leetcode-146 LRU Cache

# 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 if key exists, otherwise return -1.
# - void put(int key, int value) Update the value if key exists. Otherwise, add
#   the key-value pair. If the number of keys exceeds capacity, evict the LRU key.
#
# The functions get and put must each run in O(1) average time complexity.
#
# 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]
#
# 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 DLLNode:
    """Doubly Linked List 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 = {}  # key -> DLLNode

        # Dummy head and tail for easier manipulation
        self.head = DLLNode()
        self.tail = DLLNode()
        self.head.next = self.tail
        self.tail.prev = self.head

    def _remove(self, node: DLLNode) -> None:
        """Remove node from linked list"""
        node.prev.next = node.next
        node.next.prev = node.prev

    def _add_to_head(self, node: DLLNode) -> None:
        """Add node right after head (most recently used)"""
        node.next = self.head.next
        node.prev = self.head
        self.head.next.prev = node
        self.head.next = node

    def _move_to_head(self, node: DLLNode) -> None:
        """Move existing node to head"""
        self._remove(node)
        self._add_to_head(node)

    def _remove_tail(self) -> DLLNode:
        """Remove and return node before tail (LRU)"""
        node = self.tail.prev
        self._remove(node)
        return node

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

        node = self.cache[key]
        self._move_to_head(node)  # Mark as recently used
        return node.value

    def put(self, key: int, value: int) -> None:
        if key in self.cache:
            # Update existing key
            node = self.cache[key]
            node.value = value
            self._move_to_head(node)
        else:
            # Add new key
            node = DLLNode(key, value)
            self.cache[key] = node
            self._add_to_head(node)

            if len(self.cache) > self.capacity:
                # Evict LRU
                lru = self._remove_tail()
                del self.cache[lru.key]


# Using OrderedDict (Python 3.7+)
from collections import OrderedDict

class LRUCache2:
    def __init__(self, capacity: int):
        self.capacity = capacity
        self.cache = OrderedDict()

    def get(self, key: int) -> int:
        if key not in self.cache:
            return -1
        # Move to end (most recently used)
        self.cache.move_to_end(key)
        return self.cache[key]

    def put(self, key: int, value: int) -> None:
        if key in self.cache:
            # Move to end
            self.cache.move_to_end(key)
        self.cache[key] = value

        if len(self.cache) > self.capacity:
            # Remove first item (LRU)
            self.cache.popitem(last=False)

# Both approaches: Time O(1) for get and put, Space O(capacity)
#
# Data Structure Choice:
# - Hash Map: O(1) access to any key
# - Doubly Linked List: O(1) removal and insertion
# - Head: Most recently used
# - Tail: Least recently used
#
# Operations:
# - get(): Move accessed node to head
# - put(): Add/update node at head, evict from tail if needed
#
# Why Doubly Linked List?
# - Need to remove from middle (requires prev pointer)
# - Need to add to front (requires next pointer)
# - Singly linked wouldn't work efficiently
#
# AirBnB: Extremely common, tests system design knowledge
# Follow-up questions: Thread safety, TTL, cache invalidation