# Solving Leetcode Interviews in Seconds with AI: Implement Stack using Queues


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "225" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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"] [[], [1], [2], [], [], []] 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? 

	# Explanation
	Here's the breakdown of the problem and the Python code:

*   **Approach:**
    *   Use two queues, `q1` and `q2`. The main idea is to always maintain the stack elements in `q1`.
    *   When pushing a new element, push it into `q2`, then move all elements from `q1` to `q2`, and finally swap `q1` and `q2`. This ensures LIFO order in `q1`.
    *   `top()` and `pop()` simply access/remove from the front of `q1`.

*   **Complexity:**
    *   Runtime: push - O(n), pop - O(1), top - O(1), empty - O(1). Storage: O(n), where n is the number of elements in the stack.

	
	# Code
	```python
	from collections import deque

class MyStack:

    def __init__(self):
        self.q1 = deque()
        self.q2 = deque()

    def push(self, x: int) -> None:
        self.q2.append(x)
        while self.q1:
            self.q2.append(self.q1.popleft())
        self.q1, self.q2 = self.q2, self.q1

    def pop(self) -> int:
        return self.q1.popleft()

    def top(self) -> int:
        return self.q1[0]

    def empty(self) -> bool:
        return not self.q1


# 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()
	```