LFU Cache
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Identifying Problem Isomorphism
“LFU Cache” involves designing a cache system with a specific eviction policy, namely, least frequently used (LFU). The main crux of the problem lies in effectively managing the cache with capacity constraints and updating the frequency of use for each entry.
A simpler problem sharing this theme is “LRU Cache”. In this problem, the cache system design is similar, but the eviction policy is least recently used (LRU), which is simpler to implement compared to LFU. Here, entries need to be updated based on recency of use, rather than frequency.
A more complex problem is “Design In-Memory File System”. This problem requires design of a more complex data structure (a file system), maintaining different types of entries (files and directories) and implementing various operations.
The reasoning behind these choices:
- “LRU Cache” is simpler than “LFU Cache” as it only requires tracking of the recency of usage, which can be achieved with a doubly linked list and a hash map.
- “Design In-Memory File System” is more complex because it involves design of multiple interacting parts (files, directories, and the operations on them), making it a step further in complexity.
From simpler to more complex:
- “LRU Cache”
- “LFU Cache”
- “Design In-Memory File System”
This classification is based on the complexity of the data structure that needs to be designed and the operations that need to be supported. As always, this mapping is an approximation, and actual difficulty can vary based on individual understanding and familiarity with the concepts involved.
10 Prerequisite LeetCode Problems
“LFU Cache” involves data structure design and implementation. Specifically, hash maps, doubly-linked lists, and caching algorithms.
Here are 10 problems to build these foundational skills:
“Two Sum” (LeetCode Problem #1): This problem introduces you to the use of hash maps for efficient lookups.
“LRU Cache” (LeetCode Problem #146): This problem will help you understand the LRU caching mechanism, which is a simpler version of LFU.
“Design Doubly Linked List” (LeetCode Problem #707): This problem requires you to design a basic doubly-linked list, which is an essential part of the LFU cache.
“Implement Stack using Queues” (LeetCode Problem #225): This problem involves implementing a basic data structure using another, similar to how you might implement an LFU cache using a hash map and a doubly-linked list.
“Valid Anagram” (LeetCode Problem #242): This problem further practices the use of hash maps to count elements.
“Contains Duplicate” (LeetCode Problem #217): This problem provides practice in using hash maps to track the existence of items.
“Design HashMap” (LeetCode Problem #706): This problem requires you to design a basic hash map, which is an essential part of the LFU cache.
“Design Circular Queue” (LeetCode Problem #622): This problem introduces you to the design of more complex data structures, which is a good step up towards the LFU cache.
“Design Linked List” (LeetCode Problem #707): This problem has you design a simple linked list. Understanding linked lists is crucial for the LFU cache problem, as doubly-linked lists are a kind of linked list.
“Design Circular Deque” (LeetCode Problem #641): This problem involves the design of a more complex linked list structure, which is a great stepping stone towards the LFU cache 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. int get(int key) Gets the value of the key if the key exists in the cache. Otherwise, returns -1. void put(int key, int value) Update the value of the key if present, or inserts the key if not already present. When the cache reaches its capacity, it should invalidate and remove the least frequently used key before inserting a new item. For this problem, when there is a tie (i.e., two or more keys with the same frequency), the least recently used key would be invalidated. To determine the least frequently used key, a use counter is maintained for each key in the cache. The key with the smallest use counter is the least frequently used key.
When a key is first inserted into the cache, its use counter is set to 1 (due to the put operation). The use counter for a key in the cache is incremented either a get or put operation is called on it.
The functions get and put must each run in O(1) average time complexity.
Example 1:
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
Constraints:
1 <= capacity <= 104 0 <= key <= 105 0 <= value <= 109 At most 2 * 105 calls will be made to get and put.
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LFU Cache - The problem involves designing a cache that evicts the least frequently used item when it’s full. This is a Cache Design Problem.
Language Agnostic Coding Drills
The problem is to implement an LFU (Least Frequently Used) Cache, which requires some knowledge in data structure and algorithms. Here is the break down of the code:
Understanding Classes and Objects: In Object-Oriented Programming (OOP), objects are instances of classes. Here, there are two classes -
Node
andLFUCache
. TheNode
class represents a single node with a key, value, and frequency. TheLFUCache
class is the actual cache with several methods to manage the nodes (get, put, touch).Using Built-in Collections: The code uses Python’s built-in collections like
deque
,defaultdict
, andOrderedDict
.deque
is used for appending and removing elements efficiently,defaultdict
is used for automatically initializing missing keys, andOrderedDict
keeps track of the order in which keys are added.Implementing LFU Cache: The idea is to have the least frequently used cache system where each operation (get, put) should be as efficient as possible. The
get
method retrieves a value if it exists, andput
method inserts a value. When the cache is at full capacity, the least frequently used items are removed.Touch Method: This method updates the frequency of a node when it is accessed (either through get or put). If a frequency list becomes empty after removing a node, it is deleted from the frequency list.
Managing Frequencies: For each frequency, the program maintains a list of keys with that frequency, and also keeps track of the minimum frequency to efficiently evict the least frequently used item when necessary.
As a learning path, you’d want to first get comfortable with classes and objects in OOP, then dive into Python’s collections module. After that, understand the logic behind LFU Cache, its implementation, and how each of its operations work. Finally, review the touch
method and the management of frequencies.
To solve this problem, start by understanding what an LFU Cache is and how it works. Then, initialize your LFUCache
with the necessary data structures. Implement the get
and put
methods, making sure to update the nodes’ frequencies and remove the least frequently used item when the cache is full. Finally, make sure your touch
method correctly updates a node’s frequency and maintains the frequency lists.
Targeted Drills in Python
1. Understanding Classes and Objects:
Python Drill: Implement a simple Book
class, where each book object has a title and author. Implement a method get_info
that returns a string of the book’s title and author.
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2. Using Built-in Collections:
Python Drill: Implement a program using deque
to append values to both ends and pop values from both ends.
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Python Drill: Use defaultdict
to count the frequency of characters in a string.
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3. Managing Frequencies
Python Drill: Implement a frequency counter for a list of numbers using dictionary.
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4. Implementing LFU Cache Operations
Python Drill: Create a class MyCache
with methods get
and put
. get
retrieves a value from the cache and put
inserts a value into the cache. If the cache is at its limit, evict the least recently used item.
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The problem-specific drills would be creating a Node
class and managing frequencies. The Node
class is integral to how the LFU Cache is implemented and managing frequencies is key to the LFU Cache’s operations. These concepts are covered in the LFU Cache operations drill.