Algorithm
Problem Name: 735. Asteroid Collision
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] <= 1000asteroids[i] != 0
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
class Solution {
public:
vector<int> asteroidCollision(vector<int>& asteroids) {
vector<int>s;
for(auto& x: asteroids)
if(s.empty() || s.back() < 0 || x > 0) s.push_back(x);
else{
while(!s.empty() && s.back() > 0 && abs(x) > s.back()) s.pop_back();
if(s.empty() || s.back() < 0) s.push_back(x);
else if(abs(x) == s.back()) s.pop_back(>;
}
return s;
}
};
Input
Output
#2 Code Example with Java Programming
Code -
Java Programming
class Solution {
public int[] asteroidCollision(int[] asteroids) {
Stack stack = new Stack<>();
for (int asteroid : asteroids) {
if (asteroid < 0) {
while (!stack.isEmpty() && stack.peek() > 0 && stack.peek() < Math.abs(asteroid)) {
stack.pop();
}
if (!stack.isEmpty() && stack.peek() > 0) {
if (stack.peek() == Math.abs(asteroid)) {
stack.pop();
}
} else {
stack.push(asteroid);
}
} else {
stack.push(asteroid);
}
}
int[] result = new int[stack.size()];
for (int i = result.length - 1; i >= 0; i--) {
result[i] = stack.pop();
}
return result;
}
}
Input
Output
#3 Code Example with Javascript Programming
Code -
Javascript Programming
const asteroidCollision = function(asteroids) {
const positive = []
const res = []
for(let i = 0; i < asteroids.length; i++) {
if (asteroids[i] > 0) {
positive.push(i)
} else {
const negVal = asteroids[i];
while(positive.length > 0 && asteroids[ positive[positive.length - 1] ] + negVal < 0 ) {
asteroids[ positive[positive.length - 1] ] = undefined
positive.pop()
}
if (positive.length > 0) {
if (asteroids[ positive[positive.length - 1] ] + negVal > 0) {
asteroids[i] = undefined
} else if(asteroids[ positive[positive.length - 1] ] + negVal === 0) {
asteroids[i] = undefined
asteroids[ positive[positive.length - 1] ] = undefined
positive.pop()
}
}
}
}
return asteroids.filter(el => el !== undefined)
};
Input
Output
#4 Code Example with Python Programming
Code -
Python Programming
class Solution:
def asteroidCollision(self, asteroids):
"""
:type asteroids: List[int]
:rtype: List[int]
"""
stack = []
for asteroid in asteroids:
stack.append(asteroid)
while len(stack) > 1 and stack[-2] > 0 and stack[-1] < 0:
if stack[-2] < abs(stack[-1]): stack[-2] = stack[-1]
elif stack[-2] == abs(stack[-1]): stack.pop()
stack.pop()
return stack
Input
Output
#5 Code Example with C# Programming
Code -
C# Programming
using System.Collections.Generic;
namespace LeetCode
{
public class _0735_AsteroidCollision
{
public int[] AsteroidCollision(int[] asteroids)
{
var stack = new Stack < int>();
foreach (var asteroid in asteroids)
{
var explodeBoth = false;
while (stack.Count > 0 && asteroid < 0 && stack.Peek() > 0)
{
if (stack.Peek() + asteroid < 0)
{
stack.Pop();
continue;
}
else if (stack.Peek() + asteroid == 0)
stack.Pop();
explodeBoth = true;
break;
}
if (!explodeBoth)
stack.Push(asteroid);
}
var result = new int[stack.Count];
for (int i = stack.Count - 1; i >= 0; i--)
result[i] = stack.Pop();
return result;
}
}
}
Input
Output