735. Asteroid Collision LeetCode Solution

Last updated: February 19, 2025

In this guide, you will get 735. Asteroid Collision LeetCode Solution with the best time and space complexity. The solution to Asteroid Collision problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Asteroid Collision solution in C++
Asteroid Collision solution in Java
Asteroid Collision solution in Python
Additional Resources

735. Asteroid Collision LeetCode Solution image

Problem Statement of Asteroid Collision

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 explode. Two asteroids moving in the same direction will never meet.

Example 1:

Input: asteroids = [5,10,-5]
Output: [5,10]
Explanation: The 10 and -5 collide resulting in 10. The 5 and 10 never collide.

Example 2:

Input: asteroids = [8,-8]
Output: []
Explanation: The 8 and -8 collide exploding each other.

Example 3:

Input: asteroids = [10,2,-5]
Output: [10]
Explanation: The 2 and -5 collide resulting in -5. The 10 and -5 collide resulting in 10.

Constraints:

2 <= asteroids.length <= 104
-1000 <= asteroids[i] <= 1000
asteroids[i] != 0

Complexity Analysis

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

735. Asteroid Collision LeetCode Solution in C++

class Solution {
 public:
  vector<int> asteroidCollision(vector<int>& asteroids) {
    vector<int> stack;

    for (const int a : asteroids)
      if (a > 0) {
        stack.push_back(a);
      } else {  // a < 0
        // Destroy the previous positive one(s).
        while (!stack.empty() && stack.back() > 0 && stack.back() < -a)
          stack.pop_back();
        if (stack.empty() || stack.back() < 0)
          stack.push_back(a);
        else if (stack.back() == -a)
          stack.pop_back();  // Both asteroids explode.
        else                 // stack[-1] > the current asteroid.
          ;                  // Destroy the current asteroid, so do nothing.
      }

    return stack;
  }
};
/* code provided by PROGIEZ */

735. Asteroid Collision LeetCode Solution in Java

class Solution {
  public int[] asteroidCollision(int[] asteroids) {
    Stack<Integer> stack = new Stack<>();

    for (final int a : asteroids)
      if (a > 0) {
        stack.push(a);
      } else { // a < 0
        // Destroy the previous positive one(s).
        while (!stack.isEmpty() && stack.peek() > 0 && stack.peek() < -a)
          stack.pop();
        if (stack.isEmpty() || stack.peek() < 0)
          stack.push(a);
        else if (stack.peek() == -a)
          stack.pop(); // Both asteroids explode.
        else           // stack[-1] > the current asteroid.
          ;            // Destroy the current asteroid, so do nothing.
      }

    return stack.stream().mapToInt(Integer::intValue).toArray();
  }
}
// code provided by PROGIEZ

735. Asteroid Collision LeetCode Solution in Python

class Solution:
  def asteroidCollision(self, asteroids: list[int]) -> list[int]:
    stack = []

    for a in asteroids:
      if a > 0:
        stack.append(a)
      else:  # a < 0
        # Destroy the previous positive one(s).
        while stack and stack[-1] > 0 and stack[-1] < -a:
          stack.pop()
        if not stack or stack[-1] < 0:
          stack.append(a)
        elif stack[-1] == -a:
          stack.pop()  # Both asteroids explode.
        else:  # stack[-1] > the current asteroid.
          pass  # Destroy the current asteroid, so do nothing.

    return stack
 # code by PROGIEZ

Additional Resources

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