Implement an LRU Cache

Problem Statement

Design and implement a data structure for a Least Recently Used (LRU) Cache. It should support the following operations: get and put.

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 get and put operations must each run in average O(1) time complexity.

Interface

class LRUCache:
    def __init__(self, capacity: int):
        pass

    def get(self, key: int) -> int:
        pass

    def put(self, key: int, value: int) -> None:
        pass

Example

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);    // returns 1, cache is {2=2, 1=1}
lRUCache.put(3, 3); // evicts key 2, cache is {1=1, 3=3}
lRUCache.get(2);    // returns -1 (not found)
lRUCache.put(4, 4); // evicts key 1, cache is {3=3, 4=4}
lRUCache.get(1);    // returns -1 (not found)
lRUCache.get(3);    // returns 3
lRUCache.get(4);    // returns 4

Constraints

• 1 <= capacity <= 3000
• 0 <= key <= 10,000
• 0 <= value <= 10^5
• At most 2 * 10^5 calls will be made to get and put.