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Leetcode #2126: Destroying Asteroids

In this guide, we solve Leetcode #2126 Destroying Asteroids in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

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 ith asteroid.

Quick Facts

  • Difficulty: Medium
  • Premium: No
  • Tags: Greedy, Array, Sorting

Intuition

A locally optimal choice leads to a globally optimal result for this structure.

That means we can commit to decisions as we scan without backtracking.

Approach

Sort or preprocess if needed, then repeatedly take the best available local choice.

Maintain the minimal state necessary to validate the greedy decision.

Steps:

  • Sort or preprocess as needed.
  • Iterate and pick the best local option.
  • Track the current solution.

Example

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.

Python Solution

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

Complexity

The time complexity is O(n×log⁡n)O(n \times \log n)O(n×logn), and the space complexity is O(log⁡n)O(\log n)O(logn). The space complexity is O(log⁡n)O(\log n)O(logn).

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.


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