LRU cache

Problem

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.

The functions get and put must each run in O(1) average 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

Constraints:

  • 1 <= capacity <= 3000
  • 0 <= key <= 104
  • 0 <= value <= 105
  • At most 2 * 105 calls will be made to get and put.

Solution

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

    int size;
    Node head;
    Node tail;

    // mapping of key: node
    // allows O(1) delete
    Map<Integer, Node> m = new HashMap<>();

    public LRUCache(int capacity) {
        this.size = capacity;
        this.head = new Node(-1, -1);
        this.tail = new Node(-2, -2);
        head.next = tail;
        tail.prev = head;
    }

    void add(int key, int value) {
        var n = new Node(key ,value);
        m.put(key, n);
        n.prev = tail.prev;
        n.next = tail;
        n.prev.next = n;
        n.next.prev = n;
    }

    void remove(int key) {
        var n = m.get(key);
        m.remove(key);
        n.prev.next = n.next;
        n.next.prev = n.prev;
    }

    public int get(int key) {
        if (m.containsKey(key)) {
            var value = m.get(key).val;
            remove(key);
            add(key, value);
            return value;
        } else {
            return -1;
        }
    }

    public void put(int key, int value) {
        if (m.containsKey(key)) {
            remove(key);
        }
        add(key, value);
        if (m.size() > size) {
            remove(head.next.key);
        }
    }
}

/**
 * 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);
 */

Recent posts from blogs that I like

Test iCloud Drive using Cirrus

Problems syncing files with iCloud Drive? Don't just turn it off and back on again. Use Cirrus instead to upload a test file, and check whether syncing that works. Full details.

via The Eclectic Light Company

Concurrent Servers: Part 8 - Go

This is part 8 in a series of posts on writing concurrent network servers. In this part, we'll switch to Go and see how it tackles the challenges described earlier in the series. All posts in the series: Part 1 - Introduction Part 2 - Threads Part 3 - Event-driven Part 4 - libuv Part 5 - Redis case ...

via Eli Bendersky

You should never be angry at work

I try not to give a lot of prescriptive advice about working in tech companies1. There are many ways to be successful, and every company works differently. If you’re shipping projects and your management chain is happy, it doesn’t really matter how you’ve accomplished it. However, there’s one thing ...

via Sean Goedecke