LRU Cache

Identifying Problem Isomorphism

“LRU Cache” is about designing a cache with a specific policy, the Least Recently Used (LRU) policy. The challenge is in efficiently managing the cache with capacity constraints and maintaining the order of usage of each entry.

A simpler problem with similar concepts is “Two Sum”. Although it doesn’t involve designing a cache, it does require usage of a hash map for efficient lookup, which is a key part of the LRU Cache design.

A more complex problem is “LFU Cache”. This problem builds on the cache design problem by changing the eviction policy to Least Frequently Used (LFU). Here, the frequency of use must be tracked along with the order of use, making it more complex than the LRU Cache problem.

The reasoning behind this selection is as follows:

“Two Sum” is a simpler problem as it only involves the usage of a hash map for fast lookups.
“LFU Cache” is more complex as it not only involves designing a cache (similar to the “LRU Cache”) but also requires maintaining a frequency count for each entry.

So, if we arrange these problems from simpler to more complex, it would be:

“Two Sum”
“LRU Cache”
“LFU Cache”

These mappings are approximate. These are linked by their use of hash maps and the design of data structures to meet specific requirements.

10 Prerequisite LeetCode Problems

The following problems can provide you with a foundation of concepts in solving the LRU Cache problem:

20. Valid Parentheses: This problem introduces the concept of stack, which is crucial to understand when it comes to many data structure problems.
155. Min Stack: The concepts of stack and auxiliary space usage will be practiced here.
70. Climbing Stairs: It is a good problem to understand the basics of dynamic programming.
206. Reverse Linked List: A basic problem on linked list operations, an essential data structure.
141. Linked List Cycle: This problem gives you practice on handling pointers in linked lists.
160. Intersection of Two Linked Lists: Another problem for handling pointers in linked lists and finding intersection points.
283. Move Zeroes: This is a great problem to work on manipulating arrays in-place.
217. Contains Duplicate: This problem uses hash tables (sets in Python), which is also a key data structure for the LRU Cache problem.
448. Find All Numbers Disappeared in an Array: This problem is a great practice for understanding in-place modifications and array indexing.
560. Subarray Sum Equals K: Another problem that uses hash tables and introduces the concept of prefix sum, which is a common strategy in array-related problems.

These cover array manipulation, linked lists, hash tables, and basic dynamic programming, which will be useful in solving the LRU Cache problem.

Problem Analysis and Key Insights

Data Structure Choice: The problem explicitly states the need to implement an LRU Cache, a specific type of data structure. This immediately hints that the solution should effectively combine elements of a queue (to keep track of the ‘recency’ of usage) and a hashmap (for O(1) access to key-value pairs).
Operations: Two operations need to be supported - ‘get’ and ‘put’. Both operations need to be implemented such that they operate in O(1) time complexity. This indicates that we need to design our data structure to support these operations efficiently, which further supports the idea of using a hashmap along with some form of a linked list or queue.
Capacity Management: There is a capacity limit to the cache, and the LRU item needs to be evicted when the capacity is exceeded. This indicates that we not only need to keep track of the order of usage of items but also efficiently identify the least recently used item for eviction.
Item Updates: When using ‘put’, if a key already exists in the cache, its value needs to be updated. Also, this key now needs to be considered the most recently used item. This implies that updating an item in the cache also affects the ordering of items.
Error Handling: When using ‘get’, if a key does not exist in the cache, the operation should return -1. This shows that we need to handle situations where a key is not present in the cache.
Large Scale: The constraints indicate that the solution should be able to handle a large number of operations (up to 2 * 10^5 calls to ‘get’ and ‘put’). This underlines the importance of the O(1) time complexity requirement for the ‘get’ and ‘put’ operations, as a less efficient solution may not be acceptable for such a large scale.

From these insights, the problem demands an efficient, well-structured solution that effectively uses principles of data structures and algorithm design.

Problem Boundary

The scope of this problem encompasses several areas:

Data Structure Design: The primary focus is on designing a data structure that fits the specifications of a Least Recently Used (LRU) cache. This involves creating a structure that efficiently maintains and manipulates order based on the usage of elements, while also allowing for quick access and updates.
Time Complexity: Another significant aspect of this problem’s scope is achieving the desired time complexity. Both the ‘get’ and ‘put’ operations must execute in O(1) time on average. Thus, the solution must be optimized to handle these operations efficiently, regardless of the cache’s size or the number of operations performed.
Cache Management: Managing the cache’s capacity is a critical part of the problem. The data structure must handle the removal of the least recently used item when the cache’s capacity limit is reached.
Error Handling: The ‘get’ operation needs to return a specific value (-1) when a requested key is not found in the cache. This aspect of the problem scope involves implementing a specific response to an error condition.
Large-Scale Operations: The problem scope involves dealing with a significant number of calls to the ‘get’ and ‘put’ functions (up to 2 * 10^5), suggesting that the solution must be scalable and efficient even under heavy usage.

The problem does not include concerns about concurrency or persistent storage of the cache, which are important in real-world cache implementations but are outside this problem’s scope.

The boundary of this problem can be established by clearly defining its input, output, constraints, and the operations that the data structure needs to support:

Input: The input involves initializing the cache with a certain capacity. Further inputs include calls to the get and put operations, with each call specifying the key, and in the case of put, the value to be associated with the key.
Output: The output of the get operation is the value associated with the provided key if it exists in the cache, or -1 if it does not. The put operation doesn’t return a value, but it updates the cache by either adding a new key-value pair or updating an existing one, and possibly evicting the least recently used key-value pair if the cache is at capacity.
Constraints: The constraints specify the range of possible values for the cache’s capacity, the keys, and the values. They also limit the maximum number of calls to get and put operations.
Operations: The operations that the data structure needs to support include get, put, and implicitly, eviction of the least recently used key-value pair when the cache reaches its capacity.

The boundary of this problem does not extend to handling concurrent operations on the cache or persisting the cache to disk, which might be relevant in a real-world application but are not mentioned in the problem statement. Furthermore, the problem is bounded by the requirement for get and put operations to have an average time complexity of O(1), which imposes a limit on the types of data structures and algorithms that can be used to solve this problem.

Problem Classification

The problem belongs to Cache data structure with a Least Recently Used (LRU) eviction policy.

‘What’ Components:

A custom data structure, specifically a cache, with a specified maximum capacity.
A ‘get’ operation that retrieves the value of a key if it exists in the cache, or returns -1 if it doesn’t.
A ‘put’ operation that adds a key-value pair to the cache if the key does not already exist, or updates the value of an existing key. If adding a new key-value pair exceeds the cache’s capacity, the least recently used key-value pair needs to be evicted.
The ‘get’ and ‘put’ operations must each have an average time complexity of O(1).
The system will handle at most 2*10^5 calls to ‘get’ and ‘put’.

