[Uber][LeetCode] Asteroid Collision

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:

Input: asteroids = [5,10,-5]
Output: [5,10]
Explanation: The 10 and -5 collide resulting in 10. The 5 and 10 never collide.

Input: asteroids = [8,-8]
Output: []
Explanation: The 8 and -8 collide exploding each other.

Input: asteroids = [10,2,-5]
Output: [10]
Explanation: The 2 and -5 collide resulting in -5. The 10 and -5 collide resulting in 10.

Approach: We can use stack here to solve the problem easily. Here are the steps which we need to take:

Push the current number in stack if stack is empty or current number is positive or number at the top of the stack is negative. (No collision)

In case of current number is negative:

1. Pop from stack until stack is empty and top of the stack is positive and current number's absolute value is more than the top of the stack. (Resolve all the collisions)
2. If stack is empty or top of the stack is negative then push the current number. (All the collisions are resolved, we can safely put the current number)
3. Else if top of the stack is equal to absolute value of current number than perform a pop.(In case of equal size, asteroids explode each other)

That's all!

Implementation in C#:

    public int[] AsteroidCollision(int[] asteroids)

{

int length = asteroids?.Length ?? 0;

if (length == 0)

{

return new int[]{};

}

Stack<int> stack = new Stack<int>();

foreach (int size in asteroids)

{

if (size >= 0 || stack.Count == 0 || stack.Peek() < 0)

{

stack.Push(size);

}

else

{

int prev = stack.Peek();

while (stack.Count > 0 &&

stack.Peek() > 0 &&

stack.Peek() < -size)

{

prev = stack.Pop();

}

if (stack.Count == 0 || stack.Peek() < 0)

{

stack.Push(size);

}

else if (stack.Peek() == -size)

{

stack.Pop();

}

}

}

int[] answer = new int[stack.Count];

int i = stack.Count - 1;

while (stack.Count > 0)

{

answer[i--] = stack.Pop();

}

return answer;

}

Complexity: O(n)