# 225-implement-stack-using-queues


Try it on <a href='https://leetcode.com/problems/225-implement-stack-using-queues'>leetcode</a>

## Description
<div class="description">
<div><p>Implement a last-in-first-out (LIFO) stack using only two queues. The implemented stack should support all the functions of a normal stack (<code>push</code>, <code>top</code>, <code>pop</code>, and <code>empty</code>).</p>

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

<ul>
	<li><code>void push(int x)</code> Pushes element x to the top of the stack.</li>
	<li><code>int pop()</code> Removes the element on the top of the stack and returns it.</li>
	<li><code>int top()</code> Returns the element on the top of the stack.</li>
	<li><code>boolean empty()</code> Returns <code>true</code> if the stack is empty, <code>false</code> otherwise.</li>
</ul>

<p><b>Notes:</b></p>

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

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

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

<strong>Explanation</strong>
MyStack myStack = new MyStack();
myStack.push(1);
myStack.push(2);
myStack.top(); // return 2
myStack.pop(); // return 2
myStack.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> calls will be made to <code>push</code>, <code>pop</code>, <code>top</code>, and <code>empty</code>.</li>
	<li>All the calls to <code>pop</code> and <code>top</code> are valid.</li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow-up:</strong> Can you implement the stack using only one queue?</p>
</div>
</div>

## Solution(Python)
```Python
class MyStack:
    def __init__(self):
        self._queue = collections.deque()

    def push(self, x):
        q = self._queue
        q.append(x)
        for _ in range(len(q) - 1):
            q.append(q.popleft())

    def pop(self):
        return self._queue.popleft()

    def top(self):
        return self._queue[0]

    def empty(self):
        return not len(self._queue)


# 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()

```