This problem is a System Design problem, specifically focusing on Cache Design. The emphasis is on designing a data structure that can handle specific operations with specified constraints and performance characteristics. This type of problem is common in designing systems that require efficient access to data and could be a common question in interviews focused on system design or data structures.

The problem also involves elements of Memory Management (through the implementation of the LRU policy) and Optimization (given the requirement for O(1) operations).

This can be viewed as an Algorithm Design problem, as it involves designing the ‘get’ and ‘put’ operations to behave in a certain way and meet specific time complexity requirements.

Distilling the Problem to Its Core Elements

Fundamental Concept: The fundamental concept that this problem is based on is the Least Recently Used (LRU) caching policy, a type of cache algorithm that manages the cache capacity by removing the least recently used items first. Additionally, it involves knowledge of data structures that can perform insert, remove and search operations in constant time.
Simplified Description: Imagine you have a limited amount of space to store items (like a small shelf). You also want to be able to quickly find these items when needed. Now, when the shelf is full and you need to put a new item on it, you would remove the item that you haven’t used for the longest time to make space. Also, every time you use an item, you put it in the front so that it’s easy to find next time. We are trying to build a system that does exactly this but with numbers, not physical items.
Core Problem: The core problem here is to design a data structure that supports fast insertion, removal, and retrieval of items while maintaining the order of their usage (LRU policy). When the storage capacity is reached, the least recently used item should be discarded.
Key Components: The problem can be broken down into the following components:
- Maintaining an order of the elements based on their usage.
- Inserting an element into the structure.
- Removing an element from the structure.
- Retrieving the value associated with a specific element.
- Ensuring the size of the structure does not exceed the specified capacity.
Minimal Set of Operations: The minimal set of operations needed to solve this problem are:
- Initialization of the cache with a specified capacity.
- Insertion/Updating of a key-value pair.
- Retrieval of a value associated with a specific key.
- Removal of the least recently used key-value pair when the capacity is reached.
- Keeping track of the usage order of the key-value pairs.

Visual Model of the Problem

This problem can be visualized using a real-world analogy: a bookshelf.

Imagine a bookshelf that can only hold a certain number of books.
When you want to read a book that is already on the shelf, you take it off the shelf and after reading, you place it back at the beginning of the shelf (not the place where you took it from).
Now, if the shelf is full and you want to add a new book, you would remove the book that has been read the longest time ago (the one at the end of the shelf, as books move towards the end each time a different book is read).
You then place the new book at the beginning of the shelf.

This is similar to the LRU Cache problem. Here, the bookshelf is the cache, the capacity is the number of books it can hold, books are the key-value pairs, and reading a book is equivalent to calling get or put method. The operation of removing the least recently read book to make space for a new book is the core of the LRU cache strategy.

To solve this problem, we maintain a data structure (cache) that keeps track of the items (key-value pairs) and the order of their usage. The least recently used items are removed from the cache when it exceeds its capacity.

Here is an illustration of an LRU Cache with capacity = 3:

Step 1: (Put 1,1)

Cache: {1=1}

Step 2: (Put 2,2)

Cache: {1=1, 2=2}

Step 3: (Get 1)

Cache: {2=2, 1=1}

Step 4: (Put 3,3)

Cache: {1=1, 3=3} (2=2 has been removed as it was least recently used)

Step 5: (Get 2)

Returns: -1 (2 is not found)

Step 6: (Put 4,4)

Cache: {3=3, 4=4} (1=1 has been removed as it was least recently used)

Step 7: (Get 1)

Returns: -1 (1 is not found)

Step 8: (Get 3)

Cache: {4=4, 3=3}

Step 9: (Get 4)

Cache: {3=3, 4=4}

Each entry is pushed to the beginning of the cache when it is accessed or added, and thus the entries that have not been accessed for a while get pushed towards the end of the cache. When the cache reaches its capacity, the entries at the end (least recently used ones) get evicted.

Problem Restatement

This problem requires us to design and implement a specialized data structure called a Least Recently Used (LRU) Cache. The cache has a fixed size, or capacity, and stores pairs of keys and their corresponding values.

The LRU Cache should have the following operations:

Initialize: When the cache is created, we specify its capacity. This is the maximum number of key-value pairs it can hold at any given time.
Get: This operation receives a key and returns the value associated with that key if it exists in the cache. If the key does not exist in the cache, it returns -1. Additionally, every time we access a key-value pair using the get operation, it is considered “used” and must be moved to be the most recently used within the cache.
Put: This operation is used to add a new key-value pair to the cache or update the value of an existing key. If the key already exists, its value is updated, and it is considered as “used”, therefore, it should be moved to be the most recently used. If the key does not exist, it is added to the cache as the most recently used. However, if the cache has reached its capacity, we need to make room for the new key-value pair by removing the least recently used one.

These operations (get and put) need to be implemented such that they run in O(1) average time complexity, which means the operations should be completed in constant time regardless of the cache size.

Constraints for the problem are:

The capacity of the cache is at least 1 and at most 3000.
Keys range from 0 to 10,000.
Values range from 0 to 100,000.
The total number of get and put operations would not exceed 200,000.

The problem thus becomes one of maintaining an ordered list of key-value pairs, where the order is determined by the usage of the keys, all within the constraints of the fixed capacity of the cache and constant time operation requirements.

Abstract Representation of the Problem

We are tasked to design and implement a data structure which obeys specific rules of access and modification, with an added layer of capacity limitation. The data structure should be able to hold pairs of unique keys and their associated values, with each key-value pair having a distinct state of “recency” associated with it.

There are three operations associated with this data structure:

Initialization: The data structure is created with a pre-defined capacity which sets the limit on the number of unique key-value pairs it can hold.
Access (get operation): An access operation on a key updates the “recency” state of the key-value pair to be the most recent. If the key does not exist, it should return an error signal.
Modification (put operation): A modification either updates the value of an existing key and makes it the most recent, or adds a new key-value pair as the most recent. If adding a new pair breaches the capacity, it should remove the least recent key-value pair before the addition.

The challenge here is to design this data structure in such a way that both access and modification operations can be performed in constant time, irrespective of the capacity of the data structure.

The problem is subject to the constraints of the maximum capacity, the range of possible keys, the range of possible values, and the total number of operations performed.

At the core, this is a problem of data organization and management where the emphasis is on recency of access or modification, with the added complexity of ensuring efficiency of key operations.

