Asteroid collision

Time: O(N); Space: O(N); medium

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.

Example 4:

Input: asteroids = [-2, -1, 1, 2]

Output: [-2, -1, 1, 2]

Explanation:

  • The -2 and -1 are moving left, while the 1 and 2 are moving right.

  • Asteroids moving the same direction never meet, so no asteroids will meet each other.

Constraints:

  • The length of asteroids will be at most 10000.

  • Each asteroid will be a non-zero integer in the range [-1000, 1000]..

Hint:

  1. Say a row of asteroids is stable. What happens when a new asteroid is added on the right?

1. Stack

Intuition

A row of asteroids is stable if no further collisions will occur. After adding a new asteroid to the right, some more collisions may happen before it becomes stable again, and all of those collisions (if they happen) must occur right to left. This is the perfect situation for using a stack.

Algorithm

Say we have our answer as a stack with rightmost asteroid top, and a new asteroid comes in. If new is moving right (new > 0), or if top is moving left (top < 0), no collision occurs.

Otherwise, if abs(new) < abs(top), then the new asteroid will blow up; if abs(new) == abs(top) then both asteroids will blow up; and if abs(new) > abs(top), then the top asteroid will blow up (and possibly more asteroids will, so we should continue checking.)

[1]:

class Solution1(object):
    """
    Time: O(n), where N is the number of asteroids. Our stack pushes and pops each asteroid at most once.
    Space: O(n), the size of result.
    """
    def asteroidCollision(self, asteroids):
        """
        :type asteroids: List[int]
        :rtype: List[int]
        """
        result = []
        for asteroid in asteroids:
            while result and asteroid < 0 < result[-1]:
                if result[-1] < -asteroid:
                    result.pop()
                    continue
                elif result[-1] == -asteroid:
                    result.pop()
                break
            else:
                result.append(asteroid)
        return result

[2]:

s = Solution1()
asteroids = [5, 10, -5]
assert s.asteroidCollision(asteroids) == [5, 10]

asteroids = [8, -8]
assert s.asteroidCollision(asteroids) == []

asteroids = [10, 2, -5]
assert s.asteroidCollision(asteroids) == [10]

asteroids = [-2, -1, 1, 2]
assert s.asteroidCollision(asteroids) == [-2, -1, 1, 2]