[LeetCode] 460. LFU Cache

Problem

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 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

Solution

class LFUCache {
    Map valMap;
    Map freqMap;
    Map> kSetMap;
    int size;
    int min;
    
    public LFUCache(int capacity) {
        min = 0;
        size = capacity;
        valMap = new HashMap<>();
        freqMap = new HashMap<>();
        kSetMap = new HashMap<>();
        kSetMap.put(1, new LinkedHashSet<>());
    }
    
    public int get(int key) {
        if (!valMap.containsKey(key)) return -1;
        
        //get frequency, then update freqMap, kSetMap, min
        int frequency = freqMap.get(key);
        freqMap.put(key, frequency+1);
        kSetMap.get(frequency).remove(key);
        if (!kSetMap.containsKey(frequency+1)) {
            kSetMap.put(frequency+1, new LinkedHashSet<>());
        }
        kSetMap.get(frequency+1).add(key);
        if (min == frequency && kSetMap.get(frequency).size() == 0) {
            min++;
        }
        
        return valMap.get(key);
        
    }
    
    public void put(int key, int value) {
        if (size <= 0) return;
        //when key is existed, just update valMap value, 
        //and call get(key) to update freqmap, kSetMap and min
        if (valMap.containsKey(key)) {
            valMap.put(key, value);
            get(key);
            return;
        }
        
        //when reached capacity, remove min in all 3 maps
        if (valMap.size() == size) {
            int minKey = kSetMap.get(min).iterator().next();
            valMap.remove(minKey);
            freqMap.remove(minKey);
            kSetMap.get(min).remove(minKey);
        }
        
        //add the fresh k-v pair to all 3 maps
        valMap.put(key, value);
        freqMap.put(key, 1);
        kSetMap.get(1).add(key);
    }
}