Terminology

Here are the technical concepts and terms crucial to understanding this problem:

Data Structure: It is a specific way of organizing and storing data in a computer so that it can be used efficiently. Different types of data structures are suited to different types of applications, and some are highly specialized to specific tasks. In this problem, we are designing a custom data structure to manage the data according to LRU Cache rules.
Least Recently Used (LRU): LRU is a type of cache replacement policy. When the cache reaches its capacity limit, the LRU policy evicts the least recently used items first. In the context of this problem, it refers to the policy of removing the least recently accessed or modified key-value pair when the capacity of the data structure is exceeded.
Cache: A cache is a hardware or software component that stores data so that future requests for that data can be served faster. The data stored in a cache might be the result of an earlier computation or a copy of data stored elsewhere. In this problem, we’re implementing a type of cache data structure.
Key-Value Pair: A set of two linked data items: a key, which is a unique identifier for some item of data, and the value, which is either the data that is identified or a pointer to the location of that data.
Time Complexity: It is a computational complexity that describes the amount of computational time taken by an algorithm to run, as a function of the size of the input to the program. In this problem, get and put operations must run in O(1) average time complexity, i.e., they should take constant time irrespective of the input size.
Initialization: It is the assignment of initial values to variables or data structures. In the context of this problem, we are initializing the LRU cache with a given capacity.

Understanding these concepts is crucial for defining the problem and identifying potential solutions.

Problem Simplification and Explanation

Let’s break down the problem into simpler terms and discuss the key concepts involved.

You are asked to design a special type of “container” or “store” known as an LRU Cache. The main elements of this problem are keys, values, and the capacity of the LRU Cache.

Here are the key actions you can perform on this “store”:

Create the Store (LRUCache(capacity)): You open a new store with a limited capacity.
Add Items to the Store (put(key, value)): You can add a new item, represented by a key-value pair, to your store. If your store is full (i.e., has reached its capacity), you need to remove an item to make space for the new one. But not just any item. You remove the item that was least recently used - maybe it’s out of fashion or not in demand anymore.
Get an Item from the Store (get(key)): You can fetch an item from your store using its key. If the item is present, you return the value. If it’s not in the store, you return -1, indicating that the item is not available.

In terms of an analogy, think of the LRU Cache like a small, trendy boutique shop that only has enough room to display a certain number of items. When a customer (function call) wants to add a new item to the store, the shopkeeper (you) must make room by removing the least recently looked-at or purchased item. If a customer wants to buy an item (get), you, the shopkeeper, will provide the item if it’s in the shop or say that it’s not available if it isn’t.

The key challenge here is that you need to perform these tasks in the most efficient manner possible (O(1) time complexity), which is like saying that no matter how many items are in the shop or how many customers are in the line, each transaction (put/get) should always take the same amount of time.

Constraints

Here are some specific characteristics from the problem statement and constraints that we could exploit for an efficient solution:

Limited Cache Size (Capacity): The LRU Cache size or capacity is limited and known at the time of creation. Knowing the capacity can help design an efficient data structure to hold the key-value pairs.
Distinct Keys: Each key is unique. This characteristic could be exploited by using data structures like hash maps, which can store key-value pairs and provide constant-time complexity for insertions, deletions, and access.
Least Recently Used (LRU) Eviction Policy: This cache follows an LRU eviction policy. This policy can be efficiently implemented using a doubly linked list along with a hash map. The hash map provides O(1) access to keys, and the doubly linked list enables O(1) addition and removal of nodes. This way, the least recently used items can be identified quickly (from one end of the list) and new items can be added to the other end of the list easily.
Fixed Range of Keys and Values: The problem constraints specify that keys and values lie within a fixed range. However, in this case, the specific numerical ranges do not provide an apparent advantage due to the nature of the problem - an LRU cache is more about the sequence of operations (put/get) rather than the specific values of keys or values.
Frequent Calls to get and put Operations: The problem statement suggests that the get and put operations could be called many times. Therefore, optimizing these operations to have a time complexity of O(1) would be crucial for an efficient solution.

In conclusion, the key to an efficient solution here is selecting the appropriate data structures that can handle the specific requirements of the LRU Cache - quick access to elements (hashmap) and easy identification of least recently used items (doubly linked list).

Analyzing the constraints provides us with critical insights that help us in designing an efficient solution for the problem. Here are the key insights derived from the constraints:

Capacity Constraint: The constraint that the capacity of the LRU cache is between 1 and 3000 implies that the cache size is moderate, allowing us to use data structures like hashmaps and linked lists without worrying too much about space complexity.
Key and Value Ranges: The range of keys (0 to 10^4) and values (0 to 10^5) is quite large. However, since the keys need to be unique, and there’s no ordering requirement for the keys or values, we don’t need to worry about the specific values in the implementation.
Number of Operations: The constraint that at most 2 * 10^5 calls will be made to get and put hints at a high frequency of operations. This highlights the importance of optimizing these operations to be as efficient as possible, ideally with a time complexity of O(1).
Time Complexity Requirement: The requirement for the get and put operations to run in O(1) average time complexity is a critical constraint. It suggests that we need to use data structures that allow quick access, insertion, and deletion. For instance, a combination of a doubly linked list (for efficient removal of least recently used items) and a hashmap (for quick access to any item) could meet these requirements.

In essence, the constraints guide us towards a solution that requires a careful selection of data structures, ensuring efficient handling of operations in the LRU cache.

Case Analysis

Let’s break down a few additional test cases into different categories, namely Normal Cases, Edge Cases, and Special Cases.

Normal Cases

Case 1.1: Inputs with Non-Repeated Numbers

Input: [“LRUCache”, “put”, “put”, “get”, “put”, “get”, “put”, “get”, “get”, “get”] [[2], [1, 1], [2, 2], [1], [3, 3], [2], [4, 4], [1], [3], [4]]

Output: [null, null, null, 1, null, -1, null, -1, 3, 4]

This is the problem’s given example where keys and values are distinct. The result can be understood as explained in the problem statement.

Edge Cases

Case 2.1: Empty Cache

Input: [“LRUCache”, “get”, “get”] [[2], [1], [2]]

Output: [null, -1, -1]

In this case, the cache is empty, so all get operations return -1. This case tests the ability of the implementation to handle empty cache.

Case 2.2: Overfilling the Cache

Input: [“LRUCache”, “put”, “put”, “put”] [[2], [1, 1], [2, 2], [3, 3]]

Output: [null, null, null, null]

Here, we insert more items than the capacity of the cache, which forces the cache to evict the least recently used item. This case tests the cache’s LRU eviction mechanism.

Special Cases

Case 3.1: Repeated Keys

Input: [“LRUCache”, “put”, “put”, “get”, “put”, “get”] [[2], [1, 1], [1, 2], [1], [3, 3], [1]]

Output: [null, null, null, 2, null, -1]

In this case, we’re inserting a key that already exists in the cache. The cache should update the key’s value and also update its usage. So when the key 3 is inserted, the key 1 is evicted (since it’s least recently used) and not the key 2. This case tests whether the cache correctly handles key updates and LRU policy.

