Problem

Design a data structure that supports get(key) and put(key, value) in O(1) time, with a fixed capacity that evicts the least recently used entry when it’s full.

Key Insight

O(1) get needs a hash map (key → node). O(1) eviction of the least-recently-used entry needs a structure that can move an arbitrary node to the “most recent” end in O(1) — a doubly linked list does that, because removing and re-inserting a node only touches its neighbors.

Combine both: a hash map from key to a node in a doubly linked list, ordered by recency.

Approach

1. Maintain a doubly linked list with head (most recently used) and tail (least recently used) sentinel nodes.
2. Maintain a hash map from key → list node.
3. get(key): if present, unlink the node and re-insert it at the head, return its value.
4. put(key, value): if the key exists, update its value and move it to the head. Otherwise insert a new node at the head; if capacity is exceeded, remove the node just before the tail sentinel and delete it from the map.

Complexity

• Time: O(1) for both get and put.
• Space: O(capacity) for the map and the linked list nodes.

Implementation

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
        self.tail.prev = self.head

    def _remove(self, node: Node) -> None:
        node.prev.next = node.next
        node.next.prev = node.prev

    def _insert_at_head(self, node: Node) -> None:
        node.next = self.head.next
        node.prev = self.head
        self.head.next.prev = node
        self.head.next = node

    def get(self, key: int) -> int:
        if key not in self.cache:
            return -1
        node = self.cache[key]
        self._remove(node)
        self._insert_at_head(node)
        return node.value

    def put(self, key: int, value: int) -> None:
        if key in self.cache:
            self._remove(self.cache[key])

        node = Node(key, value)
        self.cache[key] = node
        self._insert_at_head(node)

        if len(self.cache) > self.capacity:
            lru = self.tail.prev
            self._remove(lru)
            del self.cache[lru.key]

Walkthrough

For capacity = 2: put(1, 1), put(2, 2), get(1) → 1 (moves key 1 to the head), put(3, 3) evicts key 2 (now the least recently used), get(2) → -1.

Edge Cases

• capacity == 0: every put should immediately be a no-op or evict itself; guard for this.
• Updating an existing key’s value should still mark it as most recently used.
• Repeated get calls on the same key should not affect other keys’ order.

Alternative Solutions

• Language built-ins: Python’s OrderedDict supports move_to_end and popitem(last=False), giving O(1) operations with far less code — good to mention, but interviewers usually want the hash map + linked list approach to confirm you understand why it’s O(1).

Interview Follow-ups

• How would you make this thread-safe?
• How would you implement an LFU (least-frequently-used) cache instead?
• How would you shard this cache across multiple machines?