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. Follow up: Could you do get and put in O(1) 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 <= 3000 0 <= value <= 104 At most 3 * 104 calls will be made to get and put.