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
use std::collections::VecDeque;

struct MyQueue {
    stk1: Vec<i32>,
    stk2: Vec<i32>,
}

impl MyQueue {
    fn new() -> Self {
        MyQueue {
            stk1: Vec::new(),
            stk2: Vec::new(),
        }
    }

    fn push(&mut self, x: i32) {
        self.stk1.push(x);
    }

    fn pop(&mut self) -> i32 {
        self.move_elements();
        self.stk2.pop().unwrap()
    }

    fn peek(&mut self) -> i32 {
        self.move_elements();
        *self.stk2.last().unwrap()
    }

    fn empty(&self) -> bool {
        self.stk1.is_empty() && self.stk2.is_empty()
    }

    fn move_elements(&mut self) {
        if self.stk2.is_empty() {
            while let Some(element) = self.stk1.pop() {
                self.stk2.push(element);
            }
        }
    }
}
