Asteroid Collision Leetcode Problem 735 [Python Solution]

In this blog post, we’ll explore the Asteroid Collision problem from LeetCode, categorized under the Stack section.

We’ll provide you with a detailed problem overview, constraints, a Python solution, time and space complexity analysis, and the reasoning behind our approach.

By the end of this post, you’ll have a clear understanding of how to solve this problem efficiently.

Problem Overview

The Asteroid Collision problem involves a scenario where we are given an array of asteroids represented by integers.

Each asteroid’s absolute value denotes its size, and the sign represents its direction: positive for right and negative for left.

All asteroids move at the same speed.

Our task is to determine the state of the asteroids after all potential collisions have occurred.

When two asteroids collide, the smaller one is destroyed.

If two asteroids have the same size, both are destroyed.

Importantly, asteroids moving in the same direction will never collide.

Let’s dive into the problem with an example:

Example 1:

Input: asteroids = [5, 10, -5]

Output: [5, 10]

Explanation: In this example, the asteroid with a value of 10 and the one with -5 will collide, resulting in the remaining asteroids [5, 10].

The asteroid with a value of 5 and 10 will never collide since they are both moving in the same direction.

Understanding Constraints

Before we explore the efficient Python solution for this problem, let’s discuss some essential constraints:

• The length of the ‘asteroids’ array ranges from 2 to 10^4.
• The values of the asteroids are within the range of -1000 to 1000.
• The asteroids are never equal to 0. These constraints are crucial to keep in mind while designing our solution.

Asteroid Collision LeetCode Problem Solution

Now, let’s delve into the solution to this problem.

We’ll provide both a Python code solution and explain the reasoning behind it.

Efficient Python Code Solution

def asteroidCollision(self, asteroids: List[int]) -> List[int]:
        
        stack = []
        for ass in asteroids:
            while stack and (ass < 0 and stack[-1] > 0):
                diff = ass + stack[-1]
                if diff > 0:
                    ass = 0
                elif diff &lt; 0:
                    stack.pop()
                else:
                    ass = 0
                    stack.pop()
            if ass:
                stack.append(ass)
        return stack

Now, let’s break down the solution step by step.

1. We initialize an empty stack to keep track of the remaining asteroids.
2. We iterate through the 'asteroids' array element by element.
3. Inside the loop, we check for potential collisions. For a collision to occur, the following conditions must be met:
   • The current asteroid 'ass' is negative, indicating that it’s moving to the left.
   • The top element in the stack is positive, indicating that it’s moving to the right.
   • The stack is not empty.
4. If these conditions are met, we calculate the 'diff' variable, which represents the sum of the current asteroid 'ass' and the top element of the stack. This 'diff' value helps us determine the outcome of the collision.

5. We handle three cases based on the 'diff' value:

   • If 'diff' is greater than 0, it means that the current asteroid 'ass' wins, so we set 'ass' to 0, effectively destroying it.
   • If 'diff' is less than 0, it means that the asteroid on the stack wins, so we remove it from the stack using 'stack.pop()'.
   • If 'diff' is exactly 0, both asteroids are destroyed, so we set ‘a’ to 0 and remove the asteroid from the stack.
6. After dealing with potential collisions, we check if 'ass' is still non-zero. If 'ass' survived all potential collisions, we append it to the stack.
7. Finally, after iterating through all the asteroids, we return the elements remaining in the stack, representing the state of the asteroids after all collisions.

This efficient approach leverages a stack data structure to handle asteroid collisions and guarantees a time complexity of O(n), where ‘n’ is the number of elements in the ‘asteroids’ array.

The space complexity is also O(n) due to the usage of the stack.

Reasoning Behind Our Approach

Our approach is based on the observation that asteroid collisions can be efficiently handled using a stack.

We consider each asteroid one by one, and if we detect a potential collision, we use the stack to determine the outcome.

By simulating the collisions and removing destroyed asteroids from the stack, we can find the state of the remaining asteroids accurately.

The key insight is that asteroids moving in the same direction will never collide, and we only need to focus on collisions between asteroids moving in opposite directions.

By keeping track of the direction and magnitude of asteroids in the stack, we can decide which asteroid wins in a collision and handle it accordingly.

This approach is efficient and meets the constraints of the problem, making it a suitable solution for the Asteroid Collision problem on LeetCode.

Related Interview Questions By Company:

• Sum Of Two Integers LeetCode
• Minimum Path Sum LeetCode
• Copy List With Random Pointer LeetCode

Related Interview Questions By Difficulty:

• Decode String LeetCode
• Simplify Path LeetCode
• Online Stock Span LeetCode

Conclusion

In this blog post, we explored the Asteroid Collision problem on LeetCode, providing a detailed problem overview, constraints, an efficient Python solution, and an explanation of the reasoning behind our approach.

By using a stack to simulate asteroid collisions, we can determine the state of the remaining asteroids accurately.

If you found this explanation helpful and would like to explore more coding challenges and solutions, please like and engage to support our our platform.

Feel free to ask questions, make suggestions, and share this content with others.

Happy coding!

Question Link