Leetcode #460: LFU Cache
In this guide, we solve Leetcode #460 LFU Cache in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Design, Hash Table, Linked List, Doubly-Linked List
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input
["LFUCache", "put", "put", "get", "put", "get", "get", "put", "get", "get", "get"]
[[2], [1, 1], [2, 2], [1], [3, 3], [2], [3], [4, 4], [1], [3], [4]]
Output
[null, null, null, 1, null, -1, 3, null, -1, 3, 4]
Explanation
// cnt(x) = the use counter for key x
// cache=[] will show the last used order for tiebreakers (leftmost element is most recent)
LFUCache lfu = new LFUCache(2);
lfu.put(1, 1); // cache=[1,_], cnt(1)=1
lfu.put(2, 2); // cache=[2,1], cnt(2)=1, cnt(1)=1
lfu.get(1); // return 1
// cache=[1,2], cnt(2)=1, cnt(1)=2
lfu.put(3, 3); // 2 is the LFU key because cnt(2)=1 is the smallest, invalidate 2.
// cache=[3,1], cnt(3)=1, cnt(1)=2
lfu.get(2); // return -1 (not found)
lfu.get(3); // return 3
// cache=[3,1], cnt(3)=2, cnt(1)=2
lfu.put(4, 4); // Both 1 and 3 have the same cnt, but 1 is LRU, invalidate 1.
// cache=[4,3], cnt(4)=1, cnt(3)=2
lfu.get(1); // return -1 (not found)
lfu.get(3); // return 3
// cache=[3,4], cnt(4)=1, cnt(3)=3
lfu.get(4); // return 4
// cache=[4,3], cnt(4)=2, cnt(3)=3
Python Solution
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
The time complexity is O(n). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.