What is the time complexity of Asteroid Collision?

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

What data structures are used to solve Asteroid Collision?

This problem uses Array, Stack, Simulation. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Asteroid Collision?

Insight 1: The core idea is to use a stack to simulate the collision process. Positive asteroids are pushed onto the stack, representing asteroids moving to the right. Insight 2: When a negative asteroid (moving left) encounters positive asteroids in the stack (moving right), we need to simulate the collisions until the stack is empty, the negative asteroid is destroyed, or a larger positive asteroid is encountered.

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Asteroid Collision - LeetCode 735 Solution

Q: Which companies ask LeetCode 735. Asteroid Collision in interviews?

This problem is frequently asked at Accolite, Citadel, Deutsche Bank, DoorDash, Dream11 and 15 more companies.

Asteroid Collision - Complete Solution Guide

Asteroid Collision is LeetCode problem 735, 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

We are given an array asteroids of integers representing asteroids in a row. The indices of the asteriod in the array represent their relative position in space. For each asteroid, the absolute value represents its size, and the sign represents its direction (positive meaning right, negative meaning left). Each asteroid moves at the same speed. Find out the state of the asteroids after all collisions. If two asteroids meet, the smaller one will explode. If both are the same size, both will explo

Detailed Explanation

The Asteroid Collision problem simulates the collision of asteroids moving in a single line. Each asteroid is represented by an integer, where the absolute value is its size and the sign represents its direction (positive: right, negative: left). The task is to determine the final state of the asteroids after all collisions have occurred. Asteroids moving in the same direction will never collide. When asteroids collide, the smaller one explodes. If they are the same size, both explode. The input is an array of integers, and the output is an array representing the remaining asteroids after all collisions.

Solution Approach

The solution uses a stack to keep track of the asteroids that have not yet collided. It iterates through the input array. If an asteroid is moving to the right (positive), it's pushed onto the stack. If an asteroid is moving to the left (negative), it checks for collisions with the asteroids on top of the stack (those moving to the right). If a collision occurs, the smaller asteroid is destroyed. If they are the same size, both are destroyed. The process continues until the stack is empty or the negative asteroid is destroyed or encounters a larger positive asteroid. If the negative asteroid survives, it's pushed onto the stack.

Step-by-Step Algorithm

Step 1: Initialize an empty stack to store the surviving asteroids.
Step 2: Iterate through the input array `asteroids`.
Step 3: If the current asteroid is positive (moving right), push it onto the stack.
Step 4: If the current asteroid is negative (moving left), enter a `while` loop that continues as long as the stack is not empty and the top of the stack is a positive asteroid (moving right).
Step 5: Inside the `while` loop, compare the absolute value of the negative asteroid with the top of the stack:
- If the top of the stack is smaller, pop it from the stack (it explodes).
- If the top of the stack is equal, pop it from the stack (both explode), and break out of the `while` loop.
- If the top of the stack is larger, break out of the `while` loop (the negative asteroid explodes).
Step 6: If the `while` loop completes without the negative asteroid exploding (i.e., the stack is empty or the top of the stack is negative or 0), push the negative asteroid onto the stack.
Step 7: After processing all asteroids, convert the stack to an array and return it.
Step 8: In C specifically, remember to allocate memory for the return array and free the stack's memory.

Key Insights

Insight 1: The core idea is to use a stack to simulate the collision process. Positive asteroids are pushed onto the stack, representing asteroids moving to the right.
Insight 2: When a negative asteroid (moving left) encounters positive asteroids in the stack (moving right), we need to simulate the collisions until the stack is empty, the negative asteroid is destroyed, or a larger positive asteroid is encountered.
Insight 3: The edge case where all asteroids are moving in the same direction or there are no collisions can be handled directly by using the stack to collect surviving asteroids.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Stack, Simulation.

Companies

Asked at: Accolite, Citadel, Deutsche Bank, DoorDash, Dream11, EPAM Systems, Flipkart, IMC, Lyft, Myntra, OpenAI, PhonePe, Qualtrics, Roku, Salesforce, ServiceNow, SoFi, Sprinklr, Zoho, Zynga.

Asteroid Collision | LeetCode 735

Medium

Array Stack Simulation

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

Solution Code

Available in:

import java.util.Stack;

class Solution {
    public int[] asteroidCollision(int[] asteroids) {
        Stack<Integer> stack = new Stack<>();
        for (int asteroid : asteroids) {
            if (asteroid > 0) {
                stack.push(asteroid);
            } else {
                while (!stack.isEmpty() && stack.peek() > 0) {
                    if (stack.peek() < -asteroid) {
                        stack.pop();
                        continue;
                    } else if (stack.peek() == -asteroid) {
                        stack.pop();
                        break;
                    } else {
                        break;
                    }
                }
                if (stack.isEmpty() || stack.peek() < 0) {
                    stack.push(asteroid);
                }
            }
        }

        int[] result = new int[stack.size()];
        for (int i = result.length - 1; i >= 0; i--) {
            result[i] = stack.pop();
        }
        return result;
    }
}

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Problem Info

Description

Problem 735: Asteroid Collision - Detailed explanation and solution approaches.

