We can implement this using a single queue. When pushing an element, we add it to the queue and then rotate the queue (pop and re-append) n-1 times so that the newest element is at the front.

1. push(x):
   • Add x to the queue.
   • For size - 1 times, dequeue the front element and enqueue it again.
2. pop(): Dequeue from the queue.
3. top(): Peek at the front of the queue.
4. empty(): Check if queue is empty.

• Time Complexity:
  • Push: O(N)
  • Pop: O(1)
• Space Complexity: O(N).

from collections import deque

class MyStack:
    def __init__(self):
        self.queue = deque()

    def push(self, x: int) -> None:
        self.queue.append(x)
        for _ in range(len(self.queue) - 1):
            self.queue.append(self.queue.popleft())

    def pop(self) -> int:
        return self.queue.popleft()

    def top(self) -> int:
        return self.queue[0]

    def empty(self) -> bool:
        return not self.queue