# 232-implement-queue-using-stacks


Try it on <a href='https://leetcode.com/problems/232-implement-queue-using-stacks'>leetcode</a>

## Description
<div class="description">
<div><p>Implement a first in first out (FIFO) queue using only two stacks. The implemented queue should support all the functions of a normal queue (<code>push</code>, <code>peek</code>, <code>pop</code>, and <code>empty</code>).</p>

<p>Implement the <code>MyQueue</code> class:</p>

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

<p><strong>Notes:</strong></p>

<ul>
	<li>You must use <strong>only</strong> standard operations of a stack, which means only <code>push to top</code>, <code>peek/pop from top</code>, <code>size</code>, and <code>is empty</code> operations are valid.</li>
	<li>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.</li>
</ul>

<p>&nbsp;</p>
<p><strong>Example 1:</strong></p>

<pre><strong>Input</strong>
["MyQueue", "push", "push", "peek", "pop", "empty"]
[[], [1], [2], [], [], []]
<strong>Output</strong>
[null, null, null, 1, 1, false]

<strong>Explanation</strong>
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
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= x &lt;= 9</code></li>
	<li>At most <code>100</code>&nbsp;calls will be made to <code>push</code>, <code>pop</code>, <code>peek</code>, and <code>empty</code>.</li>
	<li>All the calls to <code>pop</code> and <code>peek</code> are valid.</li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow-up:</strong> Can you implement the queue such that each operation is <strong><a href="https://en.wikipedia.org/wiki/Amortized_analysis" target="_blank">amortized</a></strong> <code>O(1)</code> time complexity? In other words, performing <code>n</code> operations will take overall <code>O(n)</code> time even if one of those operations may take longer.</p>
</div>
</div>

## Solution(Python)
```Python
class MyQueue:
    def __init__(self):
        self.stack = []
        self.queue = []

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

    def pop(self) -> int:
        self.move()
        if self.queue:
            popped = self.queue.pop()
        return popped

    def peek(self) -> int:
        self.move()
        return self.queue[-1]

    def empty(self) -> bool:
        return not self.stack and not self.queue

    def move(self) -> None:
        if not self.queue:
            while self.stack:
                self.queue.append(self.stack.pop())


# 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()

```