460. LFU Cache (Hard)

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 inO(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(2);       // returns -1 (not found)
cache.get(3);       // returns 3.
cache.put(4, 4);    // evicts key 1.
cache.get(1);       // returns -1 (not found)
cache.get(3);       // returns 3
cache.get(4);       // returns 4

Solution 1: Hash Map, Linked List

//minFreq is the smallest frequency so far //The main idea is to put all keys with the same frequency to a linked list so the least recent one can be evicted; //mIter stored the key's position in the linked list;

class LFUCache {
public:
    LFUCache(int capacity): cap(capacity), size(0), minFreq(0) {

    }

    int get(int key) {
        if (m.count(key) == 0) return -1; // not found
        int freq = ++m[key].second;
        fk[freq-1].erase(kl[key]);
        fk[freq].push_back(key);
        kl[key] = --fk[freq].end();
        if (fk[minFreq].size() == 0) ++minFreq;
        return m[key].first;
    }

    void put(int key, int value) {
        if (cap <= 0) return; // no space to put
        int storedValue = get(key);
        if (storedValue != -1) {
            m[key].first = value; return;
        }
        if (size >= cap) {
            int k = fk[minFreq].front(); // erase the least freq entry
            m.erase(k); kl.erase(k);
            fk[minFreq].pop_front();
            --size;
        }
        m[key] = {value, 1};
        fk[1].push_back(key);
        kl[key] = --fk[1].end();
        minFreq = 1;
        ++size;
    }
private:
    int cap, size, minFreq;
    unordered_map<int, pair<int,int>> m; // key: <value, freq>
    unordered_map<int, list<int>::iterator> kl; // key to list iter
    unordered_map<int, list<int>> fk; // freq to key list
};

/**
 * 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);
 */

Solution 2: PrioeityQueue