Asteroid Collision

1. Clarify the problem: Before diving into solving the problem, let's clarify the requirements:

• We are given an array of integers representing asteroids moving in the same direction.
• The positive integers represent asteroids moving to the right, and the negative integers represent asteroids moving to the left.
• Asteroids collide when they meet. When two asteroids collide, the smaller asteroid is destroyed, and the larger asteroid continues moving in the same direction.
• We need to simulate the collisions and return the state of the asteroids after all collisions are resolved.

2. Analyze the problem: To solve this problem, we can use a stack data structure to simulate the collisions between asteroids.

3. Design an algorithm: Here is the algorithm to solve the problem:

1. Initialize an empty stack to store the asteroids.
2. Iterate through the given array of asteroids.
   • If the stack is empty or the current asteroid is moving to the right (positive), push it onto the stack.
   • If the current asteroid is moving to the left (negative):
     • While the stack is not empty and the top asteroid on the stack is moving to the right (positive) and its absolute value is less than the absolute value of the current asteroid, pop the top asteroid from the stack (collision occurs, destroying the smaller asteroid).
       • If the absolute values are equal, pop the top asteroid as well as the current asteroid (both asteroids are destroyed).
     • If the stack is empty or the top asteroid on the stack is moving to the left (negative), push the current asteroid onto the stack.
3. After iterating through all the asteroids, the stack will contain the remaining asteroids after all collisions are resolved. Return the asteroids in the stack as the result.

4. Explain your approach: The approach involves simulating the collisions between asteroids using a stack. We iterate through the given array of asteroids and handle the cases where asteroids collide based on their directions and sizes. We push and pop asteroids from the stack accordingly, resulting in the final state of the asteroids after all collisions are resolved.

5. Write clean and readable code:

python

def asteroidCollision(asteroids):
    stack = []

    for asteroid in asteroids:
        if asteroid > 0 or not stack:
            stack.append(asteroid)
        else:
            while stack and stack[-1] > 0 and abs(stack[-1]) < abs(asteroid):
                stack.pop()

            if not stack or stack[-1] < 0:
                stack.append(asteroid)
            elif stack[-1] == -asteroid:
                stack.pop()

    return stack

# Test case
asteroids = [5, 10, -5]
result = asteroidCollision(asteroids)
print(result)

6. Test your code: Let's test the code with the given test case:

python

asteroids = [5, 10, -5]
result = asteroidCollision(asteroids)
print(result)

The expected output is [5, 10] since the first and second asteroids collide, and the second asteroid survives.

7. Optimize if necessary: The current solution has a time complexity of O(n) since we iterate through the asteroids once. There are no further optimizations needed for this problem.

8. Handle error cases: The code assumes that the input array asteroids is not None and contains integers. If the array is empty or None, the code will return an empty list.

9. Discuss complexity analysis:

• The time complexity of the solution is O(n), where n is the number of asteroids in the input array.
• The space complexity is O(n) since the stack can store up to n asteroids in the worst case, where n is the number of asteroids.