# 460. LFU Cache

### Problem:

Design and implement a data structure for Least Frequently Used (LFU) 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 reaches its capacity, it should invalidate the least frequently used item before inserting a new item. For the purpose of this problem, when there is a tie (i.e., two or more keys that have the same frequency), the least recently used key would be evicted.

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

Example:

``````LFUCache cache = new LFUCache( 2 /* capacity */ );

cache.put(1, 1);
cache.put(2, 2);
cache.get(1);       // returns 1
cache.put(3, 3);    // evicts key 2
cache.get(3);       // returns 3.
cache.put(4, 4);    // evicts key 1.
cache.get(3);       // returns 3
cache.get(4);       // returns 4
``````

### Solutions:

``````public class LFUCache {
private class Node {
public Node pre;
public Node next;
public int key;
public int val;
public int freq;
public Node (int key, int val) {
this.key = key;
this.val = val;
this.pre = null;
this.next = null;
this.freq = 0;
}
}
private HashMap<Integer, Node> heads;
private HashMap<Integer, Node> tails;
private HashMap<Integer, Node> data;
private PriorityQueue<Integer> freqs;
private int capa;
public LFUCache(int capacity) {
this.capa = capacity;
heads = new HashMap<Integer, Node>();
tails = new HashMap<Integer, Node>();
heads.put(0, new Node(-1, -1));
tails.put(0, new Node(-1, -1));
data = new HashMap<Integer, Node>();
freqs = new PriorityQueue<Integer>();
}

public int get(int key) {
if (!data.containsKey(key) || capa == 0) {
return -1;
}
Node n = data.get(key);
increase(n);
return n.val;
}
private void increase(Node n) {
freqs.remove(n.freq);
n.freq ++;
remove(n);
heads.put(n.freq, new Node(-1, -1));
tails.put(n.freq, new Node(-1, -1));
}
Node tail = tails.get(n.freq);
n.pre = tail.pre;
tail.pre = n;
n.pre.next = n;
n.next = tail;
}
private void remove(Node n) {
Node pre = n.pre;
Node next = n.next;
pre.next = next;
next.pre = pre;
}
public void put(int key, int value) {
if (capa == 0) {
return;
}
if (data.containsKey(key)) {
data.get(key).val = value;
increase(data.get(key));
return;
}
if (data.size() == capa) {
int freq = freqs.poll();
Node n = heads.get(freq).next;
remove(n);
data.remove(n.key);
}
Node n = new Node(key, value);
data.put(key, n);
tails.get(0).pre = n;
n.next = tails.get(0);
increase(n);
}
}

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