### LeetCode Challenge #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] <= 1000`
• `asteroids[i] != 0`
##### Video Solution
###### C++ Solution
```				```
class Solution {
public:
vector<int> asteroidCollision(vector<int>& asteroids) {

stack<int> positive;
stack<int> negative;

for(int i=asteroids.size()-1; i>=0; i--)
{
if(asteroids[i]>0)
{
if(negative.empty()) positive.push(asteroids[i]);

else
{
while(!negative.empty() and (asteroids[i]- abs(negative.top()))>0)negative.pop();

if(negative.empty()) positive.push(asteroids[i]);
else
{
if(asteroids[i]==abs(negative.top())) negative.pop();
}
}
}
else negative.push(asteroids[i]);
}

vector<int> ans;

while(!negative.empty())
{
ans.push_back(negative.top());
negative.pop();
}
while(!positive.empty())
{
ans.push_back(positive.top());
positive.pop();
}
return ans;
}
};
```
```
###### Code Explanation

The purpose of this function is to stimulate the collisions of asteroids moving to the right and determine the resulting state of the asteroids after collisions.

1. We will first initialise two stacks, ‘positive’ and ‘negative’, to represent the asteroids moving to the right (positive direction) and those moving to the left (negative direction), respectively.
2. We will make a for loop, for iterating on the array of asteroids in reverse order, starting from the last asteroid in the input vector.
• If the current asteroid is moving to the right (‘asteroids [i] > 0), the algorithm handles collisions with asteroids moving to the left (‘negative’ stack).
• And if the ‘negative’ stack is empty, the current positive asteroid is pushed onto the ‘positive’ stack.
1. If the ‘negative’ stack is not empty, a collision check is performed. If the absolute value is greater then will pop the asteroids until a collision occurs or the ‘negative’ stack becomes empty.
• if the ‘negative’ stack becomes empty, the current positive asteroid is pushed onto the ‘positive’ stack.
• if a collision occurs, and the current positive asteroid is not destroyed, it is also pushed onto the ‘positive’ stack.
1. Now, we have handled the positive and negative cases successfully, after this, we just have to push the answer.
• First, we have pushed the answer to ‘negative’ and popped it
• similarly, with ‘positive’ also, we will push back (positive.top)
1. Finally, the answer is returned