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# 460. LFU Cache

Design and implement a data structure for [Least Frequently Used (LFU)](https://en.wikipedia.org/wiki/Least_frequently_used) 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(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
```

```cpp
class LFUCache {
public:
    LFUCache(int capacity) : cap(capacity), minFreq(0) {
        
    }
    
    int get(int key) {
        if (!m.count(key)) return -1;
        freq[m[key].second].erase(iter[key]);
        ++m[key].second; // increment the frequency of the key
        freq[m[key].second].push_back(key);
        iter[key] = --freq[m[key].second].end();
        if (freq[minFreq].empty()) ++minFreq;
        return m[key].first;
    }
    
    void put(int key, int value) {
        if (cap <= 0) return;
        if (get(key) != -1) {
            m[key].first = value;
            return;
        }
        if (m.size() >= cap) {
            m.erase(freq[minFreq].front());
            iter.erase(freq[minFreq].front());
            freq[minFreq].pop_front();
        }
        m[key] = {value, 1};
        freq[1].push_back(key);
        iter[key] = --freq[1].end();
        minFreq = 1;
    }
private:
    int cap;
    int minFreq;
    unordered_map<int, pair<int, int> > m; // key -> <value, freq>
    unordered_map<int, list<int> > freq; // freq -> list of keys with the same frequency
    unordered_map<int, list<int>::iterator> iter; // key -> iterator pointing to the place in the freq list
};
```