# 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
