146. LRU Cache

Problem

Design and implement a data structure for Least Recently Used (LRU) cache. It should support the following operations: get and put.

get(key) - Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1. put(key, value) - Set or insert the value if the key is not already present. When the cache reached its capacity, it should invalidate the least recently used item before inserting a new item.

Follow up: Could you do both operations in O(1) time complexity?

Example:

LRUCache cache = new LRUCache( 2 /* capacity */ );

cache.put(1, 1);
cache.put(2, 2);
cache.get(1);       // returns 1
cache.put(3, 3);    // evicts key 2
cache.get(2);       // returns -1 (not found)
cache.put(4, 4);    // evicts key 1
cache.get(1);       // returns -1 (not found)
cache.get(3);       // returns 3
cache.get(4);       // returns 4

Solution (Java)

import java.util.HashMap;
import java.util.Map;

public class LRUCache {
    private static class LruCacheNode {
        int key;
        int value;
        LruCacheNode prev;
        LruCacheNode next;

        public LruCacheNode(int k, int v) {
            key = k;
            value = v;
        }
    }

    private int capacity;
    private Map<Integer, LruCacheNode> cacheMap = new HashMap<>();
    // insert here
    private LruCacheNode head;
    // remove here
    private LruCacheNode tail;

    public LRUCache(int cap) {
        capacity = cap;
    }

    public int get(int key) {
        LruCacheNode val = cacheMap.get(key);
        if (val == null) {
            return -1;
        }
        moveToHead(val);
        return val.value;
    }

    /*
     * Scenarios :
     * 1. Value key is already present.
     *          update
     *          move node to head
     * 2. cache is not full
     *          cache is empty (create node and assign head and tail)
     *          cache is partially empty (add node to head and update head pointer)
     * 3. cache is full
     *          remove node at tail, update head, tail pointers
     *          Recursively call put
     *
     *
     * move node to head Scenarios
     * 1. node is at head
     * 2. node is at tail
     * 3. node is in middle
     *
     */
    public void put(int key, int value) {
        LruCacheNode valNode = cacheMap.get(key);
        if (valNode != null) {
            valNode.value = value;
            moveToHead(valNode);
        } else {
            if (cacheMap.size() < capacity) {
                if (cacheMap.size() == 0) {
                    LruCacheNode node = new LruCacheNode(key, value);
                    cacheMap.put(key, node);
                    head = node;
                    tail = node;
                } else {
                    LruCacheNode node = new LruCacheNode(key, value);
                    cacheMap.put(key, node);
                    node.next = head;
                    head.prev = node;
                    head = node;
                }
            } else {
                // remove from tail
                LruCacheNode last = tail;
                tail = last.prev;
                if (tail != null) {
                    tail.next = null;
                }
                cacheMap.remove(last.key);
                if (cacheMap.size() == 0) {
                    head = null;
                }
                // Call recursively
                put(key, value);
            }
        }
    }

    /*
     * check for 3 conditions
     * 1. node is already at head
     * 2. Node is tail node
     * 3. Node in middle node
     */
    private void moveToHead(LruCacheNode node) {
        if (node == head) {
            return;
        }
        if (node == tail) {
            tail = node.prev;
        }
        // node is not head, it should have some valid prev node
        LruCacheNode prev = node.prev;
        LruCacheNode next = node.next;
        prev.next = next;
        if (next != null) {
            next.prev = prev;
        }
        node.prev = null;
        node.next = head;
        head.prev = node;
        head = 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);
 */

Solution (Javascript)

var List = function (key, val) {
  this.key = key;
  this.val = val;
  this.next = null;
  this.prev = null;
};

/**
 * @param {number} capacity
 */
var LRUCache = function(capacity) {
  this.capacity = capacity;
  this.length = 0;
  this.map = {};
  this.head = null;
  this.tail = null;
};

/** 
 * @param {number} key
 * @return {number}
 */
LRUCache.prototype.get = function(key) {
  var node = this.map[key];
  if (node) {
    this.remove(node);
    this.insert(node.key, node.val);
    return node.val;
  } else {
    return -1;
  }
};

/** 
 * @param {number} key 
 * @param {number} value
 * @return {void}
 */
LRUCache.prototype.put = function(key, value) {
  if (this.map[key]) {
    this.remove(this.map[key]);
    this.insert(key, value);
  } else {
    if (this.length === this.capacity) {
      this.remove(this.head);
      this.insert(key, value);
    } else {
      this.insert(key, value);
      this.length++;
    }
  }
};

/** 
 * Your LRUCache object will be instantiated and called as such:
 * var obj = Object.create(LRUCache).createNew(capacity)
 * var param_1 = obj.get(key)
 * obj.put(key,value)
 */

LRUCache.prototype.remove = function (node) {
  var prev = node.prev;
  var next = node.next;
  if (next) next.prev = prev;
  if (prev) prev.next = next;
  if (this.head === node) this.head = next;
  if (this.tail === node) this.tail = prev;
  delete this.map[node.key];
};

LRUCache.prototype.insert = function (key, val) {
  var node = new List(key, val);
  if (!this.tail) {
    this.tail = node;
    this.head = node;
  } else {
    this.tail.next = node;
    node.prev = this.tail;
    this.tail = node;
  }
  this.map[key] = node;
};

Explain:

nope.

Complexity:

• Time complexity : O(1).
• Space complexity : O(n).