735. Asteroid Collision (Python)

Related Topic

Stack.

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

Sample I/O

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.

Note

1. The length of asteroids will be at most 10000.
2. Each asteroid will be a non-zero integer in the range [-1000, 1000]..

Methodology

Positive meaning left->right and negative meaning right->left. The collision only happend when current astero right->left and the last astero left->right. This means the last value have to be positive and the current value have to be negative then we can start to identify the collision.

    def asteroidCollision(self, asteroids: List[int]) -> List[int]:
        stack = []
        for id in asteroids:
            if not stack or stack[-1]*id>0 or stack[-1]<0:
                stack.append(id)
            elif abs(id)>=abs(stack[-1]):
                while stack and stack[-1]>0 and abs(id) > abs(stack[-1]):
                    stack.pop()
                if not stack or stack[-1]<0:
                    stack.append(id)
                elif stack[-1] == -id:
                    stack.pop()
        return stack

BigO

We iterate the list once so the total cost will be O(n)

python (easy way) def asteroidCollision(self, asteroids): ans = [] for id in asteroids: while ans and id < 0 < ans[-1]: if ans[-1] < -id: ans.pop() continue elif ans[-1] == -id: ans.pop() break else: ans.append(id) return ans

BigO

We iterate the list once so the total cost will be O(n)