LFU Cache

Problem Statement

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.

Leetcode link

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

Constaints:

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

Code

Python

from collections import defaultdict
from collections import OrderedDict

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

class LFUCache(object):
    def __init__(self, capacity):
        """
        :type capacity: int
        """
        self.cap = capacity
        self.key2node = {}
        self.count2node = defaultdict(OrderedDict)
        self.minCount = None

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

        node = self.key2node[key]
        del self.count2node[node.count][key]

        # clean memory
        if not self.count2node[node.count]:
            del self.count2node[node.count]

        node.count += 1
        self.count2node[node.count][key] = node

        # NOTICE check minCount!!!
        if not self.count2node[self.minCount]:
            self.minCount += 1


        return node.val

    def put(self, key, value):
        """
        :type key: int
        :type value: int
        :rtype: void
        """
        if not self.cap:
            return

        if key in self.key2node:
            self.key2node[key].val = value
            self.get(key) # NOTICE, put makes count+1 too
            return

        if len(self.key2node) == self.cap:
            # popitem(last=False) is FIFO, like queue
            # it return key and value!!!
            k, n = self.count2node[self.minCount].popitem(last=False)
            del self.key2node[k]

        self.count2node[1][key] = self.key2node[key] = Node(key, value, 1)
        self.minCount = 1
        return