# LeetCode 232. Implement Queue using Stacks

## Description

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

Implement a first in first out (FIFO) queue using only two stacks. The implemented queue should support all the functions of a normal queue (`push``peek``pop`, and `empty`).

Implement the `MyQueue` class:

• `void push(int x)` Pushes element x to the back of the queue.
• `int pop()` Removes the element from the front of the queue and returns it.
• `int peek()` Returns the element at the front of the queue.
• `boolean empty()` Returns `true` if the queue is empty, `false` otherwise.

Notes:

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

Follow-up: Can you implement the queue 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.

Example 1:

```Input
["MyQueue", "push", "push", "peek", "pop", "empty"]
[[], [1], [2], [], [], []]
Output
```

[null, null, null, 1, 1, false]

Explanation MyQueue myQueue = new MyQueue(); myQueue.push(1); // queue is: [1] myQueue.push(2); // queue is: [1, 2] (leftmost is front of the queue) myQueue.peek(); // return 1 myQueue.pop(); // return 1, queue is [2] myQueue.empty(); // return false

Constraints:

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

## Explanation

Using two stacks, one stack is used for appending items, the other stack is used for pop() and peek().

## Python Solution

``````
class MyQueue:

def __init__(self):
"""
"""
self.stack1 = []
self.stack2 = []

def push(self, x: int) -> None:
"""
Push element x to the back of queue.
"""

self.stack1.append(x)

def pop(self) -> int:
"""
Removes the element from in front of queue and returns that element.
"""
if self.stack2:
return self.stack2.pop()
else:
while self.stack1:
self.stack2.append(self.stack1.pop())

return self.stack2.pop()

def peek(self) -> int:
"""
Get the front element.
"""
if self.stack2:
return self.stack2[-1]
else:
while self.stack1:
self.stack2.append(self.stack1.pop())

return self.stack2[-1]

def empty(self) -> bool:
"""
Returns whether the queue is empty.
"""

return not self.stack1 and not self.stack2

# Your MyQueue object will be instantiated and called as such:
# obj = MyQueue()
# obj.push(x)
# param_2 = obj.pop()
# param_3 = obj.peek()
# param_4 = obj.empty()``````
• Time complexity: O(N).
• Space complexity: O(1).