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.
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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.
불필요한 전문용어 없이 동작 방식 이해하기
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.
알고리즘을 연속된 의사결정으로 읽기
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Do not quote this complexity mechanically.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.
적합한 경우
- Choose it when its dominant operations match your workload.
- Compare memory overhead as well as Big-O.
보통 어디에서 문제가 생기는가
- Memorizing complexity without understanding representation.
- Ignoring worst-case shape or resizing/rebalancing costs.
전문적인 답변 구조
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.
복잡도에 대해 무엇을 언급해야 하나요?
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.
가장 흔한 실수는 무엇인가요?
Memorizing complexity without understanding representation.