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2126. Destroying Asteroids

Description

You are given an integer mass, which represents the original mass of a planet. You are further given an integer array asteroids, where asteroids[i] is the mass of the ith asteroid.

You can arrange for the planet to collide with the asteroids in any arbitrary order. If the mass of the planet is greater than or equal to the mass of the asteroid, the asteroid is destroyed and the planet gains the mass of the asteroid. Otherwise, the planet is destroyed.

Return true if all asteroids can be destroyed. Otherwise, return false.

 

Example 1:

Input: mass = 10, asteroids = [3,9,19,5,21]
Output: true
Explanation: One way to order the asteroids is [9,19,5,3,21]:
- The planet collides with the asteroid with a mass of 9. New planet mass: 10 + 9 = 19
- The planet collides with the asteroid with a mass of 19. New planet mass: 19 + 19 = 38
- The planet collides with the asteroid with a mass of 5. New planet mass: 38 + 5 = 43
- The planet collides with the asteroid with a mass of 3. New planet mass: 43 + 3 = 46
- The planet collides with the asteroid with a mass of 21. New planet mass: 46 + 21 = 67
All asteroids are destroyed.

Example 2:

Input: mass = 5, asteroids = [4,9,23,4]
Output: false
Explanation: 
The planet cannot ever gain enough mass to destroy the asteroid with a mass of 23.
After the planet destroys the other asteroids, it will have a mass of 5 + 4 + 9 + 4 = 22.
This is less than 23, so a collision would not destroy the last asteroid.

 

Constraints:

  • 1 <= mass <= 105
  • 1 <= asteroids.length <= 105
  • 1 <= asteroids[i] <= 105

Solutions

Solution 1: Sorting + Greedy

According to the problem description, we can sort the asteroids by mass in ascending order, and then iterate through the asteroids. If the planet's mass is less than the asteroid's mass, the planet will be destroyed, and we return false. Otherwise, the planet will gain the mass of the asteroid.

If all asteroids can be destroyed, return true.

The time complexity is \(O(n \times \log n)\), and the space complexity is \(O(\log n)\). Where \(n\) is the number of asteroids.

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class Solution:
    def asteroidsDestroyed(self, mass: int, asteroids: List[int]) -> bool:
        asteroids.sort()
        for x in asteroids:
            if mass < x:
                return False
            mass += x
        return True

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class Solution {
    public boolean asteroidsDestroyed(int mass, int[] asteroids) {
        Arrays.sort(asteroids);
        long m = mass;
        for (int x : asteroids) {
            if (m < x) {
                return false;
            }
            m += x;
        }
        return true;
    }
}

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class Solution {
public:
    bool asteroidsDestroyed(int mass, vector<int>& asteroids) {
        ranges::sort(asteroids);
        long long m = mass;
        for (int x : asteroids) {
            if (m < x) {
                return false;
            }
            m += x;
        }
        return true;
    }
};

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func asteroidsDestroyed(mass int, asteroids []int) bool {
    sort.Ints(asteroids)
    for _, x := range asteroids {
        if mass < x {
            return false
        }
        mass += x
    }
    return true
}

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function asteroidsDestroyed(mass: number, asteroids: number[]): boolean {
    asteroids.sort((a, b) => a - b);
    for (const x of asteroids) {
        if (mass < x) {
            return false;
        }
        mass += x;
    }
    return true;
}

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impl Solution {
    pub fn asteroids_destroyed(mass: i32, mut asteroids: Vec<i32>) -> bool {
        let mut mass = mass as i64;
        asteroids.sort_unstable();
        for &x in &asteroids {
            if mass < x as i64 {
                return false;
            }
            mass += x as i64;
        }
        true
    }
}

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/**
 * @param {number} mass
 * @param {number[]} asteroids
 * @return {boolean}
 */
var asteroidsDestroyed = function (mass, asteroids) {
    asteroids.sort((a, b) => a - b);
    for (const x of asteroids) {
        if (mass < x) {
            return false;
        }
        mass += x;
    }
    return true;
};

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