LRU Cache – Clean Code

Design a data structure that follows the constraints of a Least Recently Used (LRU) cache.

Implement the LRUCache class:

  • LRUCache(int capacity) Initialize the LRU cache with positive size capacity.
  • int get(int key) Return the value of the key if the key exists, otherwise return -1.
  • void put(int key, int value) Update the value of the key if the key exists. Otherwise, add the key-value pair to the cache. If the number of keys exceeds the capacity from this operation, evict the least recently used key.

Follow up:
Could you do get and put in O(1) time complexity?

Example 1:

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]

Explanation
LRUCache lRUCache = new LRUCache(2);
lRUCache.put(1, 1); // cache is {1=1}
lRUCache.put(2, 2); // cache is {1=1, 2=2}
lRUCache.get(1);    // return 1
lRUCache.put(3, 3); // LRU key was 2, evicts key 2, cache is {1=1, 3=3}
lRUCache.get(2);    // returns -1 (not found)
lRUCache.put(4, 4); // LRU key was 1, evicts key 1, cache is {4=4, 3=3}
lRUCache.get(1);    // return -1 (not found)
lRUCache.get(3);    // return 3
lRUCache.get(4);    // return 4

Java Program

class LRUCache {
    
    class Node 
    {
        int key,value;
        Node prev,next;
        public Node(int key,int value)
        {
            this.key = key;
            this.value = value;
        }
    }

    HashMap map;
    Node head = new Node(0,0);
    Node tail = new Node(0,0);
    int capacity;
    
    public LRUCache(int capacity) {
        this.capacity = capacity;
        map = new HashMap();
        head.next = tail;
        tail.prev = head;
    }
    
    public int get(int key) {
        if(map.containsKey(key))
        {
            Node node = map.get(key);
            remove(node);
            insert(node);
            return node.value;
        }
        return -1;
    }
    
    public void put(int key, int value) {
        if(map.containsKey(key))
        {
            remove(map.get(key));
        }
        if(map.size() == capacity)
        {
            remove(tail.prev);
        }
        insert(new Node(key,value));
    }
    public void remove(Node node) {
        map.remove(node.key);
        node.prev.next = node.next;
        node.next.prev =  node.prev;
    }
    public void insert(Node node)
    {
        map.put(node.key,node);
        Node headNext = head.next;
        head.next = node;
        node.prev = head;
        node.next = headNext;
        headNext.prev = node;
    }
}

/**
 * Your LRUCache object will be instantiated and called as such:
 * LRUCache obj = new LRUCache(capacity);
 * int param_1 = obj.get(key);
 * obj.put(key,value);
 */

All the operations will take place in O(1) time😊

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