These test cases together ensure the solution’s robustness by covering various scenarios. They test whether the cache correctly implements the LRU policy, handles edge cases like empty cache, and updates keys correctly.

Analyzing the different cases can provide several key insights:

Handling of Cache Capacity: It’s essential to correctly handle the capacity of the cache. When the cache is full, and a new key-value pair is to be inserted, the least recently used key-value pair should be removed from the cache to make space. This process checks the implementation of the LRU (Least Recently Used) policy.
Updating Existing Keys: When a key that already exists in the cache is inserted again with a different value, the cache should correctly update the value of that key. Moreover, the usage of that key should also be updated, meaning it should now be considered the most recently used key.
Getting Non-existing Keys: If a get operation is attempted with a key that doesn’t exist in the cache, the function should return -1. This requirement tests the cache’s ability to handle invalid get requests.
Cache Initialization and Empty State: The cache should correctly initialize with the given capacity. Moreover, if the cache is empty, all get operations should return -1. These requirements test the cache’s initialization and its behavior when it’s empty.
Time Complexity Constraints: Both get and put operations should run in O(1) average time complexity. This requirement is not directly tested by the cases, but it’s a crucial performance constraint that the solution should meet.

These insights help ensure that the cache correctly implements the necessary operations and the LRU policy, and also meets the time complexity requirements.

Identification of Applicable Theoretical Concepts

There are several mathematical and algorithmic concepts that can be applied to simplify this problem and make it more manageable. These include:

Hashing: A hash map, which provides O(1) access time, can be used to quickly look up values based on keys. In this problem, keys map to nodes of a doubly linked list, so they can be quickly accessed and removed if necessary.
Doubly Linked List: A doubly linked list can be used to easily track the order of how keys are used. Whenever a key is used, the corresponding node in the list can be quickly removed and re-inserted to the top. When the cache is full and needs to evict the least recently used key, it can simply remove the tail node of the list.
Eviction Strategy (LRU Policy): This problem specifically uses the Least Recently Used (LRU) policy, a common eviction strategy in caching algorithms. When the cache is full and a new entry needs to be added, LRU policy discards the least recently used items first. This concept is crucial to solving this problem effectively.

Applying these concepts allows us to keep track of the keys and their corresponding values, and the order of how keys are used, all in constant time, which is a requirement for this problem.

Simple Explanation

Let’s imagine you’re a librarian and you have a small table where you can only display a few of the most recently read books from your library. This table can only hold a certain number of books (which we can think of as the “capacity” of the table). Each time a person reads a book, you place that book on this table.

However, if the table is already full, you must remove a book to make room for the new one. Which book will you remove? Well, to keep the table interesting and relevant, you decide to remove the book that has been read the least recently - that is, the one that people have shown the least interest in lately.

In addition, if someone asks about a book (akin to the ‘get’ operation), you show them the book if it’s on the table and then move it to the front, indicating it has just been read. If the book is not on the table, you say you don’t have it (return -1).

And that’s what the LRU Cache problem is about: it asks you to design a system like the librarian’s table, where you can “put” books (add key-value pairs), “get” books (access values by keys), and if necessary, remove the least recently “read” book (evict the least recently used key), all in the most efficient way possible.

Problem Breakdown and Solution Methodology

I’ll explain this in the context of our librarian analogy, but keep in mind that we’ll be implementing these operations programmatically.

We can break down the process of solving this problem into several steps:

Initialize the Cache: The first step is to initialize our table (LRU Cache). The size of the table is given as capacity. We need to ensure that we don’t exceed this capacity.
Put Operation: When a person reads a book (when we ‘put’ a key-value pair into the cache), we perform the following actions:
- If the book is already on the table, we remove it (since we’re going to re-add it).
- We then add the book to the table. If adding this book would exceed the table’s capacity, we first remove the least recently read book from the table. We know this is the book at the bottom of the stack, as we always add new books on top.
- Finally, we place the new book on top of the stack.
Get Operation: When someone asks about a book (when we ‘get’ a value from the cache using a key):
- If the book is on the table, we remove it (we’ll add it back to the top), and tell the person about it (return the value).
- If the book is not on the table, we let them know we don’t have it (return -1).

The trickiest part of this problem is making sure both the ‘put’ and ‘get’ operations run in O(1) average time complexity. This can be achieved using a combination of a doubly linked list (for keeping track of the order of elements) and a hash map (for quick access to elements). The doubly linked list enables us to add, remove, and update elements in constant time, while the hash map lets us access any element in constant time.

Changes in the problem parameters would affect our solution as follows:

Increasing the Cache Capacity: A larger cache would allow more key-value pairs before eviction starts. However, the time complexity of our operations would remain O(1).
Increasing the Number of Operations: More operations would mean a longer total runtime, but the average time per operation would remain constant (O(1)).

Let’s consider an example to demonstrate this approach:

Suppose we initialize an LRU cache with a capacity of 2 (LRUCache(2)), and perform the following operations:

put(1, 1): The cache is {1=1}. The key 1 and value 1 are added to the cache.
put(2, 2): The cache is {1=1, 2=2}. The key 2 and value 2 are added.
get(1): returns 1. The key 1 is in the cache with value 1.
put(3, 3): evicts key 2, as the cache is full and key 1 was recently accessed. The cache is {1=1, 3=3}.
get(2): returns -1. The key 2 is not in the cache.
put(4, 4): evicts key 1, as the cache is full and key 3 was recently accessed. The cache is {4=4, 3=3}.
get(1): returns -1. The key 1 is not in the cache.
get(3): returns 3. The key 3 is in the cache with value 3.
get(4): returns 4. The key 4 is in the cache with value 4.

This process demonstrates how the LRU Cache maintains its capacity limit while ensuring the most recently used items are always available.

Inference of Problem-Solving Approach from the Problem Statement

Here are the key terms or concepts in the problem:

Least Recently Used (LRU): This is a type of cache replacement algorithm. In an LRU cache, the least recently used items are discarded first. In this problem, it means that when the cache is full and a new item needs to be added, the item that has not been accessed for the longest time is removed. This concept informs the need for a data structure that can efficiently keep track of the recency of access.
Cache: A cache is a hardware or software component that stores data so that future requests for that data can be served faster. Here it’s a metaphor for a data structure where certain key-value pairs are stored. It guides us towards the use of a hashmap or dictionary structure to provide O(1) access and retrieval times.
Capacity: In this problem, capacity refers to the maximum number of items that can be held in the cache at any one time. This guides us towards ensuring our implemented data structure doesn’t exceed this limit.
Get Operation: This is a function that retrieves an item from the cache using a key. If the key doesn’t exist in the cache, it returns -1. The need to retrieve items in constant time suggests the use of a data structure like a hashmap.
Put Operation: This function inserts an item into the cache. If the cache is full, it removes the least recently used item before inserting the new item. This operation guides us to design a method to track the usage of items.

