lfu_cache.py

"""
https://leetcode.com/problems/lfu-cache/

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.

 

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=[3,4], cnt(4)=2, cnt(3)=3
 

Constraints:

0 <= capacity, key, value <= 104
At most 105 calls will be made to get and put.
 

Follow up: Could you do both operations in O(1) time complexity?
"""

from collections import defaultdict, OrderedDict

class Node:
    def __init__(self, key, val, freq):
        self.val = val
        self.key = key 
        self.freq = freq

class LFUCache:

    def __init__(self, capacity: int):

        self.capacity = capacity
        self.min_freq = None 

        # hashmap for key-node
        self.key2node = dict()

        # doubly linked hashmap
        self.freq2node = defaultdict(OrderedDict)
        

    def get(self, key: int) -> int:
        if key not in self.key2node:
            return -1

        node = self.key2node[key]

        del self.freq2node[node.freq][key]

        if not self.freq2node[node.freq]:
            del self.freq2node[node.freq]

        node.freq += 1
        self.freq2node[node.freq][key] = node 

        if not self.freq2node[self.min_freq]:
            self.min_freq += 1

        return node.val            

        

    def put(self, key: int, value: int) -> None:
        if not self.capacity:
            return 

        if key in self.key2node:
            self.key2node[key].val = value 
            _ = self.get(key)
            return 

        if len(self.key2node) == self.capacity:
            k, node = self.freq2node[self.min_freq].popitem(last=False)
            del self.key2node[k]

        self.freq2node[1][key] = self.key2node[key] = Node(key, value, 1)
        self.min_freq = 1
                     

        


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