- categories: Code, Interview Question, leetcode, Medium
- source: https://leetcode.com/problems/lru-cache
- topics: Hash Table, Linked List
Design a data structure that follows the constraints of a Least Recently Used (LRU) cache.
Implement the LRUCache
class:
LRUCache(int capacity)
Initialize the LRU cache with positive sizecapacity
.int get(int key)
Return the value of thekey
if the key exists, otherwise return-1
.void put(int key, int value)
Update the value of thekey
if thekey
exists. Otherwise, add thekey-value
pair to the cache. If the number of keys exceeds thecapacity
from this operation, evict the least recently used key.
The functions get
and put
must each run in O(1)
average time complexity.
Idea
Use hashmap + counter + deque. (may be also done using linked list)
class LRUCache:
def __init__(self, capacity: int):
self.dic = {}
self.counter = defaultdict(int)
self.d = deque()
self.capacity = capacity
def get(self, key: int) -> int:
if key in self.dic:
self.d.append(key)
self.counter[key] += 1
return self.dic.get(key)
return -1
def put(self, key: int, value: int) -> None:
if key not in self.dic and len(self.counter) >= self.capacity:
while True:
left = self.d.popleft()
self.counter[left] -= 1
if not self.counter[left]:
del self.counter[left]
del self.dic[left]
break
self.dic[key] = value
self.counter[key] += 1
self.d.append(key)
Approach 2: OrderedDict
class LRUCache:
def __init__(self, capacity: int):
self.capacity = capacity
self.dic = OrderedDict()
def get(self, key: int) -> int:
if key not in self.dic:
return -1
self.dic.move_to_end(key)
return self.dic[key]
def put(self, key: int, value: int) -> None:
if key in self.dic:
self.dic.move_to_end(key)
self.dic[key] = value
if len(self.dic) > self.capacity:
self.dic.popitem(False)