The combination of these operations and constraints suggests a specific approach: a data structure that combines a hashmap (for fast access and insertion of items) and a doubly-linked list (for tracking the order of use of items). The hashmap keys can point to the nodes of the linked list, allowing us to quickly move items around as they are accessed and inserted.

Simple Explanation of the Proof

The algorithm for the LRU Cache problem can seem a bit complex because it combines two data structures: a Hashmap (also known as a Dictionary) and a Doubly-Linked List. Let’s go through it step by step:

Hashmap/Dictionary: A Hashmap allows us to store and retrieve key-value pairs in O(1) time complexity. The key in our case is the key of the item we want to store in the cache, and the value is a node in a Doubly-Linked List that contains the key-value pair and pointers to the next and previous nodes. This way, we can quickly access the linked list node containing the data associated with a given key.
Doubly-Linked List: A Doubly-Linked List lets us add nodes to the front or remove nodes from anywhere in the list in O(1) time, assuming we have a reference to the node. It’s perfect for keeping track of the order of use of the items in the cache. The head of the list represents the most recently used item, while the tail of the list represents the least recently used item.

Now, when we perform a get operation, we do the following:

a. Use the key to find the corresponding node in the Hashmap.

b. If the node exists, we move it to the front of the Doubly-Linked List (representing that it’s the most recently used).

c. Return the value of the node.

When we perform a put operation, we do the following:

a. If the key already exists in the Hashmap, we update the value of the node and move it to the front of the Doubly-Linked List.

b. If the key doesn’t exist, we create a new node, add it to the front of the list, and add an entry in the Hashmap with the key and the new node as the value.

c. If adding the new node causes the cache to exceed its capacity, we remove the tail node from the list (the least recently used item) and also remove its entry from the Hashmap.

The reason this algorithm works is that it satisfies all the conditions for an LRU cache: it always keeps the most recently used items, it evicts the least recently used items when the cache is full, and it performs both get and put operations in O(1) time complexity. The use of a Hashmap and Doubly-Linked List in tandem allows us to quickly access and update our cache while keeping track of the usage order.

Stepwise Refinement of Approach:
Step 1: Initialize a Hashmap and a Doubly-Linked List. The Hashmap will store keys and corresponding nodes in the list as its values. The list will store nodes containing key-value pairs.
Step 2: Define the get operation:
- Retrieve the node from the Hashmap using the given key.
- If the key does not exist, return -1.
- If the key exists, remove the node from its current position in the list and move it to the front of the list (since it’s the most recently used now).
- Return the value from the node.
Step 3: Define the put operation:
- Check if the key exists in the Hashmap.
- If it does, remove the corresponding node from the list, update its value, and move it to the front of the list.
- If it does not, create a new node with the key-value pair and add it to the front of the list and Hashmap.
- If adding the new node causes the cache to exceed its capacity, remove the tail node from the list (the least recently used item) and also remove its entry from the Hashmap.
Granular, Actionable Steps:
- Implement the Doubly-Linked List with operations for adding a node to the front, removing a node from anywhere, and removing the last node.
- Implement the Hashmap using a built-in data structure.
- Implement the get and put methods as per the stepwise refinement above.
- Ensure that both get and put operations run in O(1) time.
Independent Parts:
- The implementation of the Doubly-Linked List and the Hashmap can be done independently.
- get and put operations are also independent of each other.
Repeatable Patterns:
- For both get and put operations, if a key exists, the corresponding node in the list is moved to the front. This is a repeatable pattern in the solution.
- Whenever a new node is added, a check is performed to ensure the capacity of the cache is not exceeded. If it is, the least recently used item (the tail of the list) is removed. This is another repeatable pattern.

Solution Approach and Analysis

This problem asks us to implement an LRU (Least Recently Used) Cache, which is a common type of cache used in computer systems. In simple terms, a cache is like a small memory storage that keeps frequently or recently used items handy so that we can access them quickly. You can think of it as a pocket in your bag where you keep your most used items like your phone, wallet, and keys.

Just like your pocket has a limited size, the LRU Cache has a limited capacity, and once that capacity is reached, we need to remove the least recently used item to make room for the new one. This is similar to removing the least needed item from your pocket (maybe a shopping list from two weeks ago) to make space for something more important (like a parking ticket).

Now let’s break down how we approach this problem.

Step 1: Initialization

We will use two main data structures: a HashMap and a Doubly-Linked List. The HashMap will store each item’s key and its corresponding node in the linked list. The linked list will hold the items and their order of usage.

Step 2: Implement ‘get’ operation

When we want to get an item from the cache, we look for it in the HashMap. If it’s not there, we return -1, signifying that it’s not present. If it is, we return its value and move it to the front of the linked list, showing that it has been recently used.

Step 3: Implement ‘put’ operation

When we want to put an item in the cache, we first check if the item already exists in the HashMap. If it does, we update its value and move it to the front of the list. If it doesn’t exist, we add a new node at the front of the list and add the item to the HashMap. If this addition causes the cache to go over its capacity, we remove the node at the tail of the list and its entry in the HashMap.

Let’s look at an example to demonstrate how this works. Let’s say we initialize an LRU cache with a capacity of 2:

LRUCache cache = new LRUCache(2);

This creates an empty cache with a capacity of 2.

Now let’s add some items:

cache.put(1,1); // The cache is now {1=1}
cache.put(2,2); // The cache is now {1=1, 2=2}

If we try to get key 1:

cache.get(1); // Returns 1, and the cache becomes {2=2, 1=1}

Now, let’s try to add a new item to the cache:

cache.put(3,3); // The cache is {1=1, 3=3}, as 2 has been evicted because it was the least recently used

Getting key 2 now should return -1:

cache.get(2); // Returns -1, as 2 does not exist in the cache anymore

If we add another item:

cache.put(4,4); // The cache is {3=3, 4=4}, as 1 has been evicted

Getting key 1 now should also return -1, while getting keys 3 and 4 should return 3 and 4 respectively.

cache.get(1); // Returns -1
cache.get(3); // Returns 3
cache.get(4); // Returns 4

This detailed explanation shows how the operations of an LRU Cache work. The primary objective is to ensure that the most recently used items are easily accessible, and the least recently used ones are removed when the cache reaches its capacity.

