Implement an LRU cache

Question: Implement an LRU cache

Companies: Google, Meta, Amazon

Difficulty: Medium

Problem statement

Implement a least-recently-used cache with get(key) and put(key, value), both in O(1) time, with a fixed capacity.

Approach

Keep a hash map from key to node, and a doubly linked list ordered by recency (most recent at the head). get moves the node to the head; put inserts at the head and evicts from the tail when over capacity. Both are O(1).

Sentinel head and tail nodes remove the empty-list edge cases.

Code

class Node:
    __slots__ = ("key", "value", "prev", "next")

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


class LRUCache:
    def __init__(self, capacity: int):
        self.capacity = capacity
        self.map = {}
        self.head, self.tail = Node(), Node()
        self.head.next, self.tail.prev = self.tail, self.head

    def _remove(self, node):
        node.prev.next, node.next.prev = node.next, node.prev

    def _push_front(self, node):
        node.prev, node.next = self.head, self.head.next
        self.head.next.prev = node
        self.head.next = node

    def get(self, key: int) -> int:
        node = self.map.get(key)
        if not node:
            return -1
        self._remove(node)
        self._push_front(node)
        return node.value

    def put(self, key: int, value: int) -> None:
        if key in self.map:
            self._remove(self.map[key])
        node = Node(key, value)
        self.map[key] = node
        self._push_front(node)
        if len(self.map) > self.capacity:
            lru = self.tail.prev
            self._remove(lru)
            del self.map[lru.key]

Complexity

• get and put: O(1) time.
• Space: O(capacity).

Test cases

Follow-ups interviewers ask: make it thread-safe (a lock around both operations, or striped locks), add TTL expiry, or use OrderedDict.move_to_end in Python.