Lesson 2 of 27
In Progress

LRU Cache

August 31, 2020

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.

Least Recently Used (LRU) Cache organizes items in order of use, allowing you to quickly identify which item hasn’t been used for the longest amount of time.

Under the hood, an LRU cache is often implemented by pairing a doubly linked list with a HashMap / ConcurrentHashMap (For Thread Safe).

We’ll set up our linked list with the most-recently used item at the head of the list and the least-recently used item at the tail. This lets us access the LRU element in O(1) time by looking at the tail of the list.

In genral finding an element in LinkedList will take O(n) time. Inorder to optimize the search we add HashMap to the it which and that lets us find an element in our cache’s linked list in O(1) time, instead of O(n).

LRUCache cache = new LRUCache( 2 /* capacity */ );
cache.put(1, 1)
cache.put(2, 2)
cache.put(3, 3)
cache.put(4, 4)

cache.get(1): 1
cache.get(2): -1
cache.get(1): -1
cache.get(3): 3
cache.get(4): 4

cache.put(1, 1)
cache.put(2, 2)
cache.get(1): 1
cache.put(3, 3)
cache.get(2): -1
cache.put(4, 4)
cache.get(1): -1
cache.get(3): 3
cache.get(4): 4
import java.util.HashMap;
import java.util.Map;
import java.util.concurrent.ConcurrentHashMap;
public class LRUCache {
public static void main(String[] args) {
LRUCache cache = new LRUCache(2);
System.out.println("cache.put(1, 1)");
cache.put(1, 1);
System.out.println("cache.put(2, 2)");
cache.put(2, 2);
System.out.println("cache.get(1): " + cache.get(1));
System.out.println("cache.put(3, 3)");
cache.put(3, 3);
System.out.println("cache.get(2): " + cache.get(2));
System.out.println("cache.put(4, 4)");
cache.put(4, 4);
System.out.println("cache.get(1): " + cache.get(1));
System.out.println("cache.get(3): " + cache.get(3));
System.out.println("cache.get(4): " + cache.get(4));
private Map<Integer, Node> cache;
private Node head;
private Node tail;
private int capacity;
public LRUCache(int capacity) {
this.capacity = capacity;
// Here I am using ConcurrentHashMap just to be thread safe.
cache = new ConcurrentHashMap<>();
head = new Node(0, 0);
tail = new Node(0, 0);
head.next = tail;
tail.pre = head;
head.pre = null;
tail.next = null;
public int get(int key) {
// if cache does not contains key return -1
if (!cache.containsKey(key)) {
return -1;
// if it contains key get the linkedlist node
Node node = cache.get(key);
int result = node.value;
// delete the node from that linkedlist and add it to the head
// adding node to head as it is recently used.
return result;
public void put(int key, int value) {
// check if it there in cache or not if it is in cache,
// add to the head of linkedlist
if (cache.get(key) != null) {
Node node = cache.get(key);
node.value = value;
} else { // element is not in the cache so we have to add in cache and also to head of linkedlist
Node node = new Node(key, value);
cache.put(key, node);
// if cache has capacity add it
if (cache.size() <= capacity) {
} else { // cache does not have capacity remove last recently used element which is at tail
// here tail points to numm and so last element is tail.pre
public void deleteNode(Node node) {
node.pre.next = node.next;
node.next.pre = node.pre;
public void addToHead(Node node) {
node.next = head.next;
node.next.pre = node;
node.pre = head;
head.next = node;
// ------- Double Linked List Node --------
private static class Node {
int key;
int value;
Node pre;
Node next;
public Node(int key, int value) {
this.key = key;
this.value = value;
public String toString() {
return " Node{" +
"k=" + key +
", v=" + value +
view raw LRUCache.java hosted with ❤ by GitHub

Question Source: LeetCode – 146

  • Time Complexity:
    • get: O(1)
    • set: O(1)
  • Space Complexity: O(N) where N is number of elements in cache