LeetCode #735 — MEDIUM

Asteroid Collision

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

The Problem

Problem Statement

We are given an array asteroids of integers representing asteroids in a row. The indices of the asteroid in the array represent their relative position in space.

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.

Example 4:

Input: asteroids = [3,5,-6,2,-1,4]
Output: [-6,2,4]
Explanation: The asteroid -6 makes the asteroid 3 and 5 explode, and then continues going left. On the other side, the asteroid 2 makes the asteroid -1 explode and then continues going right, without reaching asteroid 4.

Constraints:

2 <= asteroids.length <= 10⁴
-1000 <= asteroids[i] <= 1000
asteroids[i] != 0

Step 01

Brute Force Baseline

Problem summary: We are given an array asteroids of integers representing asteroids in a row. The indices of the asteroid in the array represent their relative position in space. 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.

Baseline thinking

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

Pattern signal: Array · Stack

Example 1

[5,10,-5]

Example 2

[8,-8]

Example 3

[10,2,-5]

Core Insight

What unlocks the optimal approach

Say a row of asteroids is stable. What happens when a new asteroid is added on the right?

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 #735: Asteroid Collision
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();
    }
}

// Accepted solution for LeetCode #735: Asteroid Collision
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
}

# Accepted solution for LeetCode #735: Asteroid Collision
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

// Accepted solution for LeetCode #735: Asteroid Collision
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
    }
}

// Accepted solution for LeetCode #735: Asteroid Collision
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;
}

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)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan left (or right) to find the next greater/smaller element. The inner scan can visit up to n elements per outer iteration, giving O(n²) total comparisons. No extra space needed beyond loop variables.

MONOTONIC STACK

O(n) time

O(n) space

Each element is pushed onto the stack at most once and popped at most once, giving 2n total operations = O(n). The stack itself holds at most n elements in the worst case. The key insight: amortized O(1) per element despite the inner while-loop.

Shortcut: Each element pushed once + popped once → O(n) amortized. The inner while-loop does not make it O(n²).

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.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.