0460.LFU_Cache.py

"""
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:

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

"""
1.好的思路是用2个dict去存，key->freq and freq -> (key, value),以及要存储最小评率
因为我们既要根据key查freq,又要根据freq查key,注意题目中很多细节问题
"""
class LFUCache:

    def __init__(self, capacity: int):
        self.capacity = capacity
        self.key_to_freq = defaultdict(int)
        self.freq_to_orddict = defaultdict(OrderedDict)
        self.min_freq = 0

    def get(self, key: int) -> int:
        if key not in self.key_to_freq:
            return -1
        freq = self.key_to_freq[key]
        val = self.freq_to_orddict[freq][key]
        self.update(key, val)
        return val

    def put(self, key: int, value: int) -> None:
        if self.capacity == 0:
            return
    
        #if key exist in key_to_freq, 更新
        if key in self.key_to_freq:
            self.update(key, value)
        
        #if key not exist in key_to_freq 更新最小频率，加入key_to_freq， freq_to_orddict。如果长度大于capacity要更新capacity
        else:
            if len(self.key_to_freq) == self.capacity:  
                min_key, min_val = self.freq_to_orddict[self.min_freq].popitem(last=False)
                del self.key_to_freq[min_key]  # del the least used key from the cache
            
            self.min_freq = 1
            self.key_to_freq[key] = 1
            self.freq_to_orddict[1][key] = value

    """
    1.先更新freq
    2.更新value：
    3.更新min_freq
    """
    def update(self, key: int, val: int):
        freq = self.key_to_freq[key]
        self.freq_to_orddict[freq].pop(key)    # delete key,value in freq_to_orddict.value中
        if self.min_freq == freq and len(self.freq_to_orddict[freq]) == 0:
            self.min_freq += 1
        
        freq += 1
        self.key_to_freq[key] = freq
        self.freq_to_orddict[freq][key] = val
        

# Your LFUCache object will be instantiated and called as such:
# obj = LFUCache(capacity)
# param_1 = obj.get(key)
# obj.put(key,value)