Design a queue-like data structure that moves the most recently used element to the end of the queue.

Implement the MRUQueue class:

• MRUQueue(int n) constructs the MRUQueue with n elements: [1,2,3,...,n].
• int fetch(int k) moves the kth element (1-indexed) to the end of the queue and returns it.

 

Example 1:

Input:
["MRUQueue", "fetch", "fetch", "fetch", "fetch"]
[[8], [3], [5], [2], [8]]
Output:
[null, 3, 6, 2, 2]

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

 

Constraints:

• 1 <= n <= 2000
• 1 <= k <= n
• At most 2000 calls will be made to fetch.

 

Follow up: Finding an O(n) algorithm per fetch is a bit easy. Can you find an algorithm with a better complexity for each fetch call?