leetcode - 735. Asteroid Collision

Description

We are given an array asteroids of integers representing asteroids in a row.

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 <= 10^4
-1000 <= asteroids[i] <= 1000
asteroids[i] != 0

Solution

Recursive

Every time, find the positive that has a negative at the right, and then collide, then recursively solve the rest of the list.
Time complexity: $o(n^2)$
Space complexity: $o(n)$

Stack

Time complexity: $o(n)$
Space complexity: $o(n)$

Code

Recursive

class Solution:
    def asteroidCollision(self, asteroids: List[int]) -> List[int]:
        # find collisions
        i = 1
        while i < len(asteroids):
            if asteroids[i - 1] > 0 and asteroids[i] < 0:
                break
            i += 1
        if i >= len(asteroids):
            return asteroids
        if asteroids[i - 1] + asteroids[i] > 0:
            return self.asteroidCollision(asteroids[:i] + asteroids[i+1:])
        elif asteroids[i - 1] + asteroids[i] < 0:
            return self.asteroidCollision(asteroids[:i-1] + asteroids[i:])
        else:
            return self.asteroidCollision(asteroids[:i-1] + asteroids[i+1:])

Stack

class Solution:
    def asteroidCollision(self, asteroids: List[int]) -> List[int]:
        stack = []
        i = 0
        while i < len(asteroids):
            while stack and stack[-1] > 0 and i < len(asteroids) and asteroids[i] < 0:
                if stack[-1] + asteroids[i] > 0:
                    i += 1
                elif stack[-1] + asteroids[i] < 0:
                    stack.pop()
                else:
                    stack.pop()
                    i += 1
            if i < len(asteroids):
                stack.append(asteroids[i])
                i += 1
        return stack