460. Lfu Cache¶

Difficulty: Hard

460. LFU Cache

Hard

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=[4,3], cnt(4)=2, cnt(3)=3

Constraints:

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

Solution¶

class LFUCache {
    private TreeSet<Node> set;
    private HashMap<Integer, Integer> map;
    private int currTime;
    private HashMap<Integer, Integer> freq;
    private HashMap<Integer, Integer> time;
    private int totalCap, currCap;

    static class Node {
        int key, value, count, currTime;
        public Node(int key, int value, int count, int currTime) {
            this.key = key;
            this.value = value;
            this.count = count;
            this.currTime = currTime;
        }
        @Override
        public String toString() {
            return "(" + key + " " + value + " " + count + " " + currTime + ")";
        }

        @Override
        public boolean equals(Object obj) {
            if (this == obj)
                return true;
            if (obj == null || getClass() != obj.getClass())
                return false;
            Node current = (Node)(obj);
            return current.key == key && current.value == value && current.count == count && current.currTime == currTime;
        }

        @Override
        public int hashCode() {
            return Objects.hash(key, value, count, currTime);
        }
    }

    static class customSort implements Comparator<Node> {
        @Override
        public int compare(Node first, Node second) {
            int op1 = Integer.compare(second.count, first.count);
            if (op1 != 0)
                return op1;
            return Integer.compare(second.currTime, first.currTime);
        }
    }

    public LFUCache(int capacity) {
        set = new TreeSet<>(new customSort()); 
        map = new HashMap<>();
        currTime = 0;
        freq = new HashMap<>();
        time = new HashMap<>();
        currCap = 0;
        totalCap = capacity;
    }

    public int get(int key) {
        if (map.containsKey(key)) {
            Node prev = new Node(key, map.get(key), freq.get(key), time.get(key));
            int res = map.get(key);
            set.remove(prev);
            time.put(key, currTime);
            set.add(new Node(key, map.get(key), freq.get(key) + 1, currTime));
            freq.put(key, freq.get(key) + 1);
            currTime++;
            return res;
        }
        return -1;
    }

    public void put(int key, int value) {
        if (map.containsKey(key)) {
            Node prev = new Node(key, map.get(key), freq.get(key), time.get(key));
            set.remove(prev);
            map.put(key, value);
            freq.put(key, freq.get(key) + 1);
            time.put(key, currTime);
            set.add(new Node(key, value, freq.get(key), currTime));
            currTime++;
        } else {
            if (currCap < totalCap) {
                map.put(key, value);
                freq.put(key, 1);
                time.put(key, currTime);
                set.add(new Node(key, value, 1, currTime));
                currTime++;
                currCap++; 
            } else {
                Node last = set.pollLast();
                map.remove(last.key);
                map.put(key, value);
                freq.put(key, 1);
                time.put(key, currTime);
                set.add(new Node(key, value, 1, currTime));
                currTime++;
                currCap++; 
            }
        }
    }
}

/**
 * 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);
 */

Complexity Analysis¶

Time Complexity: O(?)
Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here