"""
146. LRU Cache
Difficulty: Medium
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 <= 10^4

        • 0 <= value <= 10^5

        • At most 2 * 10^5 calls will be made to get and put.
"""


class Node:
    def __init__(self, key=0, value=0):
        self.key = key
        self.value = value
        self.prev = None
        self.next = None


class LRUCache:
    def __init__(self, capacity: int):
        self.capacity = capacity
        self.cache = {}
        self.head = Node()
        self.tail = Node()
        self.head.next = self.tail  # pyright: ignore[reportAttributeAccessIssue]
        self.tail.prev = self.head  # pyright: ignore[reportAttributeAccessIssue]

    def _remove_node(self, node):
        prev_node = node.prev
        next_node = node.next
        prev_node.next = next_node
        next_node.prev = prev_node

    def _add_node(self, node):
        # Insert just before the tail (most recent at end)
        node.next = self.tail
        node.prev = self.tail.prev
        self.tail.prev.next = node  # pyright: ignore[reportAttributeAccessIssue]
        self.tail.prev = node

    def get(self, key: int) -> int:
        if key not in self.cache:
            return -1

        node = self.cache[key]
        self._remove_node(node)
        self._add_node(node)
        return node.value

    def put(self, key: int, value: int) -> None:
        if key in self.cache:
            node = self.cache[key]
            node.value = value
            self._remove_node(node)
            self._add_node(node)
        else:
            new_node = Node(key, value)
            self.cache[key] = new_node
            self._add_node(new_node)

            if len(self.cache) > self.capacity:
                lru = self.head.next  # Oldest node is next to the head
                self._remove_node(lru)
                del self.cache[lru.key]  # pyright: ignore[reportAttributeAccessIssue]


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