460. LFU Cache

HardLeetCode

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) updates 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

• 1 <= capacity <= 10^4
• 0 <= key <= 10^5
• 0 <= value <= 10^9
• At most 2 * 10^5 calls will be made to get and put.

How to solve the problem

• Double HashMap + LinkedHashMap (OrderedDict) Approach

class LFUCache:
    def __init__(self, capacity: int):
        self.capacity = capacity
        self.size = 0
        self.minFreq = 0
        self.keyMap = {}  # key -> (value, frequency)
        self.freqMap = defaultdict(OrderedDict)  # frequency -> OrderedDict(key -> value)

    def get(self, key: int) -> int:
        if key not in self.keyMap:
            return -1
        val, freq = self.keyMap[key]
        # Remove the key from the old frequency bucket
        del self.freqMap[freq][key]
        if not self.freqMap[freq]:
            del self.freqMap[freq]
            # If the old frequency was the min frequency and now empty, increase minFreq
            if self.minFreq == freq:
                self.minFreq += 1
        # Insert the key into the new frequency bucket
        self.freqMap[freq+1][key] = val
        self.keyMap[key] = (val, freq+1)
        return val

    def put(self, key: int, value: int) -> None:
        if self.capacity == 0:
            return
        if key in self.keyMap:
            # Equivalent to get(key), then update the value
            self.keyMap[key] = (value, self.keyMap[key][1])
            self.get(key)  # Increase frequency
            return
        if self.size >= self.capacity:
            # Evict the LRU key from the minFreq bucket
            k, v = self.freqMap[self.minFreq].popitem(last=False)
            del self.keyMap[k]
            if not self.freqMap[self.minFreq]:
                del self.freqMap[self.minFreq]
        else:
            self.size += 1
        # Insert the new key with frequency = 1
        self.keyMap[key] = (value, 1)
        self.freqMap[1][key] = value
        self.minFreq = 1


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

Complexity

OrderedDict Approach

• Time Complexity: O(1) for both get and put operations
  • get: O(1) for key lookup in keyMap + O(1) for frequency bucket operations
  • put: O(1) for dictionary operations + O(1) for frequency bucket management
• Space Complexity: O(capacity) for storing the cache entries and frequency buckets

The approach achieves the required O(1) time complexity for all operations by using a combination of hash maps and ordered dictionaries to efficiently track both key-value pairs and their frequencies.