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 BinaryIndexedTree {
    private int n;
    private int[] c;

    public BinaryIndexedTree(int n) {
        this.n = n;
        this.c = new int[n + 1];
    }

    public void update(int x, int v) {
        while (x <= n) {
            c[x] += v;
            x += x & -x;
        }
    }

    public int query(int x) {
        int s = 0;
        while (x > 0) {
            s += c[x];
            x -= x & -x;
        }
        return s;
    }
}

class MRUQueue {
    private int n;
    private int[] q;
    private BinaryIndexedTree tree;

    public MRUQueue(int n) {
        this.n = n;
        q = new int[n + 2010];
        for (int i = 1; i <= n; ++i) {
            q[i] = i;
        }
        tree = new BinaryIndexedTree(n + 2010);
    }

    public int fetch(int k) {
        int l = 1, r = n;
        while (l < r) {
            int mid = (l + r) >> 1;
            if (mid - tree.query(mid) >= k) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        int x = q[l];
        q[++n] = x;
        tree.update(l, 1);
        return x;
    }
}

/**
 * Your MRUQueue object will be instantiated and called as such:
 * MRUQueue obj = new MRUQueue(n);
 * int param_1 = obj.fetch(k);
 */