LFU Cache — LeetCode 460 Python Solution
HardDesignHash TableLinked ListDoubly-Linked List
- Problem
- #460
- Pattern
- Linked List
- Reading time
- 16 min
- Source
- leetcode.com
The problem
Design and implement a data structure for a Least Frequently Used (LFU) cache. Implement the LFUCache class: LFUCache(int capacity) Initializes the object with the capacity of the data structure.
Example
- Input
- ["LFUCache", "put", "put", "get", "put", "get", "get", "put", "get", "get", "get"]
- Output
- [null, null, null, 1, null, -1, 3, null, -1, 3, 4]
- Explanation
- // cnt(x) = the use counter for key x
Python solution
Python
class Node:
def __init__(self, key: int, value: int) -> None:
self.key = key
self.value = value
self.freq = 1
self.prev = None
self.next = None
class DoublyLinkedList:
def __init__(self) -> None:
self.head = Node(-1, -1)
self.tail = Node(-1, -1)
self.head.next = self.tail
self.tail.prev = self.head
def add_first(self, node: Node) -> None:
node.prev = self.head
node.next = self.head.next
self.head.next.prev = node
self.head.next = node
def remove(self, node: Node) -> Node:
node.next.prev = node.prev
node.prev.next = node.next
node.next, node.prev = None, None
return node
def remove_last(self) -> Node:
return self.remove(self.tail.prev)
def is_empty(self) -> bool:
return self.head.next == self.tail
class LFUCache:
def __init__(self, capacity: int):
self.capacity = capacity
self.min_freq = 0
self.map = defaultdict(Node)
self.freq_map = defaultdict(DoublyLinkedList)
def get(self, key: int) -> int:
if self.capacity == 0 or key not in self.map:
return -1
node = self.map[key]
self.incr_freq(node)
return node.value
def put(self, key: int, value: int) -> None:
if self.capacity == 0:
return
if key in self.map:
node = self.map[key]
node.value = value
self.incr_freq(node)
return
if len(self.map) == self.capacity:
ls = self.freq_map[self.min_freq]
node = ls.remove_last()
self.map.pop(node.key)
node = Node(key, value)
self.add_node(node)
self.map[key] = node
self.min_freq = 1
def incr_freq(self, node: Node) -> None:
freq = node.freq
ls = self.freq_map[freq]
ls.remove(node)
if ls.is_empty():
self.freq_map.pop(freq)
if freq == self.min_freq:
self.min_freq += 1
node.freq += 1
self.add_node(node)
def add_node(self, node: Node) -> None:
freq = node.freq
ls = self.freq_map[freq]
ls.add_first(node)
self.freq_map[freq] = ls
# Your LFUCache object will be instantiated and called as such:
# obj = LFUCache(capacity)
# param_1 = obj.get(key)
# obj.put(key,value)Complexity
| Measure | Complexity |
|---|---|
| Time | O(n) |
| Space | O(n) auxiliary |
Pattern: Linked List
Rewire pointers in place, with a dummy head and a saved next to keep it safe. LeetCode 460. LFU Cache is filed here because LeetCode tags it Linked List and Doubly-Linked List, which is the vocabulary this hub collects.
The linked list guide has the Python template for the pattern and the 75 LeetCode problems that use it.
Related problems
Frequently asked questions
- How hard is LeetCode 460. LFU Cache?
- LeetCode 460. LFU Cache is rated Hard on LeetCode.
- What is the time complexity of LeetCode 460. LFU Cache?
- The Python solution on this page runs in O(n).
- What is the space complexity of LeetCode 460. LFU Cache?
- The Python solution on this page uses O(n) auxiliary space.
- What topics does LeetCode 460. LFU Cache cover?
- LeetCode 460. LFU Cache is tagged Design, Hash Table, Linked List and Doubly-Linked List on LeetCode.