[link](https://leetcode.com/problems/lru-cache/)

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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.

The functions get and put must each run in O(1) average 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

#### Constraints:

- 1 <= capacity <= 3000
- 0 <= key <= 104
- 0 <= value <= 105
- At most 2 * 105 calls will be made to get and put.

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The Node class is defined to represent individual nodes in the doubly linked list. Each node contains a key and a val (value) attribute, as well as references to the previous and next nodes in the list.

The LRUCache class is defined to implement the LRU cache. It takes the capacity as an input parameter during initialization. The __init__ method initializes the cache's capacity, an empty dictionary cache, and two sentinel nodes left and right. The left and right nodes are initially connected to each other.

The remove method takes a node as input and removes it from the doubly linked list by updating the references of the previous and next nodes. The insert method takes a node as input and inserts it at the end of the doubly linked list, just before the right sentinel node.

The get method takes a key as input and retrieves the corresponding value from the cache. If the key is present in the cache, it removes the corresponding node from its current position in the linked list and inserts it at the end (to mark it as the most recently used). It then returns the value. If the key is not present, it returns -1.

The put method takes a key and a value as input and inserts them as a new key-value pair in the cache. If the key is already present, it removes the corresponding node from the linked list and inserts the new node at the end. If the cache has reached its capacity, it evicts the least recently used node (the node next to the left sentinel node) by removing it from the linked list and deleting it from the dictionary.

#### Solution 1

class Node:

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

class LRUCache:

    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.prev = self.right, self.left

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


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

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

get
O(T): O(1)

put
O(T): O(1)

O(S): O(cap)