LRU Cache
An LRU cache evicts the least recently used item when capacity is full. The classic O(1) design combines a hash map with a doubly linked list.
Beginne mit diesem mentalen Modell
An LRU cache evicts the least recently used item when capacity is full. The classic O(1) design combines a hash map with a doubly linked list.
So funktioniert es – ohne unnötigen Jargon
The map gives O(1) key lookup.
The doubly linked list stores recency order so an accessed item can move to the front and the oldest item can be removed from the tail.
Lies den Algorithmus als Folge von Entscheidungen
Identify the invariant that must always remain true.
Follow the read/update path through the structure.
After a mutation, restore the invariant before returning.
Read a tiny input by hand and write down the state after every important operation. For LRU Cache, the goal is to make the invariant visible rather than to memorize code.
get/put O(1)
Do not quote this complexity mechanically. Ask which operation or input dimension it describes, then derive the cost from the loops, recursion depth, data-structure operations and any preprocessing assumptions.
Gut geeignet
- Choose it when its dominant operations match your workload.
- Compare memory overhead as well as Big-O.
Was typischerweise schiefgeht
- Memorizing complexity without understanding representation.
- Ignoring worst-case shape or resizing/rebalancing costs.
Struktur einer professionellen Antwort
The key insight is why two structures are needed: map for lookup, doubly linked list for O(1) recency updates and eviction.
What problem does LRU Cache solve?
An LRU cache evicts the least recently used item when capacity is full. The classic O(1) design combines a hash map with a doubly linked list.
Was sollte ich zur Komplexität erwähnen?
Its typical cost here is get/put O(1), but a complete answer should connect that cost to the representation, input constraints and the dominant operations.
Was ist der häufigste Fehler?
Memorizing complexity without understanding representation.