Identify Invariant

In this LRU Cache problem, the invariant, which is the condition that remains unchanged as the program executes, is that the capacity of the cache remains the same once it is instantiated.

Another significant invariant is the property of the LRU Cache that ensures the “most recently used” items are always near the front of the doubly linked list (or whatever data structure is used to maintain the order of elements), and the “least recently used” items are near the end. This property is maintained through every operation (get and put) of the LRU Cache, ensuring that we always evict the least recently used item when the cache is full and a new item needs to be added.

Finally, the synchronization between the HashMap and the Doubly-Linked List is also an invariant. Every entry in the HashMap must correspond to a node in the Doubly-Linked List, and vice versa. This invariant ensures the O(1) time complexity for get and put operations. This condition is maintained throughout the lifetime of the cache - whenever an item is added or removed, it’s added or removed from both the HashMap and the list.

Identify Loop Invariant

In the problem of designing a Least Recently Used (LRU) Cache, there is no explicit loop stated in the problem description. However, if we consider the continual usage of the LRU Cache over time as an implicit loop, then a loop invariant would still be the properties mentioned earlier:

The capacity of the cache remains constant after initialization.
The most recently used items are at the front of the cache data structure (like a doubly linked list), and the least recently used items are at the end.
There’s always synchronization between the HashMap and the doubly linked list, ensuring O(1) time complexity for get and put operations.

These invariants would hold true before, during, and after every single iteration (i.e., each call to get or put).

However, remember that this problem doesn’t explicitly involve any loops, so discussing loop invariants might not be directly relevant in this case. Loop invariants are typically more useful in analyzing algorithms that contain explicit loops, such as sorting or searching algorithms.

Thought Process

The basic thought process and steps involved in solving this problem are as follows:

Understand the Problem: The problem statement is asking us to implement an LRU Cache. This is a cache that removes the least recently used items first. This type of cache has a limited size, and when that size is reached, it needs to make room for new items by discarding the oldest item, which is the least recently used one.
Understand the Operations: There are two main operations that our LRU Cache needs to support: get and put. Both operations must run in O(1) average time complexity.
Identify the Data Structures: An efficient way to implement an LRU Cache is by using two data structures: a doubly linked list and a hash map. The doubly linked list is used to store the keys of the cache, with the most recently used keys stored near the head, and the least recently used keys stored near the tail. The hash map, on the other hand, is used for fast access to the nodes in the linked list.
Plan the Implementation: Now that we have our data structures identified, we need to plan how we will implement the get and put operations. For the get operation, we need to retrieve the value of the key if it exists in our cache, and also move the corresponding node in the linked list to the head. For the put operation, we need to insert a new key-value pair into our cache, or update the value if the key already exists. In either case, we need to move the corresponding node to the head of the linked list. If the cache is at capacity, we need to remove the least recently used item, which is at the tail of the linked list.

Given these insights, here’s a Python3 implementation of the LRU Cache:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
class Node:
    def __init__(self, key, value):
        self.key = key
        self.value = value
        self.prev = None
        self.next = None


class LRUCache:

    def __init__(self, capacity: int):
        self.capacity = capacity
        self.hash_map = {}
        self.head = Node(0, 0)
        self.tail = Node(0, 0)
        self.head.next = self.tail
        self.tail.prev = self.head

    def get(self, key: int) -> int:
        if key in self.hash_map:
            node = self.hash_map[key]
            self._remove(node)
            self._add(node)
            return node.value
        return -1

    def put(self, key: int, value: int) -> None:
        if key in self.hash_map:
            self._remove(self.hash_map[key])
        node = Node(key, value)
        self._add(node)
        self.hash_map[key] = node
        if len(self.hash_map) > self.capacity:
            node = self.head.next
            self._remove(node)
            del self.hash_map[node.key]

    def _remove(self, node):
        p = node.prev
        n = node.next
        p.next = n
        n.prev = p

    def _add(self, node):
        p = self.tail.prev
        p.next = node
        self.tail.prev = node
        node.prev = p
        node.next = self.tail

In this code, the _remove and _add helper functions manage the updating of nodes in the doubly linked list. When we get a value, we remove the node from its current position and add it to the tail (to indicate it has been recently used). When we put a value, we add it to the tail, and if the cache is over capacity, we remove the node next to the head (the least recently used item).

Establishing Preconditions and Postconditions

Let’s answer these questions based on the provided problem and the implemented solution in Python:

Parameters:
- There are two classes: Node and LRUCache. Node takes two parameters - key and value which are of integer type. They represent the key-value pairs to be stored in the cache. LRUCache takes one parameter - capacity, which is also an integer, representing the maximum size of the cache.
Preconditions:
- Before any method is called, a LRUCache object must be initialized with a capacity. The capacity must be a positive integer.
- The keys must be within the range 0 <= key <= 10^4 and the values must be within the range 0 <= value <= 10^5.
- The LRUCache object must be in a state that respects these constraints.
Method Functionality:
- The get method is expected to return the value of a given key if it exists in the cache, and -1 otherwise. If the key exists, the corresponding node is moved to the end of the doubly linked list to indicate it is recently used.
- The put method inserts a new key-value pair to the cache or updates the value if the key already exists. If the cache reaches its capacity, the least recently used item, which is at the beginning of the doubly linked list, is removed.
Postconditions:
- After get or put has been called, the state of the cache has been updated reflecting these operations. The return value of get represents the value of the key in the cache or -1 if not found. The put operation has no return value but modifies the state of the cache.
- If the cache is at capacity and a put operation is invoked, the least recently used key-value pair is removed from the cache.
Error Handling:
- The provided solution does not explicitly handle errors. If a get operation is called with a key that doesn’t exist, it returns -1. For put, if a key is inserted that already exists, it updates the value. If the preconditions are not met, e.g., if a non-integer key is used, Python will raise an appropriate exception.

Problem Decomposition

Problem Understanding:
- The problem is about implementing a cache system which uses a Least Recently Used (LRU) policy to evict items when it’s full. The cache should support two operations: getting the value of a key, and putting a new key-value pair into the cache. When putting a new pair, if the cache is full, it should remove the least recently used item.
Initial Breakdown:
- Major parts of the problem are implementing get and put operations, maintaining the LRU order, and handling cache capacity.
Subproblem Refinement:
- get operation can be further broken down into: searching for the key in the cache and updating the LRU order.
- put operation can be broken down into: inserting a new key-value pair, updating an existing pair, removing the LRU item when cache is full, and updating the LRU order.
Task Identification:
- Searching for a key in the cache and updating the LRU order are tasks common to both get and put operations.
Task Abstraction:
- The task of searching for a key in the cache is well abstracted: given a key, return the corresponding value if it exists.
- Updating the LRU order is also abstracted: given a key, move the corresponding item to the end of the list (most recently used position).
Method Naming:
- The get and put operations are named as per their functionality. For the abstract tasks, move_to_end can be a method for moving an item to the end of the list, and pop_head for removing the least recently used item.
Subproblem Interactions:
- The get and put operations can be performed independently.
- Within each operation, the tasks must be performed in a specific order: for get, search for the key, then update the LRU order. For put, check if the key exists and update it or insert new key-value pair, check the cache capacity and remove LRU item if needed, then update the LRU order.
- There are dependencies in the tasks, as the updating of the LRU order depends on the key to be updated, and removing LRU item depends on the cache being at its full capacity.

