735. Asteroid Collision

Description of Problem

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

Tags: Stack

Explanation

Construct a stack that store only stable asteroids (which means no collision i.e. no pair of (-->,<--) in the stack).

If there is collision when appending new asteroid, keep colliding until the stack is stable.

Code (Rust)

impl Solution {
    pub fn asteroid_collision(asteroids: Vec<i32>) -> Vec<i32> {
        let mut stack : Vec<i32> = vec![];

        for incoming in asteroids.into_iter(){
            let mut incoming = incoming;
            // Keep collide when (rightmost)--> <--(incoming)
            while let Some(&rightmost) = stack.last() {
                if rightmost > 0 && incoming < 0 && rightmost.abs() < incoming.abs() {
                    // `rightmost` is destroyed, remain `incoming`
                    stack.pop();
                }else if rightmost > 0 && incoming < 0 && rightmost.abs() > incoming.abs(){
                    // `incoming` is destroyed, remain `rightmost`
                    stack.pop();
                    incoming = rightmost
                }else if rightmost > 0 && incoming < 0 && rightmost.abs() == incoming.abs(){
                    // Both are destroyed
                    stack.pop();
                    incoming = 0;
                    break;
                }else{
                    // Collision does not happen
                    break;
                }
            }

            // The incoming asteroid still exist, append into the stack
            if incoming != 0 {
                stack.push(incoming);
            }
            
        }

        return stack;
    }
}

Complexity

n is length of asteroids

Time Complexity

\( T(n) = \Theta(n) \)

Auxiliary Space

\( S(n) = O(n) \)

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