Problem URL: LRU cache

We will use a hashmap to lookup the key in out cache and a link list to keep track of most and least used cache item.

# Definition for dubly-linked list.
class Node:
    def __init__(self, key, val):
        self.key = key
        self.val = val
        self.prev = self.next = None

class LRUCache:
    def __init__(self, capacity: int):
        self.capacity = capacity
        self.cache = {}

        self.left = Node(0, 0)  # LRU
        self.right = Node(0, 0)  # MRU
        self.left.next = self.right
        self.right.prev = self.left

    def remove(self, node):
        prev, nxt = node.prev, node.next
        prev.next, nxt.prev = nxt, prev

    def insert(self, node):
        prev, nxt = self.right.prev, self.right
        prev.next = nxt.prev = node
        node.next, node.prev = nxt, prev

    def get(self, key: int) -> int:
        if key in self.cache:
            self.remove(self.cache[key])
            self.insert(self.cache[key])
            return self.cache[key].val
        return -1

    def put(self, key: int, value: int) -> None:
        if key in self.cache:
            self.remove(self.cache[key])
        self.cache[key] = Node(key, value)
        self.insert(self.cache[key])

        if len(self.cache) > self.capacity:
            lru = self.left.next
            self.remove(lru)
            del self.cache[lru.key]

# Your LRUCache object will be instantiated and called as such:
# obj = LRUCache(capacity)
# param_1 = obj.get(key)
# obj.put(key,value)

Time Complexity: O(1) for each operation
Space Complexity: O(k), k is the size of the cache