Leetcode 146. LRU Cache

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

get(key) – Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1.
put(key, value) – Set or insert the value if the key is not already present. When the cache reached its capacity, it should invalidate the least recently used item before inserting a new item.

The cache is initialized with a positive capacity.

Follow up:
Could you do both operations in O(1) time complexity?

Example:

LRUCache cache = new LRUCache( 2 /* capacity */ );

cache.put(1, 1);
cache.put(2, 2);
cache.get(1);       // returns 1
cache.put(3, 3);    // evicts key 2
cache.get(2);       // returns -1 (not found)
cache.put(4, 4);    // evicts key 1
cache.get(1);       // returns -1 (not found)
cache.get(3);       // returns 3
cache.get(4);       // returns 4

Difficulty: Medium

Solution:

class LRUCache:

    def __init__(self, capacity: int):
        self.d = OrderedDict()
        self.cap = capacity

    def get(self, key: int) -> int:
        if key not in self.d:
            return -1
        # make sure we put this key to the very end of OrderedDict
        val = self.d.pop(key)
        self.d[key] = val
        return val

    def put(self, key: int, value: int) -> None:
        if key in self.d:
            self.d.pop(key)
        if len(self.d) == self.cap:
            self.d.popitem(last=False)
        self.d[key] = value