LRU Cache

Design a data structure that follows the constraints of a **Least Recently Used (LRU) cache**. Implement the `LRUCache` class: - `LRUCache(int capacity)` Initialize the LRU cache with **positive** size `capacity`. - `int get(int key)` Return the value of the `key` if the key exists, otherwise return `-1`. - `void put(int key, int value)` Update the value of the `key` if the `key` exists. Otherwise, add the `key-value` pair to the cache. If the number of keys exceeds the `capacity` from this operation, **evict** the least recently used key. The functions `get` and `put` must each run in `O(1)` average time complexity.   **Example 1:** **Input** ["LRUCache", "put", "put", "get", "put", "get", "put", "get", "get", "get"] [[2], [1, 1], [2, 2], [1], [3, 3], [2], [4, 4], [1], [3], [4]] **Output** [null, null, null, 1, null, -1, null, -1, 3, 4] **Explanation** LRUCache lRUCache = new LRUCache(2); lRUCache.put(1, 1); // cache is {1=1} lRUCache.put(2, 2); // cache is {1=1, 2=2} lRUCache.get(1); // return 1 lRUCache.put(3, 3); // LRU key was 2, evicts key 2, cache is {1=1, 3=3} lRUCache.get(2); // returns -1 (not found) lRUCache.put(4, 4); // LRU key was 1, evicts key 1, cache is {4=4, 3=3} lRUCache.get(1); // return -1 (not found) lRUCache.get(3); // return 3 lRUCache.get(4); // return 4 ```   **Constraints:** - `1 <= capacity <= 3000` - `0 <= key <= 104` - `0 <= value <= 105` - At most `2 * 105` calls will be made to `get` and `put`.