Went with a hashmap plus a doubly linked list, which is the standard answer, and I knew it going in.
Start by clarifying requirements (capacity N, O(1) get/put, eviction policy). Then propose a hash map combined with a doubly linked list to achieve O(1) operations, and walk through the implementation details, including edge cases. Finally, analyze time and space complexity and discuss potential optimizations or trade-offs.
Pro tip: Mention that using a doubly linked list allows O(1) removal and insertion, and that the hash map stores references to list nodes. Also, discuss thread-safety if the cache might be accessed concurrently, as Intuit values production-ready thinking.
Confirm the capacity N, that get and put must be O(1), and the eviction policy (least recently used). Ask about concurrency requirements and whether null values are allowed.
Propose a hash map (for O(1) key lookup) and a doubly linked list (for O(1) insertion/removal and maintaining recency order). Explain that the hash map maps keys to nodes in the list.
Describe get: if key exists, move node to front (most recently used) and return value; else return -1. Describe put: if key exists, update value and move to front; else create new node, add to front, and if capacity exceeded, remove least recently used (tail) and delete from map.
State that both get and put are O(1) time due to hash map and linked list operations. Space complexity is O(N) for storing up to N entries.
Mention handling of capacity 0 or 1, updating existing keys, and potential thread-safety (e.g., using locks or concurrent data structures). Optionally, compare with alternative implementations like using LinkedHashMap in Java.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Start by explaining the core data structures: a frequency map where each frequency points to a doubly linked list of nodes (LRU order), and a node-to-frequency mapping. Then describe how get and put update frequencies by moving nodes between lists, and how to find the minimum frequency in O(1) using a min_freq pointer. Finally, explain tie-breaking by maintaining recency order within each frequency list.
Pro tip: Mention that LFU with recency tie-breaking is essentially a combination of LFU and LRU, and that using a min_freq pointer avoids scanning for the least frequency. Also, note that amortized O(1) is achieved because each operation moves a node at most once.
Confirm that operations must be O(1) or amortized O(1), and that ties are broken by recency (LRU). Ask about cache size and whether frequencies can be updated on both get and put.
Use a hash map for key-to-node lookup, a frequency map mapping frequency to a doubly linked list of nodes (ordered by recency), and maintain a min_freq variable. Each node stores key, value, freq, and pointers for the linked list.
On get, if key exists, retrieve node, increment its frequency, move it from its current frequency list to the next frequency list (or create new), update min_freq if necessary, and return value. If not, return -1.
On put, if key exists, update value and increment frequency similarly to get. If new, insert with frequency 1; if cache is full, evict the least frequently used node (from min_freq list, least recently used among them), then insert new node. Update min_freq to 1.
Within each frequency list, maintain recency order by adding new nodes at the head and moving accessed nodes to the head of the next frequency list. This ensures ties are broken by LRU. All operations are O(1) amortized because each node moves at most once per operation.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.