LRU cache design with hash map and doubly linked list in Data Structures Theory - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to define the main data structure used to store cache entries for fast access.

Data Structures Theory

cache = [1]()

Drag options to blanks, or click blank then click option'

Aset

Bdict

Clist

Dqueue

Attempts:

3 left

💡 Hint

Common Mistakes

Using a list instead of a dictionary causes slow lookups.

Using a set does not store key-value pairs.

The cache uses a hash map (dictionary) to store keys and their corresponding nodes for O(1) access.

2fill in blank

medium

Complete the code to add a new node to the front of the doubly linked list representing the most recently used item.

Data Structures Theory

def add_to_front(node):
    node.prev = None
    node.next = [1]
    if head:
        head.prev = node
    head = node

Drag options to blanks, or click blank then click option'

Ahead

Btail

Cnode

DNone

Attempts:

3 left

💡 Hint

Common Mistakes

Setting next to tail instead of head.

Setting next to None breaks the list linkage.

To add a node to the front, its next pointer should point to the current head.

3fill in blank

hard

Fix the error in the code that removes the least recently used node from the tail of the doubly linked list.

Data Structures Theory

def remove_tail():
    if not tail:
        return
    prev_node = tail.prev
    if prev_node:
        prev_node.next = [1]
    tail = prev_node

Drag options to blanks, or click blank then click option'

Ahead

Bprev_node

CNone

Dtail

Attempts:

3 left

💡 Hint

Common Mistakes

Setting next to tail creates a cycle.

Setting next to prev_node is incorrect.

When removing the tail, the previous node's next pointer should be set to None to mark the new end.

4fill in blank

hard

Fill both blanks to update the cache when a key is accessed: move the node to the front and update pointers.

Data Structures Theory

def access_key(key):
    node = cache[key]
    # Remove node from current position
    node.prev.next = [1]
    node.next.prev = [2]

Drag options to blanks, or click blank then click option'

Anode.next

Bnode.prev

Cnode

DNone

Attempts:

3 left

💡 Hint

Common Mistakes

Linking nodes incorrectly causing broken list.

Setting pointers to None prematurely.

To remove the node, link its previous node's next to its next node, and its next node's prev to its previous node.

5fill in blank

hard

Fill all three blanks to implement the put operation: add new node, update cache, and remove LRU if capacity exceeded.

Data Structures Theory

def put(key, value):
    if key in cache:
        node = cache[key]
        node.value = value
        access_key(key)
    else:
        new_node = Node(key, value)
        cache[[1]] = new_node
        add_to_front(new_node)
        if len(cache) > capacity:
            lru_key = tail.key
            remove_tail()
            del cache[[2]]
            # Optional: clear tail pointer
            tail = [3]

Drag options to blanks, or click blank then click option'

Akey

Blru_key

CNone

Dnew_node

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong variable names for keys.

Not deleting the LRU key from cache.

Not updating tail pointer after removal.

When adding a new node, store it in cache with the key. When removing LRU, delete the lru_key from cache and set tail to None if empty.

Practice

1. What is the main purpose of using a doubly linked list in an LRU cache design?

easy

A. To hash keys for constant time access

B. To store key-value pairs for fast lookup

C. To sort items by their values

D. To keep track of the order of item usage for quick updates

Solution

1. Step 1: Understand the role of doubly linked list in LRU cache

   The doubly linked list maintains the order of items from most recently used to least recently used.

2. Step 2: Explain why order tracking is important

   This order allows quick removal of the least recently used item when the cache is full.

3. Final Answer:

   To keep track of the order of item usage for quick updates -> Option D

4. Quick Check:

   Doubly linked list = usage order tracking [OK]

Hint: Doubly linked list tracks usage order in LRU cache [OK]

Common Mistakes:

• Confusing hash map role with doubly linked list
• Thinking doubly linked list stores key-value pairs
• Assuming it sorts items by value

2. Which of the following correctly describes the time complexity of the get operation in an LRU cache implemented with a hash map and doubly linked list?

easy

A. O(n), where n is the number of items in the cache

B. O(1), constant time

C. O(log n), due to sorting

D. O(n^2), because of list traversal

Solution

1. Step 1: Analyze the get operation components

   The hash map provides direct access to the node in constant time.

2. Step 2: Consider the doubly linked list update

   Moving the accessed node to the front is done in constant time using pointers.

3. Final Answer:

   O(1), constant time -> Option B

4. Quick Check:

   Hash map + doubly linked list = O(1) get [OK]

Hint: Hash map lookup + list update = O(1) get [OK]

Common Mistakes:

• Assuming list traversal causes O(n) time
• Confusing sorting with LRU operations
• Thinking get requires scanning all items

3. Consider an LRU cache with capacity 2. After these operations: put(1, 1), put(2, 2), get(1), put(3, 3), what will get(2) return?

medium

A. 2

B. 1

C. -1

D. 3

Solution

1. Step 1: Track cache state after each operation

   After put(1,1) and put(2,2), cache has keys 1 and 2. get(1) makes key 1 most recently used.

2. Step 2: Insert put(3,3) and evict least recently used

   Key 2 is least recently used and gets evicted. Cache now has keys 1 and 3.

3. Step 3: Check get(2) result

   Key 2 was evicted, so get(2) returns -1.

4. Final Answer:

   -1 -> Option C

5. Quick Check:

   Evicted key 2, so get(2) = -1 [OK]

Hint: Evicted least used key returns -1 on get [OK]

Common Mistakes:

• Forgetting eviction after capacity reached
• Assuming get updates usage but doesn't evict
• Returning wrong value for evicted key

4. In an LRU cache implementation, a developer notices that the cache never evicts any items even after exceeding capacity. What is the most likely cause?

medium

A. The eviction logic is missing or not triggered when capacity is exceeded

B. The doubly linked list is not updated to move recently used items to front

C. The hash map is not updated when items are accessed

D. The cache capacity is set to zero

Solution

1. Step 1: Understand eviction role in LRU cache

   Eviction removes the least recently used item when capacity is exceeded.

2. Step 2: Identify why eviction might not happen

   If eviction logic is missing or not triggered, items will never be removed despite capacity limits.

3. Final Answer:

   The eviction logic is missing or not triggered when capacity is exceeded -> Option A

4. Quick Check:

   Missing eviction = no item removal [OK]

Hint: Eviction missing means cache never removes items [OK]

Common Mistakes:

• Blaming hash map updates for eviction issues
• Ignoring eviction step in put operation
• Assuming capacity zero means no eviction

5. You want to extend an LRU cache to also support a peekLeastUsed() operation that returns the least recently used key without removing it. Which data structure property allows this operation to be efficient?

hard

A. The tail pointer of the doubly linked list points to the least recently used item

B. The hash map stores keys in sorted order

C. The doubly linked list is singly linked for fast traversal

D. The cache uses a priority queue to track usage

Solution

1. Step 1: Identify how least recently used item is tracked

   The doubly linked list keeps most recently used at head and least recently used at tail.

2. Step 2: Explain how to access least used without removal

   Using the tail pointer, we can directly access the least recently used key efficiently.

3. Final Answer:

   The tail pointer of the doubly linked list points to the least recently used item -> Option A

4. Quick Check:

   Tail pointer = least recently used access [OK]

Hint: Tail of doubly linked list = least recently used key [OK]

Common Mistakes:

• Assuming hash map stores sorted keys
• Thinking singly linked list is used
• Confusing priority queue with LRU structure