LFU Cache — LeetCode 460 Python Solution

HardDesignHash TableLinked ListDoubly-Linked List
Problem
#460
Reading time
16 min

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

MeasureComplexity
TimeO(n)
SpaceO(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.

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