Topics

Array Stack Simulation

Companies

AccoliteCitadelDeutsche BankDoorDashDream11EPAM SystemsFlipkartIMC

Stats

Difficulty:Medium

Languages:4

Companies:20

Problem #:735

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

11 steps

1
Step 1: Initialize an empty stack to store the surviving asteroids.
2
Step 2: Iterate through the input array `asteroids`.
3
Step 3: If the current asteroid is positive (moving right), push it onto the stack.
4
Step 4: If the current asteroid is negative (moving left), enter a `while` loop that continues as long as the stack is not empty and the top of the stack is a positive asteroid (moving right).
5
Step 5: Inside the `while` loop, compare the absolute value of the negative asteroid with the top of the stack:
6
- If the top of the stack is smaller, pop it from the stack (it explodes).
7
- If the top of the stack is equal, pop it from the stack (both explode), and break out of the `while` loop.
8
- If the top of the stack is larger, break out of the `while` loop (the negative asteroid explodes).
9
Step 6: If the `while` loop completes without the negative asteroid exploding (i.e., the stack is empty or the top of the stack is negative or 0), push the negative asteroid onto the stack.
10
Step 7: After processing all asteroids, convert the stack to an array and return it.
11
Step 8: In C specifically, remember to allocate memory for the return array and free the stack's memory.

💡

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 algorithm iterates through the input array once (O(n)). In the worst case, all asteroids could be positive, and thus all would be pushed onto the stack. Therefore, the stack can grow up to the size of the input array, resulting in O(n) space complexity.

📖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

Two Pointers

An alternative approach might use two pointers to simulate the collisions in-place. However, this would be more complex to implement and might not be as efficient due to the need to shift elements in the array frequently.

Time: O(n^2) in the worst case due to shifting elements.Space: O(1)

Trade-offs: The two-pointer approach uses less space, but the time complexity is worse, and implementation is more difficult. Therefore, the stack-based approach is generally preferred.

✨

Key Insights

1Insight 1: The core idea is to use a stack to simulate the collision process. Positive asteroids are pushed onto the stack, representing asteroids moving to the right.
2Insight 2: When a negative asteroid (moving left) encounters positive asteroids in the stack (moving right), we need to simulate the collisions until the stack is empty, the negative asteroid is destroyed, or a larger positive asteroid is encountered.
3Insight 3: The edge case where all asteroids are moving in the same direction or there are no collisions can be handled directly by using the stack to collect surviving asteroids.

📚

Key Terms Explained

For Students

Array

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

Stack

LIFO structure - Last In, First Out. Like a stack of plates!

Edge Cases

Edge case 1: All asteroids are moving to the right (all positive). In this case, all asteroids will be added to the stack, and the stack will be returned as the result. The `while` loop in the negative asteroid case will never execute.
Edge case 2: All asteroids are moving to the left (all negative). In this case, all asteroids will be added to the stack because the `while` loop condition (stack not empty and top is positive) will always be false.
Edge case 3: Alternating positive and negative asteroids that cancel each other out. The algorithm correctly handles the case where positive and negative asteroids of equal size collide and both are destroyed.
Edge case 4: Empty input array: The algorithm should handle the empty array case gracefully, returning an empty array.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the case where the top of the stack is equal to the negative of the current asteroid (both explode).

✗Common mistake 2: Not correctly handling the case where the negative asteroid survives after colliding with multiple positive asteroids.

✗Common mistake 3: Not considering the case where the stack becomes empty during collision processing. Ensure proper handling of the `stack.isEmpty()` check.

✗Common mistake 4: In C, forgetting to allocate memory for the returned array and/or forgetting to free the dynamically allocated stack.

✅

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: Simulating particle interactions in physics, where particles with opposing charges or velocities collide and annihilate each other.

🏦Real-world scenario 2: Managing resource allocation where competing processes request and release resources, and conflicting requests may lead to cancellation of operations (e.g., database transactions).

💡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

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

⚠️ Watch Out

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

👨‍💻 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: June 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

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 a monotonic stack and when should I use it?

A monotonic stack maintains elements in sorted order (increasing or decreasing). Use it for "next greater element," "largest rectangle," or "stock span" problems. It achieves O(n) time by avoiding redundant comparisons through maintaining order invariants.

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 "Asteroid Collision"?

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

🤖 Need help with this problem?