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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 <= 104
  • -1000 <= asteroids[i] <= 1000
  • asteroids[i] != 0

Solutions

Solution 1: Stack

We traverse each asteroid $x$ from left to right. Since each asteroid may collide with multiple asteroids before it, we consider using a stack to store.

  • For the current asteroid, if $x>0$, it will definitely not collide with the previous asteroid, and we can directly push $x$ into the stack.
  • Otherwise, if the stack is not empty and the top element of the stack is greater than $0$, and the top element of the stack is less than $-x$, then the top element of the stack corresponds to the asteroid will explode, we loop to the top element of the stack Pop out until the condition is not satisfied. At this time, if the top element of the stack is equal to $-x$, then the two asteroids will explode, and we only need to pop the top element of the stack; if the stack is empty, or the top element of the stack is less than $0$, then the current asteroid will not collide, we will push $x$ into the stack.

Finally, we return the elements in the stack as the answer.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the length of the array $asteroids$.

  • class Solution {
    public int[] asteroidCollision(int[] asteroids) {
        Deque<Integer> stk = new ArrayDeque<>();
        for (int x : asteroids) {
            if (x > 0) {
                stk.offerLast(x);
            } else {
                while (!stk.isEmpty() && stk.peekLast() > 0 && stk.peekLast() < -x) {
                    stk.pollLast();
                }
                if (!stk.isEmpty() && stk.peekLast() == -x) {
                    stk.pollLast();
                } else if (stk.isEmpty() || stk.peekLast() < 0) {
                    stk.offerLast(x);
                }
            }
        }
        return stk.stream().mapToInt(Integer::valueOf).toArray();
    }
}

  • class Solution {
public:
    vector<int> asteroidCollision(vector<int>& asteroids) {
        vector<int> stk;
        for (int x : asteroids) {
            if (x > 0) {
                stk.push_back(x);
            } else {
                while (stk.size() && stk.back() > 0 && stk.back() < -x) {
                    stk.pop_back();
                }
                if (stk.size() && stk.back() == -x) {
                    stk.pop_back();
                } else if (stk.empty() || stk.back() < 0) {
                    stk.push_back(x);
                }
            }
        }
        return stk;
    }
};

  • class Solution:
    def asteroidCollision(self, asteroids: List[int]) -> List[int]:
        stk = []
        for x in asteroids:
            if x > 0:
                stk.append(x)
            else:
                while stk and stk[-1] > 0 and stk[-1] < -x:
                    stk.pop()
                if stk and stk[-1] == -x:
                    stk.pop()
                elif not stk or stk[-1] < 0:
                    stk.append(x)
        return stk


  • func asteroidCollision(asteroids []int) (stk []int) {
	for _, x := range asteroids {
		if x > 0 {
			stk = append(stk, x)
		} else {
			for len(stk) > 0 && stk[len(stk)-1] > 0 && stk[len(stk)-1] < -x {
				stk = stk[:len(stk)-1]
			}
			if len(stk) > 0 && stk[len(stk)-1] == -x {
				stk = stk[:len(stk)-1]
			} else if len(stk) == 0 || stk[len(stk)-1] < 0 {
				stk = append(stk, x)
			}
		}
	}
	return
}

  • function asteroidCollision(asteroids: number[]): number[] {
    const stk: number[] = [];
    for (const x of asteroids) {
        if (x > 0) {
            stk.push(x);
        } else {
            while (stk.length && stk.at(-1) > 0 && stk.at(-1) < -x) {
                stk.pop();
            }
            if (stk.length && stk.at(-1) === -x) {
                stk.pop();
            } else if (!stk.length || stk.at(-1) < 0) {
                stk.push(x);
            }
        }
    }
    return stk;
}


  • impl Solution {
    #[allow(dead_code)]
    pub fn asteroid_collision(asteroids: Vec<i32>) -> Vec<i32> {
        let mut stk = Vec::new();
        for &x in &asteroids {
            if x > 0 {
                stk.push(x);
            } else {
                while !stk.is_empty() && *stk.last().unwrap() > 0 && *stk.last().unwrap() < -x {
                    stk.pop();
                }
                if !stk.is_empty() && *stk.last().unwrap() == -x {
                    stk.pop();
                } else if stk.is_empty() || *stk.last().unwrap() < 0 {
                    stk.push(x);
                }
            }
        }
        stk
    }
}

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