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]. 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?