Implement Queue using Stacks
<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&#39;s standard operations.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input</strong>
[&quot;MyQueue&quot;, &quot;push&quot;, &quot;push&quot;, &quot;peek&quot;, &quot;pop&quot;, &quot;empty&quot;]
[[], [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>
Stack; Design; Queue
class MyQueue {
    private Deque<Integer> stk1 = new ArrayDeque<>();
    private Deque<Integer> stk2 = new ArrayDeque<>();

    public MyQueue() {
    }

    public void push(int x) {
        stk1.push(x);
    }

    public int pop() {
        move();
        return stk2.pop();
    }

    public int peek() {
        move();
        return stk2.peek();
    }

    public boolean empty() {
        return stk1.isEmpty() && stk2.isEmpty();
    }

    private void move() {
        while (stk2.isEmpty()) {
            while (!stk1.isEmpty()) {
                stk2.push(stk1.pop());
            }
        }
    }
}

/**
 * Your MyQueue object will be instantiated and called as such:
 * MyQueue obj = new MyQueue();
 * obj.push(x);
 * int param_2 = obj.pop();
 * int param_3 = obj.peek();
 * boolean param_4 = obj.empty();
 */