# Implement Stack using Queues

Implement a last-in-first-out (LIFO) stack using only two queues. The implemented stack should support all the functions of a normal stack (`push`, `top`, `pop`, and `empty`).

Implement the `MyStack` class:

• `void push(int x)` Pushes element x to the top of the stack.
• `int pop()` Removes the element on the top of the stack and returns it.
• `int top()` Returns the element on the top of the stack.
• `boolean empty()` Returns `true` if the stack is empty, `false` otherwise.

Notes:

• You must use only standard operations of a queue, which means that only `push to back`, `peek/pop from front`, `size` and `is empty` operations are valid.
• Depending on your language, the queue may not be supported natively. You may simulate a queue using a list or deque (double-ended queue) as long as you use only a queue's standard operations.

Example 1:

Input
["MyStack", "push", "push", "top", "pop", "empty"]
[[], , , [], [], []]
Output
[null, null, null, 2, 2, false]

Explanation
MyStack myStack = new MyStack();
myStack.push(1);
myStack.push(2);
myStack.top(); // return 2
myStack.pop(); // return 2
myStack.empty(); // return False

Constraints:

• `1 <= x <= 9`
• At most `100` calls will be made to `push`, `pop`, `top`, and `empty`.
• All the calls to `pop` and `top` are valid.

Follow-up: Can you implement the stack using only one queue?

SOLUTION:

``````class MyStack:

def __init__(self):
self.stack = []

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

def pop(self) -> int:
if not self.empty():
return self.stack.pop()

def top(self) -> int:
if not self.empty():
return self.stack[-1]

def empty(self) -> bool:
return len(self.stack) == 0

# Your MyStack object will be instantiated and called as such:
# obj = MyStack()
# obj.push(x)
# param_2 = obj.pop()
# param_3 = obj.top()
# param_4 = obj.empty()
``````