Leetcode 146. LRU Cache

Problem

Leetcode 146 - LRU Cache

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);    // 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

Approach

Here is the Python code for the solution:

class Node:
    def __init__(self, key, val):
        self.key = key
        self.val = val
        self.prev = self.next = None

class LRUCache:
    # initialize variables
    def __init__(self, capacity:int):
        self.cap = capacity
        self.cache = {}
        self.left, self.right = Node(0, 0), Node(0, 0)
        self.left.next = self.right
        self.right.prev = self.left
    
    # removing node
    def remove(self, node):
        prev, nxt = node.prev, node.next
        prev.next, nxt.prev = nxt, prev
    
    # inserting node to right, since it's recently used
    def insert(self, node):
        prev, nxt = self.right.prev, self.right
        prev.next = nxt.prev = node
        node.next, node.prev = nxt, prev

    # get value using key in dict, update the linked list
    def get(self, key: int):
        if key in self.cache:
            self.remove(self.cache[key])
            self.insert(self.cache[key])
            return self.cache[key].val
        
        return -1
    
    # put value in node, and dict update the linked list
    def put(self, key: int, val: int):
        if key in self.cache:
            self.remove(self.cache[key])
        self.cache[key] = Node(key, val)
        self.insert(self.cahce[key])

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

Time Complexity and Space Complexity

Time Complexity: $O(1)$ for all operations

Space Complexity: $O(n)$