# LeetCode 225. Implement Stack using Queues

## Description

https://leetcode.com/problems/implement-queue-using-stacks/

Implement a last in first out (LIFO) stack using only two queues. The implemented stack should support all the functions of a normal queue (`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 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 such that each operation is amortized`O(1)` time complexity? In other words, performing `n` operations will take overall `O(n)` time even if one of those operations may take longer. You can use more than two queues.

## Explanation

Using two queues, one queue is for getting items, the other queue is only for internal conversion to get the last item.

## Python Solution

``````import queue
class MyStack:

def __init__(self):
"""
Initialize your data structure here.
"""
self.queue1 = queue.Queue()
self.queue2 = queue.Queue()

def push(self, x: int) -> None:
"""
Push element x onto stack.
"""
self.queue1.put(x)

def pop(self) -> int:
"""
Removes the element on top of the stack and returns that element.
"""
while self.queue1.qsize() > 1:
self.queue2.put(self.queue1.get())

top = self.queue1.get()

self.queue1, self.queue2 = self.queue2, self.queue1

def top(self) -> int:
"""
Get the top element.
"""
while self.queue1.qsize() > 1:
self.queue2.put(self.queue1.get())

top = self.queue1.get()
self.queue2.put(top)

self.queue1, self.queue2 = self.queue2, self.queue1

def empty(self) -> bool:
"""
Returns whether the stack is empty.
"""
return not self.queue1.qsize()

# 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()``````
• Time complexity: O(N).
• Space complexity: O(1).