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The following asteroid collision problem is from Leetcode.

### The problem

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:**

1 2 3 4 | 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:**

1 2 3 4 | Input: asteroids = [8,-8] Output: [] Explanation: The 8 and -8 collide exploding each other. |

**Example 3:**

1 2 3 4 | 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:**

1 2 3 4 | Input: asteroids = [-2,-1,1,2] Output: [-2,-1,1,2] Explanation: The -2 and -1 are moving left, while the 1 and 2 are moving right. Asteroids moving the same direction never meet, so no asteroids will meet each other. |

**Constraints:**

`1 <= asteroids <= 10`

^{4}`-1000 <= asteroids[i] <= 1000`

`asteroids[i] != 0`

## Solution

We process from left to the right each integer value:

- We create a list to store the result
`res`

- If a new asteroid value is positive, we add to the list as it will never collide with ones in
`res`

- Else, we remove all the positive asteroids that smaller than the new value (negative)
- Then add the new value to
`res`

if`res`

is empty or the last element is also negative (as it already processed in the previous run) - If we have two opposite asteroids of the same size, then remove the last one.

Python code for asteroid collision problem:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 | #Python code class Solution: def asteroidCollision(self, asteroids: List[int]) -> List[int]: res = [] for a in asteroids: if a > 0: res.append(a) else: while len(res) != 0 and res[-1] > 0 and res[-1] < abs(a): res.pop() if len(res) == 0 or res[-1] < 0: res.append(a) elif res[-1] == abs(a): res.pop() return res |

Java code for asteroid collision problem:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 | class Solution { public int[] asteroidCollision(int[] asteroids) { Stack<Integer> res = new Stack(); for(int a: asteroids){ if(a > 0) { res.push(a);//never collide with any in the stack } else { //check collision with all Positive asteroids that have higher asb value from the right while(!res.empty() && res.peek() > 0 && res.peek() < Math.abs(a)){ res.pop();//destroy all a* < a } //Add a if (A has max abs or empty stack or last element is neg) if(res.empty() || res.peek() < 0) { res.push(a); } else if(res.peek() == Math.abs(a)){ res.pop(); } } } return res.stream().mapToInt(i->i).toArray();//convert stack to int[] } } |

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