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LRU Cache Implementation

Certification: Intermediate Level Accuracy: 0% Submissions: 0 Points: 10

Design a data structure that follows the constraints of a Least Recently Used (LRU) cache. Implement the LRUCache class with get(key) and put(key, value) methods. The cache should have a fixed capacity and when the capacity is reached, it should invalidate the least recently used item before inserting a new item.

Example 1
    
      
  • Input: capacity = 2, operations = ["put(1,1)", "put(2,2)", "get(1)", "put(3,3)", "get(2)", "put(4,4)", "get(1)", "get(3)", "get(4)"]
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  • Output: [null, null, 1, null, -1, null, -1, 3, 4]
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  • Explanation: 
    • LRUCache lru = new LRUCache(2) - Create cache with capacity 2. 
    • lru.put(1, 1) - Cache: {1=1}. lru.put(2, 2) - Cache: {1=1, 2=2}. 
    • lru.get(1) returns 1 - Cache: {2=2, 1=1} (1 becomes most recent). 
    • lru.put(3, 3) - Cache: {1=1, 3=3} (2 is evicted as LRU). lru.get(2) returns -1 (not found). 
    • lru.put(4, 4) - Cache: {3=3, 4=4} (1 is evicted as LRU).
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Example 2
    
      
  • Input: capacity = 1, operations = ["put(2,1)", "get(2)", "put(3,2)", "get(2)", "get(3)"]
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  • Output: [null, 1, null, -1, 2]
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  • Explanation: 
    • LRUCache lru = new LRUCache(1) - Create cache with capacity 1. 
    • lru.put(2, 1) - Cache: {2=1}. lru.get(2) returns 1 - Cache: {2=1}. 
    • lru.put(3, 2) - Cache: {3=2} (2 is evicted as capacity is 1). 
    • lru.get(2) returns -1 (not found). lru.get(3) returns 2 - Cache: {3=2}.
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Constraints
    
      
  • 1 ≤ capacity ≤ 3000
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  • 0 ≤ key ≤ 10^4
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  • 0 ≤ value ≤ 10^5
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  • At most 2 * 10^5 calls will be made to get and put
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  • Time Complexity: O(1) for both get and put operations
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  • Space Complexity: O(capacity)
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Linked ListHash MapGoldman SachsD. E. Shaw
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Hint
Solution

Solution Hints

    
      
  • Use a combination of HashMap and Doubly Linked List for O(1) operations
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  • HashMap stores key-node pairs for O(1) access to any node
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  • Doubly Linked List maintains the order of usage (most recent to least recent)
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  • Move accessed nodes to the front of the list to mark them as most recently used
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  • Remove nodes from the tail when capacity is exceeded (least recently used)
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Steps to solve by this approach:

 Step 1: Create a Node class for doubly linked list with key, value, prev, and next pointers.
 Step 2: Initialize LRU cache with HashMap for O(1) access and dummy head/tail nodes for the doubly linked list.
 Step 3: Implement helper methods: addToHead, removeNode, moveToHead, and removeTail for list operations.
 Step 4: For get operation, check if key exists in HashMap, move node to head if found, return value or -1.
 Step 5: For put operation, check if key exists; if yes, update value and move to head.
 Step 6: If key doesn't exist, create new node; if capacity exceeded, remove tail node and its HashMap entry.
 Step 7: Add the new node to HashMap and to the head of the doubly linked list.

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