460. LFU Cache
Read the full problem statement on LeetCode.
Difficulty: hard Acceptance: 46% Topics: Hash Table, Linked List, Design, Doubly-Linked List
View full problem on LeetCode Reading material
Reference solution (spoiler · python)
# Time: O(1), per operation
# Space: O(k), k is the capacity of cache
import collections
# using OrderedDict
class LFUCache(object):
def __init__(self, capacity):
"""
:type capacity: int
"""
self.__capa = capacity
self.__size = 0
self.__min_freq = float("inf")
self.__freq_to_nodes = collections.defaultdict(collections.OrderedDict)
self.__key_to_freq = {}
def get(self, key):
"""
:type key: int
:rtype: int
"""
if key not in self.__key_to_freq:
return -1
value = self.__freq_to_nodes[self.__key_to_freq[key]][key]
self.__update(key, value)
return value
def put(self, key, value):
"""
:type key: int
:type value: int
:rtype: void
"""
if self.__capa <= 0:
return
if key not in self.__key_to_freq and self.__size == self.__capa:
del self.__key_to_freq[self.__freq_to_nodes[self.__min_freq].popitem(last=False)[0]]
if not self.__freq_to_nodes[self.__min_freq]:
del self.__freq_to_nodes[self.__min_freq]
self.__size -= 1
self.__update(key, value)
def __update(self, key, value):
freq = 0
if key in self.__key_to_freq:
freq = self.__key_to_freq[key]
del self.__freq_to_nodes[freq][key]
if not self.__freq_to_nodes[freq]:
del self.__freq_to_nodes[freq]
if self.__min_freq == freq:
self.__min_freq += 1
self.__size -= 1
freq += 1
self.__min_freq = min(self.__min_freq, freq)
self.__key_to_freq[key] = freq
self.__freq_to_nodes[freq][key] = value
self.__size += 1
# Time: O(1), per operation
# Space: O(k), k is the capacity of cache
import collections
class ListNode(object):
def __init__(self, key, value, freq):
self.key = key
self.val = value
self.freq = freq
self.next = None
self.prev = None
class LinkedList(object):
def __init__(self):
self.head = None
self.tail = None
def append(self, node):
node.next, node.prev = None, None # avoid dirty node
if self.head is None:
self.head = node
else:
self.tail.next = node
node.prev = self.tail
self.tail = node
def delete(self, node):
if node.prev:
node.prev.next = node.next
else:
self.head = node.next
if node.next:
node.next.prev = node.prev
else:
self.tail = node.prev
node.next, node.prev = None, None # make node clean
class LFUCache2(object):
def __init__(self, capacity):
"""
:type capacity: int
"""
self.__capa = capacity
self.__size = 0
self.__min_freq = float("inf")
self.__freq_to_nodes = collections.defaultdict(LinkedList)
self.__key_to_node = {}
def get(self, key):
"""
:type key: int
:rtype: int
"""
if key not in self.__key_to_node:
return -1
value = self.__key_to_node[key].val
self.__update(key, value)
return value
def put(self, key, value):
"""
:type key: int
:type value: int
:rtype: void
"""
if self.__capa <= 0:
return
if key not in self.__key_to_node and self.__size == self.__capa:
del self.__key_to_node[self.__freq_to_nodes[self.__min_freq].head.key]
self.__freq_to_nodes[self.__min_freq].delete(self.__freq_to_nodes[self.__min_freq].head)
if not self.__freq_to_nodes[self.__min_freq].head:
del self.__freq_to_nodes[self.__min_freq]
self.__size -= 1
self.__update(key, value)
def __update(self, key, value):
freq = 0
if key in self.__key_to_node:
old_node = self.__key_to_node[key]
freq = old_node.freq
self.__freq_to_nodes[freq].delete(old_node)
if not self.__freq_to_nodes[freq].head:
del self.__freq_to_nodes[freq]
if self.__min_freq == freq:
self.__min_freq += 1
self.__size -= 1
freq += 1
self.__min_freq = min(self.__min_freq, freq)
self.__key_to_node[key] = ListNode(key, value, freq)
self.__freq_to_nodes[freq].append(self.__key_to_node[key])
self.__size += 1
Solution from kamyu104/LeetCode-Solutions · MIT
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