LeetCode #2126 — MEDIUM

Destroying Asteroids

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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 i^th 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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Greedy

Example 1

10
[3,9,19,5,21]

Example 2

5
[4,9,23,4]

Core Insight

What unlocks the optimal approach

Choosing the asteroid to collide with can be done greedily.
If an asteroid will destroy the planet, then every bigger asteroid will also destroy the planet.
You only need to check the smallest asteroid at each collision. If it will destroy the planet, then every other asteroid will also destroy the planet.
Sort the asteroids in non-decreasing order by mass, then greedily try to collide with the asteroids in that order.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2126: Destroying Asteroids
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;
    }
}

// Accepted solution for LeetCode #2126: Destroying Asteroids
func asteroidsDestroyed(mass int, asteroids []int) bool {
	sort.Ints(asteroids)
	for _, x := range asteroids {
		if mass < x {
			return false
		}
		mass += x
	}
	return true
}

# Accepted solution for LeetCode #2126: Destroying Asteroids
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

// Accepted solution for LeetCode #2126: Destroying Asteroids
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
    }
}

// Accepted solution for LeetCode #2126: Destroying Asteroids
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(log n)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.