LFU Cache

Problem

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.

Examples

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

Solution

Method 1 - PriorityQueue and HashMap

Code

class LFUCache {
	long stamp;
	int capacity;
	int num;
	PriorityQueue<Pair> minHeap;
	HashMap<Integer, Pair> hashMap;
	
	// @param capacity, an integer
	public LFUCache(int capacity) {
	    // Write your code here
	    this.capacity = capacity;
	    num = 0;
	    minHeap = new PriorityQueue<Pair>();
	    hashMap = new HashMap<Integer, Pair>();
	    stamp = 0;
	}
	
	// @param key, an integer
	// @param value, an integer
	// @return nothing
	public void set(int key, int value) {
	    if (capacity == 0) {
	        return;
	    }
	    // Write your code here
	    if (hashMap.containsKey(key)) {
	        Pair old = hashMap.get(key);
	        minHeap.remove(old);
	        
	        Pair newNode = new Pair(key, value, old.times + 1, stamp++);
	        
	        hashMap.put(key, newNode);
	        minHeap.offer(newNode);
	    } else if (num == capacity) {
	        Pair old = minHeap.poll();
	        hashMap.remove(old.key);
	        
	        Pair newNode = new Pair(key, value, 1, stamp++);
	        
	        hashMap.put(key, newNode);
	        minHeap.offer(newNode);
	    } else {
	        num++;
	        Pair pair = new Pair(key, value, 1, stamp++);
	        hashMap.put(key, pair);
	        minHeap.offer(pair);
	    }
	}
	
	public int get(int key) {
	    if (capacity == 0) {
	        return -1;
	    }
	    // Write your code here
	    if (hashMap.containsKey(key)) {
	        Pair old = hashMap.get(key);
	        minHeap.remove(old);
	        
	        Pair newNode = new Pair(key, old.value, old.times + 1, stamp++);
	        
	        hashMap.put(key, newNode);
	        minHeap.offer(newNode);
	        return hashMap.get(key).value;
	    } else {
	        return -1;
	    }
	}
	
	class Pair implements Comparable<Pair> {
	    long stamp;
	    int key;
	    int value;
	    int times;
	    public Pair(int key, int value, int times, long stamp) {
	        this.key = key;
	        this.value = value;
	        this.times = times;
	        this.stamp = stamp;
	    }
	    
	    public int compareTo(Pair that) {
	        if (this.times == that.times) {
	            return (int)(this.stamp - that.stamp);
	        } else {
	            return this.times - that.times;    
	        }
	    }
	}
}

Complexity

• Time: get O(n) where n is capacity, and heap remove takes O(n) time.
• Space: O(capacity)

Method 2 - Using 3 Hashmaps and LinkedHashSet

Code

class LFUCache {
    private final Map<Integer, Integer> vals;
    private final Map<Integer, Integer> counts; // count of get's
    private final Map<Integer, LinkedHashSet<Integer>> freqToValList;
    private final int capacity;
    private int min = -1;

    public LFUCache(int capacity) {
        this.capacity = capacity;
        vals = new HashMap<>();
        counts = new HashMap<>();
        freqToValList = new HashMap<>();
        freqToValList.put(1, new LinkedHashSet<>());
    }

    public int get(int key) {
        if (!vals.containsKey(key)) {
            return -1;
        }
        int count = counts.get(key);
        counts.put(key, count + 1);
        freqToValList.get(count).remove(key);
        if (count == min && freqToValList.get(count).size() == 0) {
            min++;
        }

        freqToValList.putIfAbsent(count + 1, new LinkedHashSet<>());
        freqToValList.get(count + 1).add(key);
        return vals.get(key);
    }

    public void put(int key, int value) {
        if (capacity <= 0) {
            return;
        }
        if (vals.containsKey(key)) {
            vals.put(key, value);
            get(key);
            return;
        }
        if (vals.size() >= capacity) {
            int evicted = freqToValList.get(min).iterator().next();
            freqToValList.get(min).remove(evicted);
            vals.remove(evicted);
        }
        vals.put(key, value);
        counts.put(key, 1);
        min = 1;
        freqToValList.get(1).add(key);
    }
}

Complexity

• Time: set O(capacity) get O(capacity)
• Space: O(capacity)