Problem 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:

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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:

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

Example 3:

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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:

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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:

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

Solution:

If the stack still contains element and stack[-1] moving toward right but asteroid moving toward left, check which stays.

else:
some senario

  • same direction
  • asteroid is negative moving toward left, but no positive in the stack

we just append asteroid to stack

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class Solution:
    def asteroidCollision(self, asteroids: List[int]) -> List[int]:
        res = []
        for asteroid in asteroids:
            # We only need to resolve collisions under the following conditions:
            # - stack is non-empty
            # - current asteroid is -ve
            # - top of the stack is +ve
            while len(res) and asteroid < 0 and res[-1] > 0:
                # Both asteroids are equal, destroy both.
                if res[-1] == -asteroid: 
                    res.pop()
                    break
                # Stack top is smaller, remove the +ve asteroid 
                # from the stack and continue the comparison.
                elif res[-1] < -asteroid:
                    res.pop()
                    continue
                # Stack top is larger, -ve asteroid is destroyed.
                elif res[-1] > -asteroid:
                    break
            else:
                # -ve asteroid made it all the way to the 
                # bottom of the stack and destroyed all asteroids.
                res.append(asteroid)
        return res

time complexity: $O()$
space complexity: $O()$
reference:
related problem: