Algorithm-Day36-LRU-Cache-lc-146

🧩 Problem Description

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

Implement the LRUCache class:

• LRUCache(int capacity) initializes the cache with a positive size capacity.
• int get(int key) returns the value of the key if it exists, otherwise returns -1.
• void put(int key, int value) updates the value of the key. If the number of keys exceeds the capacity, evict the least recently used key.

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💬 Example

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]

💡 Intuition

To achieve O(1) time complexity for both get and put, use:

• A HashMap to store key → node
• A Doubly Linked List to track the usage order
  • Most recently used at the front
  • Least recently used at the end

When a key is accessed:

• Move it to the front of the list
  When inserting:
• If at capacity, remove the tail (least recently used)

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🔢 Java Code (Custom Doubly Linked List)

class LRUCache {

    class Node {
        int key, value;
        Node prev, next;
        Node(int k, int v) {
            key = k;
            value = v;
        }
    }

    private final int capacity;
    private Map<Integer, Node> cache;
    private Node head, tail;

    public LRUCache(int capacity) {
        this.capacity = capacity;
        cache = new HashMap<>();

        // Dummy head and tail
        head = new Node(0, 0);
        tail = new Node(0, 0);

        head.next = tail;
        tail.prev = head;
    }

    public int get(int key) {
        if (!cache.containsKey(key)) return -1;

        Node node = cache.get(key);
        moveToHead(node);
        return node.value;
    }

    public void put(int key, int value) {
        if (cache.containsKey(key)) {
            Node node = cache.get(key);
            node.value = value;
            moveToHead(node);
        } else {
            if (cache.size() == capacity) {
                Node lru = tail.prev;
                remove(lru);
                cache.remove(lru.key);
            }

            Node newNode = new Node(key, value);
            cache.put(key, newNode);
            addToHead(newNode);
        }
    }

    // Doubly Linked List helpers
    private void remove(Node node) {
        node.prev.next = node.next;
        node.next.prev = node.prev;
    }

    private void addToHead(Node node) {
        node.next = head.next;
        node.prev = head;
        head.next.prev = node;
        head.next = node;
    }

    private void moveToHead(Node node) {
        remove(node);
        addToHead(node);
    }
}

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⏱ Time and Space Complexity

• Time Complexity:
  • get: O(1)
  • put: O(1)
• Space Complexity: O(capacity)

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🧠 Summary

• Combines a HashMap (for O(1) access) and a Doubly Linked List (for O(1) insert/delete).
• A great example of system design + data structure synergy.
• Common interview favorite!

Know this one cold — it’s the basis for many real-world cache systems (like Redis, in-memory caching, etc.)