LFU

Design and implement a data structure for a Least Frequently Used (LFU) cache.

Implement the LFUCache class:

• LFUCache(int capacity) Initializes the object with the capacity of the data structure.
• int get(int key) Gets the value of the key if the key exists in the cache. Otherwise, returns -1.
• void put(int key, int value) Update the value of the key if present, or inserts the key if not already present. When the cache reaches its capacity, it should invalidate and remove the least frequently used key before inserting a new item. For this problem, when there is a tie (i.e., two or more keys with the same frequency), the least recently used key would be invalidated.

To determine the least frequently used key, a use counter is maintained for each key in the cache. The key with the smallest use counter is the least frequently used key.

When a key is first inserted into the cache, its use counter is set to 1 (due to the put operation). The use counter for a key in the cache is incremented either a get or put operation is called on it.

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

Example 1:

Input
["LFUCache", "put", "put", "get", "put", "get", "get", "put", "get", "get", "get"]
[[2], [1, 1], [2, 2], [1], [3, 3], [2], [3], [4, 4], [1], [3], [4]]
Output
[null, null, null, 1, null, -1, 3, null, -1, 3, 4]

Explanation
// cnt(x) = the use counter for key x
// cache=[] will show the last used order for tiebreakers (leftmost element is  most recent)
LFUCache lfu = new LFUCache(2);
lfu.put(1, 1);   // cache=[1,_], cnt(1)=1
lfu.put(2, 2);   // cache=[2,1], cnt(2)=1, cnt(1)=1
lfu.get(1);      // return 1
                 // cache=[1,2], cnt(2)=1, cnt(1)=2
lfu.put(3, 3);   // 2 is the LFU key because cnt(2)=1 is the smallest, invalidate 2.
                 // cache=[3,1], cnt(3)=1, cnt(1)=2
lfu.get(2);      // return -1 (not found)
lfu.get(3);      // return 3
                 // cache=[3,1], cnt(3)=2, cnt(1)=2
lfu.put(4, 4);   // Both 1 and 3 have the same cnt, but 1 is LRU, invalidate 1.
                 // cache=[4,3], cnt(4)=1, cnt(3)=2
lfu.get(1);      // return -1 (not found)
lfu.get(3);      // return 3
                 // cache=[3,4], cnt(4)=1, cnt(3)=3
lfu.get(4);      // return 4
                 // cache=[3,4], cnt(4)=2, cnt(3)=3

Constraints:

• 0 <= capacity <= 104
• 0 <= key <= 105
• 0 <= value <= 109
• At most 2 * 105 calls will be made to get and put.

class LFUCache {
    class Node{
        int key;
        int val;
        int freq;
        int seq;
        Node(int key,int val,int seq){
            this.key=key;
            this.val=val;
            this.seq=seq;
            this.freq=1;
        }
    }
    PriorityQueue pq;
    HashMap map;
    int currseq = 0;
    int cap=0;
    
    public LFUCache(int capacity) {
        this.cap = capacity;
        pq = new PriorityQueue((x,y)->{
            if( x.freq != y.freq ) return x.freq-y.freq;
            return x.seq-y.seq;
        });
        map = new HashMap();
    }
    
    public int get(int key) {
        if(cap==0) return -1;
        //0.get Node with key
        Node node  = map.get(key);
        //1.if key exists
        if(node!=null){
            //remove from pq
            pq.remove(node);
            //let freq++;
            node.freq++;
            node.seq=currseq++;
            //add to pq again
            pq.offer(node);
            //return val;
            return node.val;
        }
        //2.key not exists
        else
            return -1;        
    }
    
    public void put(int key, int value) {
        if(cap==0) return;
        //0.get Node with key
        Node node  = map.get(key);
        //1.if key exists
        if(node!=null){
            //set the new value
            node.val=value;
            //remove from pq
            pq.remove(node);
            //let freq++;
            node.seq=currseq++;
            node.freq++;
            //add to pq again     
            pq.offer(node);
        }
        //2.key not exists
        else{
            //new Node
            node  = new Node(key,value,currseq++);
            //if the map is full
            if(map.size()==cap){
                Node temp  = pq.poll();
                map.remove(temp.key);
            }
            //put it into pq
            pq.offer(node);
            //put it into map
            map.put(key,node);
        }
    }
}