Problem description:

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:

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

Solution:

get(key)

  1. query the node by calling self.cache[key]
  2. find the frequency by checking node.freq, assigned as f, and query the freq_bucket that this node is in, through calling self.freq_bucket[freq]
  3. pop this node
  4. update node’s frequence, append the node to the new freq_bucket with frequency freq+1
  5. if the freq_bucket is empty and self._min_freq == freq, update self._minfreq to freq+1.
  6. return node.val

put(key, value)

  • If key is already in cache, do the same thing as get(key), and update node.val as value
  • Otherwise:
    1. if the cache is full, pop the least frequenly used element (*)
    2. add new node to self._node
    3. add new node to self.freq_bucket[1]
    4. reset self.min_freq to 1

Python popitem(last): Pairs are returned in LIFO order if last is true or FIFO order if false.

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class LFUCache:
    def __init__(self, capacity: int):
        self.cache = {} # cache[ket] = [freq, value]
        self.freq_bucket = defaultdict(OrderedDict) # freq_brucket[1][key] = value
        self.min_freq = 0
        self.capacity = capacity

    def get(self, key: int, new_value = None) -> int:
        if key in self.cache:
            freq, value = self.cache.pop(key)
            if new_value:
                value = new_value
            self.freq_bucket[freq].pop(key)
            self.freq_bucket[freq+1][key] = value
            self.cache[key] = [freq+1, value]
            if self.min_freq == freq and not self.freq_bucket[freq]:
                self.min_freq += 1
            return value
        else:
            return -1

    def put(self, key: int, value: int) -> None:
        if not self.capacity:
            return 
        if key in self.cache:
            self.get(key, value)
        else:
            # new element
            if len(self.cache) == self.capacity:
                pop_key, pop_value = self.freq_bucket[self.min_freq].popitem(last=False)
                del self.cache[pop_key]
            
            self.cache[key] = [1, value]
            self.freq_bucket[1][key] = value
            self.min_freq = 1

time complexity: $O(1)$
space complexity: $O(1)$
reference:
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