Design Most Recently Used Queue
<p>Design a queue-like data structure that moves the most recently used element to the end of the queue.</p>

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

<ul>
	<li><code>MRUQueue(int n)</code> constructs the <code>MRUQueue</code> with <code>n</code> elements: <code>[1,2,3,...,n]</code>.</li>
	<li><code>int fetch(int k)</code> moves the <code>k<sup>th</sup></code> element <strong>(1-indexed)</strong> to the end of the queue and returns it.</li>
</ul>

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

<pre>
<strong>Input:</strong>
[&quot;MRUQueue&quot;, &quot;fetch&quot;, &quot;fetch&quot;, &quot;fetch&quot;, &quot;fetch&quot;]
[[8], [3], [5], [2], [8]]
<strong>Output:</strong>
[null, 3, 6, 2, 2]

<strong>Explanation:</strong>
MRUQueue mRUQueue = new MRUQueue(8); // Initializes the queue to [1,2,3,4,5,6,7,8].
mRUQueue.fetch(3); // Moves the 3<sup>rd</sup> element (3) to the end of the queue to become [1,2,4,5,6,7,8,3] and returns it.
mRUQueue.fetch(5); // Moves the 5<sup>th</sup> element (6) to the end of the queue to become [1,2,4,5,7,8,3,6] and returns it.
mRUQueue.fetch(2); // Moves the 2<sup>nd</sup> element (2) to the end of the queue to become [1,4,5,7,8,3,6,2] and returns it.
mRUQueue.fetch(8); // The 8<sup>th</sup> element (2) is already at the end of the queue so just return it.
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 2000</code></li>
	<li><code>1 &lt;= k &lt;= n</code></li>
	<li>At most <code>2000</code> calls will be made to <code>fetch</code>.</li>
</ul>

<p>&nbsp;</p>
<strong>Follow up:</strong> Finding an <code>O(n)</code> algorithm per <code>fetch</code> is a bit easy. Can you find an algorithm with a better complexity for each <code>fetch</code> call?
Stack; Design; Binary Indexed Tree; Array; Hash Table; Ordered Set
class MRUQueue:
    def __init__(self, n: int):
        self.q = list(range(1, n + 1))

    def fetch(self, k: int) -> int:
        ans = self.q[k - 1]
        self.q[k - 1 : k] = []
        self.q.append(ans)
        return ans


# Your MRUQueue object will be instantiated and called as such:
# obj = MRUQueue(n)
# param_1 = obj.fetch(k)