From Brute Force to Optimal Solution

Brute Force Solution:

The simplest, but highly inefficient, solution to this problem would be to use an array or list for storing the key-value pairs, and another list to keep track of the usage of the keys. The get and put operations would be as follows:

get(int key):
- We iterate over the array to find the key. If we find it, we return the value, and move this key to the end of the usage list.
- If the key is not found, we return -1.
put(int key, int value):
- We again iterate over the array to find if the key already exists. If it does, we update the value and move the key to the end of the usage list.
- If the key does not exist, we add the key-value pair to the end of the array and the key to the usage list. If the cache is at full capacity, we remove the first key from the usage list and its corresponding key-value pair from the array.

Inefficiencies:

The primary inefficiency of this approach is that both the get and put operations have a time complexity of O(n), where n is the number of keys in the cache. This is because in the worst-case scenario, we may have to iterate over all the keys to find a specific key or to remove a key-value pair. This does not meet the problem’s requirement of having these operations in O(1) time complexity.

Optimization Steps:

Use a Data Structure that supports Fast Lookups and Inserts:
- To reduce the time complexity of the get and put operations, we need a data structure that supports fast lookups and inserts.
- A Hash Table (in Python, a dictionary) can provide O(1) lookups and inserts.
Use a Data Structure that supports Order:
- We also need a data structure that can keep track of the order of usage of keys, and allows us to quickly move keys to the end and remove the least recently used key.
- A doubly-linked list can help us maintain the order of keys, as it supports efficient removal and addition of nodes at both ends.
Combining the Two Data Structures:
- We can use a combination of a Hash Table and a Doubly-Linked List, where the Hash Table stores the keys and pointers to the corresponding nodes in the list.
- Now, get operation can be done in O(1) by directly accessing the key in the Hash Table and returning its value from the node it points to in the list.
- The put operation can also be done in O(1) by either updating the node the key points to, or creating a new node at the end. If the cache is full, the node at the beginning of the list can be removed in O(1) and its key removed from the Hash Table.

Impact on Time and Space Complexity:

This optimization improves the time complexity of both get and put operations from O(n) to O(1). However, the space complexity increases to O(capacity), as we’re now storing a node in the list and an entry in the Hash Table for each key-value pair. This trade-off between time and space complexity is often necessary to increase efficiency.

Code Explanation and Design Decisions

Let’s discuss a detailed walk-through of a Python solution for an LRU Cache problem.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
class ListNode:
    def __init__(self, key, value):
        self.key = key
        self.value = value
        self.prev = None
        self.next = None

class LRUCache:
    def __init__(self, capacity: int):
        self.capacity = capacity
        self.size = 0
        self.hash_map = {}
        self.head, self.tail = ListNode(None, None), ListNode(None, None)
        self.head.next = self.tail
        self.tail.prev = self.head

    def get(self, key: int) -> int:
        if key in self.hash_map:
            node = self.hash_map[key]
            self._remove(node)
            self._add(node)
            return node.value
        return -1

    def put(self, key: int, value: int) -> None:
        if key in self.hash_map:
            self._remove(self.hash_map[key])
        elif self.size == self.capacity:
            del self.hash_map[self.tail.prev.key]
            self._remove(self.tail.prev)
            self.size -= 1
        self.hash_map[key] = ListNode(key, value)
        self._add(self.hash_map[key])
        self.size += 1

    def _remove(self, node):
        node.prev.next = node.next
        node.next.prev = node.prev

    def _add(self, node):
        node.next = self.head.next
        node.prev = self.head
        self.head.next.prev = node
        self.head.next = node

Initial parameters:
- capacity: The maximum number of key-value pairs the cache can hold. It is significant as it sets a limit to how many pairs we can store.
Primary Loop or Iteration:
- There isn’t a primary loop that iterates over input data in this case, but rather, the get and put methods are invoked repetitively, and each invocation constitutes a “step” in our overall process. Each step either fetches a value (get), or stores a value (put).
Conditions or Branches within the Loop:
- Inside get and put methods, there are conditional branches that check if a key exists in the hash_map. These checks help decide whether to return/update an existing value, or add a new value to the cache.
- In the put method, another condition checks if the cache is full (size equals capacity), and if so, it removes the least recently used (LRU) pair before adding a new one.
Updates or Modifications to Parameters:
- get modifies the list by moving the accessed node to the front (most recently used end).
- put modifies the list by adding a new node to the front, and if the cache is full, it also removes a node from the back (LRU end).
- These modifications are necessary to ensure the list always reflects the usage order of the keys, which is crucial for the LRU eviction policy.
Invariant:
- The invariant in this problem is that the linked list always maintains the keys in order of their usage, with the most recently used key at the head, and the least recently used key at the tail. This invariant is crucial for the LRU policy to work correctly.
Significance of the Final Output:
- The final output from get is the value associated with the given key if it exists in the cache, or -1 if it doesn’t. It satisfies the problem’s requirements by providing a way to access values in constant time.
- The final output from put is implicit. It modifies the state of the cache by adding a new key-value pair or updating an existing one, and possibly evicting the LRU pair. It satisfies the problem’s requirements by providing a way to store values with an LRU eviction policy.

