Implement Stack using Queues

Solution

There is no need to implement using 2 queues, as a queue does not change the order of elements after popping and pushing.

One Queue

There are 2 similar implementations w/ similar runtime: built-in list [] or collections.deque.

Complexity

time: push $O(1)$, pop $O(n)$
space: $O(1)$

Built-in List

class MyStack:

    def __init__(self):
        """
        Initialize your data structure here.
        """
        self._q = []

    def push(self, x: int) -> None:
        """
        Push element x onto stack.
        """
        self._q.append(x)
        for _ in range(1,len(self._q)):
            self._q.append(self._q[0])
            self._q.pop(0)

    def pop(self) -> int:
        """
        Removes the element on top of the stack and returns that element.
        """
        return self._q.pop(0)

    def top(self) -> int:
        """
        Get the top element.
        """
        return self._q[0]

    def empty(self) -> bool:
        """
        Returns whether the stack is empty.
        """
        return not self._q

Deque

from collections import deque

class MyStack:

    def __init__(self):
        """
        Initialize your data structure here.
        """
        self._q = deque()

    def push(self, x: int) -> None:
        """
        Push element x onto stack.
        """
        self._q.append(x)
        for _ in range(1,len(self._q)):
            self._q.append(self._q.popleft())

    def pop(self) -> int:
        """
        Removes the element on top of the stack and returns that element.
        """
        return self._q.popleft()

    def top(self) -> int:
        """
        Get the top element.
        """
        return self._q[0]

    def empty(self) -> bool:
        """
        Returns whether the stack is empty.
        """
        return not self._q