 0; i < n; i++) {
            Queue<Integer> q = new LinkedList<>();
            Set<Integer> visited = new HashSet<>();
            q.offer(i);
            visited.add(i);

            while (!q.isEmpty()) {
                int current_bomb = q.poll();
                for (int neighbor : adj.get(current_bomb)) {
                    if (!visited.contains(neighbor)) {
                        visited.add(neighbor);
                        q.offer(neighbor);
                    }
                }
            }

            if (visited.size() > max_bombs_detonated) {
                max_bombs_detonated = visited.size();
            }

            if (max_bombs_detonated == n) {
                return n;
            }
        }

        return max_bombs_detonated;
    }
}

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Problem Info

Description

Problem 2101: Detonate the Maximum Bombs - Detailed explanation and solution approaches.

Topics

Array Math Depth-First Search Breadth-First Search Graph Geometry

Companies

Chime

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2101

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

7 steps

1
Step 1: Build the adjacency list (graph). Iterate through all pairs of bombs (i, j). If bomb j is within the radius of bomb i, add an edge from i to j in the adjacency list.
2
Step 2: Iterate through each bomb from index 0 to n-1. For each bomb `i`, assume we start by detonating this bomb.
3
Step 3: Perform BFS starting from bomb `i`. Use a queue to keep track of bombs to detonate and a set to keep track of visited/detonated bombs.
4
Step 4: In the BFS, while the queue is not empty, dequeue a bomb and iterate through its neighbors (bombs within its range). If a neighbor has not been visited yet, add it to the visited set and enqueue it.
5
Step 5: After the BFS completes, the size of the visited set represents the number of bombs detonated starting from bomb `i`. Update the maximum number of bombs detonated found so far.
6
Step 6: Repeat steps 2-5 for each bomb in the list.
7
Step 7: Return the maximum number of bombs detonated found across all starting bombs.

💡

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^3)

💾

Space Complexity

O(n^2)

📚 Quadratic space - Uses lots of memory!

📝

Why?

The time complexity is O(n^3) because building the adjacency list takes O(n^2) time (nested loops), and we perform BFS n times, each taking O(n+m) time in the worst case, where m is the number of edges. Since the maximum number of edges is O(n^2), each BFS takes O(n^2) time. Therefore, n BFS calls take O(n^3) time. The space complexity is O(n^2) because the adjacency list can store up to O(n^2) edges. The visited set in BFS takes O(n) space, but this is dominated by the adjacency list.

📖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 traverse the graph and find connected components. The core logic remains the same, but the traversal order is different.

Time: O(n^3)Space: O(n^2)

Trade-offs: DFS and BFS have similar performance characteristics in this case. DFS might use slightly less memory in some scenarios due to its stack-like nature, but the difference is generally negligible.

Tarjan's Algorithm (Strongly Connected Components)

Tarjan's algorithm is an efficient way to find strongly connected components (SCCs) in a directed graph. Although this problem doesn't strictly require finding SCCs, it *could* be adapted. If any bombs are mutually detonating, they form an SCC. These SCCs could be collapsed to a single node. However, since we need to start the process by choosing a single bomb, this approach wouldn't be more efficient.

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

Trade-offs: Tarjan's would only slightly improve the adjacency list building, but still require testing each starting bomb to see the extent of the detonation and likely doesn't improve on O(n^3) overall in the worst case.

✨

Key Insights

1Insight 1: The problem can be modeled as a directed graph where bombs are nodes, and an edge from bomb A to bomb B exists if detonating bomb A will detonate bomb B (i.e., bomb B is within bomb A's radius).
2Insight 2: We need to try detonating each bomb as a starting point and find the connected component size (number of reachable nodes) in the graph. The largest such connected component represents the maximum number of bombs that can be detonated.
3Insight 3: Since the constraints are relatively small (n <= 100), an O(n^3) solution is acceptable. This allows for a brute-force approach of trying each bomb as a starting point and using Depth-First Search (DFS) or Breadth-First Search (BFS) to find the connected component size.

📚

Key Terms Explained

For Students

Array

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

Graph

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

Edge Cases

Edge case 1: A bomb detonates no other bombs. The BFS will only visit the starting bomb itself, correctly returning 1.
Edge case 2: All bombs are within the range of a single bomb. The BFS starting from that bomb will visit all bombs, correctly returning n.
Edge case 3: No bomb detonates any other. The algorithm will correctly return 1 for all bombs as the starting detonation point.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Integer overflow when calculating the distance between bombs. The squared distances (xi - xj)^2 + (yi - yj)^2 can be quite large, exceeding the maximum value of an integer. Using long data types is crucial to avoid this.

✗Common mistake 2: Forgetting to check if i == j in the adjacency list construction. This would lead to self-loops in the graph, which could affect the BFS result.

✗Common mistake 3: Not properly initializing the visited set for each BFS. If the visited set is not cleared before each BFS, the results will be incorrect.

✅

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: Modeling the spread of a virus or disease. Each person is a node, and if one person infects another, there's an edge. This algorithm can determine the maximum number of people that could be infected starting from a single infected person.

🏦Real-world scenario 2: Social network influence. Each user is a node, and an edge represents a follows relationship. This algorithm can determine the maximum number of users that could be influenced by a single user's post.

🎮Real-world scenario 3: Network security. Bombs can be thought of as compromised machines in a network, and detonation range can represent how far a worm or attack can spread. This determines potential network reach if a single system is compromised.

💡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: September 30, 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

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

@AlgoMasterJul 14

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

@InterviewProJul 7

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

@CodeNinjaJun 30

❓

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 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 "Detonate the Maximum Bombs"?

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

🤖 Need help with this problem?