Completed both the code

[avcode3]

No description provided.

[super30admin]

Strengths:

• Your code is simple and easy to understand.
• It correctly implements the FIFO behavior for the queue operations.

Areas for improvement:

• The problem requires using two stacks to implement the queue. Your solution does not use any stacks; instead, it uses a list and directly manipulates it. You need to adhere to the problem constraints, which mandate using only standard stack operations (push to top, pop from top, peek from top, size, and is empty).
• The pop operation in your solution has O(n) time complexity due to list slicing. This can be optimized by using two stacks as in the reference solution, which provides amortized O(1) time for pop and peek operations.
• Consider using two stacks: one for input and one for output. When you need to pop or peek, if the output stack is empty, transfer all elements from the input stack to the output stack. This way, the elements are reversed in the output stack, and you can pop from the top of the output stack to get the front of the queue.

Here is how you can improve your solution:

class MyQueue:
    def __init__(self):
        self.in_stack = []
        self.out_stack = []

    def push(self, x: int) -> None:
        self.in_stack.append(x)

    def pop(self) -> int:
        self.peek()  # Ensure out_stack has the front element
        return self.out_stack.pop()

    def peek(self) -> int:
        if not self.out_stack:
            while self.in_stack:
                self.out_stack.append(self.in_stack.pop())
        return self.out_stack[-1]

    def empty(self) -> bool:
        return not self.in_stack and not self.out_stack

This approach uses two stacks and meets the problem requirements. The pop and peek operations are efficient due to the amortized constant time complexity.

[super30admin]

Your implementation of the queue using two stacks is good overall. The push and empty methods are correct. The pop method correctly transfers elements from in_stack to out_stack when out_stack is empty, which ensures amortized O(1) time complexity.

However, there is an issue in the peek method. Currently, when out_stack is empty, you return in_stack[0]. This is not correct because the in_stack is a stack, and the bottom element (in_stack[0]) is indeed the front of the queue when out_stack is empty. But the problem requires using only standard stack operations, and accessing the bottom of the stack (in_stack[0]) is not a standard stack operation (which only allows push, pop, peek from top, size, and is empty). Instead, you should transfer all elements from in_stack to out_stack when out_stack is empty, and then return the top of out_stack. This is the standard approach and ensures that only stack operations are used.

Also, in the pop method, you should consider calling peek to ensure the out_stack is populated, which can avoid code duplication. The reference solution does this by having pop call peek first.

Here's how you can fix the peek method:

def peek(self) -> int:
    if not self.out_stack:
        while self.in_stack:
            self.out_stack.append(self.in_stack.pop())
    return self.out_stack[-1]

Similarly, you can simplify the pop method by using peek:

def pop(self) -> int:
    self.peek()  # Ensure out_stack has the front element
    return self.out_stack.pop()

This change makes the code more concise and avoids duplication.

Overall, your solution is close to correct, but the peek method needs adjustment to use only stack operations.