Coding Constructs

Let’s break down the code for the LRU Cache problem:

High-level problem-solving strategies or techniques:
- This code primarily uses a combination of a hash map (also known as a dictionary) and a doubly linked list. The hash map allows for quick access to any item by key, and the doubly linked list lets us quickly add, remove, and move items, which is key to maintaining the LRU (least recently used) order.
Purpose of the code to a non-programmer:
- This code is a model of a real-world cache, which is a place to store a limited amount of information that you need to access quickly. It’s like a small, fast-access pocket where you keep your most frequently used items. If the pocket gets full and you need to put in something new, you remove the item you used the longest time ago to make room for it.
Logical elements or constructs used in this code, independent of any programming language:
- The fundamental constructs used in this code include variables, functions (or methods), conditionals (if statements), and data structures (specifically a hash map and a doubly linked list).
Algorithmic approach used by this code in plain English:
- When you want to get the value for a key, the code checks if it has it. If it does, it returns the value and moves this item to the “front” of the pocket (meaning it’s the most recently used). If it doesn’t have it, it returns -1.
- When you want to put a new key-value pair into the cache, the code first checks if it already has this key. If it does, it removes the old key-value pair. It then checks if the pocket is full. If it is, it removes the item that was used the least recently. Then it adds the new key-value pair to the “front” of the pocket.
Key steps or operations this code is performing on the input data, and why:
- get: This function checks the hash map for the requested key. If it exists, it removes the corresponding node from the linked list and re-inserts it at the front, then returns the value. If it doesn’t exist, it returns -1. These operations ensure quick access to the value and update the usage order.
- put: This function first checks if the key already exists in the cache. If so, it removes the old node. Then it checks if the cache is full, and if so, it evicts the least recently used node (at the tail of the list). Finally, it inserts the new node at the front of the list. These operations ensure the new or updated value is added to the cache, maintaining the LRU order and not exceeding the cache capacity.
Algorithmic patterns or strategies used by this code, irrespective of the specific programming language syntax:
- The key algorithmic pattern here is the combination of a hash map for quick lookup and a doubly linked list for quick addition/removal of items. The code also uses the common technique of maintaining a “dummy” head and tail for the list to simplify edge case handling. Furthermore, the pattern of checking conditions and returning early in the get method is a good practice known as “guard clauses”.

Language Agnostic Coding Drills

Let’s break down the LRU Cache problem into the smallest possible learning units:

Dissect the Code: Here are the distinct concepts contained in the code:
- Basic Programming Constructs: This includes basic elements like variables, loops, conditionals, and functions/methods.
- Object-Oriented Programming (OOP): This is about defining classes, creating objects, and using methods.
- Hash Maps: These are data structures that provide fast access to values based on their keys.
- Doubly Linked Lists: These are data structures where each element points to the next and the previous element.
- Dummy Nodes: These are placeholder elements used in data structures to simplify handling of edge cases.
- Cache Eviction Policies: This refers to the logic of determining which elements should be removed when a cache is full, in this case, LRU (Least Recently Used).
List Concepts in Order of Increasing Difficulty:
- Basic Programming Constructs (Easy): These are the foundation of all programming. Variables store values, loops repeat actions, conditionals make decisions, and functions/methods group actions.
- Object-Oriented Programming (OOP) (Medium): This involves encapsulating data and behaviors into objects. It’s more complex than basic constructs but provides a way to manage complexity by organizing code into reusable pieces.
- Hash Maps (Medium): This data structure is used for fast lookup operations. Understanding hash maps requires knowledge of hashing and handling hash collisions.
- Doubly Linked Lists (Hard): This data structure is used for efficient addition and removal of elements. It requires understanding of pointers or references and can be tricky because operations may involve updating multiple pointers correctly.
- Dummy Nodes (Hard): This technique simplifies edge cases in data structure manipulation, especially linked lists, by providing a guaranteed non-null node to work with. It requires a good understanding of the data structure involved and the specific edge cases it has.
- Cache Eviction Policies (Hard): This concept requires knowledge of caching strategies and algorithms to decide which items to evict when a cache is full. In this case, it’s the LRU policy, which requires tracking the usage order of items.
Problem-Solving Approach:
- Start by understanding the problem and the requirements for the LRU Cache. Identify the operations needed (get and put) and the constraints (limited capacity, LRU eviction).
- Design the class structure using OOP principles. You’ll need a class for the LRU Cache and another for the list node.
- Identify the data structures needed. A hash map will hold the keys and their corresponding nodes in the list. A doubly linked list will hold the values and maintain their usage order.
- Implement the basic functionality for getting and putting values in the cache, without considering capacity or eviction. This involves hash map operations (get, put) and linked list operations (add to front, remove).
- Implement the cache capacity constraint. This involves checking the size of the cache and removing the least recently used item when the cache is full.
- Add the dummy head and tail nodes to the linked list to simplify edge case handling. This involves updating the linked list operations to work with these dummy nodes.
- Finally, test the implementation with various use cases to ensure it works correctly and satisfies all constraints.

Targeted Drills in Python

Here’s how you might create Python coding drills for each of the concepts identified:

Basic Programming Constructs:

A basic Python function and conditional statements.

1
2
3
4
5
6
def is_even(num):
    if num % 2 == 0:
        return True
    else:
        return False
print(is_even(4))  # Output: True

Object-Oriented Programming (OOP):

A simple class in Python.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
class Car:
    def __init__(self, brand, model):
        self.brand = brand
        self.model = model

    def get_info(self):
        return f"Car Brand: {self.brand}, Model: {self.model}"

car = Car("Toyota", "Corolla")
print(car.get_info())  # Output: Car Brand: Toyota, Model: Corolla

Hash Maps:

Using a dictionary in Python (Python’s built-in hash map).

1
2
3
phonebook = {'John': '123-456', 'Jane': '789-012'}
phonebook['Mary'] = '345-678'  # Adding a new item
print(phonebook.get('Jane'))  # Output: 789-012

Doubly Linked Lists:

Creating a simple doubly linked list in Python.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
class Node:
    def __init__(self, data=None):
        self.data = data
        self.next = None
        self.prev = None

class DoublyLinkedList:
    def __init__(self):
        self.head = Node()  # Dummy head

    def append(self, data):
        if self.head.data is None:
            self.head.data = data
            return
        new_node = Node(data)
        cur = self.head
        while cur.next:
            cur = cur.next
        cur.next = new_node
        new_node.prev = cur

Dummy Nodes:

Adding a dummy tail node to the doubly linked list.

1
2
3
4
5
6
class DoublyLinkedList:
    def __init__(self):
        self.head = Node()  # Dummy head
        self.tail = Node()  # Dummy tail
        self.head.next = self.tail
        self.tail.prev = self.head

Cache Eviction Policies:

A simple implementation of an LRU Cache eviction policy (without size constraint for simplicity).

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
class LRUCache:
    def __init__(self):
        self.cache = {}

    def get(self, key):
        if key in self.cache:
            value = self.cache.pop(key)
            self.cache[key] = value  # Move to end
            return value
        else:
            return -1

    def put(self, key, value):
        if key in self.cache:
            self.cache.pop(key)
        self.cache[key] = value  # Add to end

Finally, you’d assemble these concepts together to create an LRU Cache with a size constraint and using a doubly linked list with dummy head and tail nodes to efficiently add and remove nodes. Each node would represent a cache item, and you’d use a hash map to quickly access any node in the list. The LRU eviction policy would be implemented by moving recently accessed items to the front of the list and removing items from the end of the list when the cache is full.