LeetCode 0460. LFU Cache Solution in Java, C++, Python & More | Explanation + Code

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0460. LFU Cache

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:

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

 

 

Solutions

Solution 1

PythonJavaC++GoRust
class Node: def __init__(self, key: int, value: int) -> None: self.key = key self.value = value self.freq = 1 self.prev = None self.next = None class DoublyLinkedList: def __init__(self) -> None: self.head = Node(-1, -1) self.tail = Node(-1, -1) self.head.next = self.tail self.tail.prev = self.head def add_first(self, node: Node) -> None: node.prev = self.head node.next = self.head.next self.head.next.prev = node self.head.next = node def remove(self, node: Node) -> Node: node.next.prev = node.prev node.prev.next = node.next node.next, node.prev = None, None return node def remove_last(self) -> Node: return self.remove(self.tail.prev) def is_empty(self) -> bool: return self.head.next == self.tail class LFUCache: def __init__(self, capacity: int): self.capacity = capacity self.min_freq = 0 self.map = defaultdict(Node) self.freq_map = defaultdict(DoublyLinkedList) def get(self, key: int) -> int: if self.capacity == 0 or key not in self.map: return -1 node = self.map[key] self.incr_freq(node) return node.value def put(self, key: int, value: int) -> None: if self.capacity == 0: return if key in self.map: node = self.map[key] node.value = value self.incr_freq(node) return if len(self.map) == self.capacity: ls = self.freq_map[self.min_freq] node = ls.remove_last() self.map.pop(node.key) node = Node(key, value) self.add_node(node) self.map[key] = node self.min_freq = 1 def incr_freq(self, node: Node) -> None: freq = node.freq ls = self.freq_map[freq] ls.remove(node) if ls.is_empty(): self.freq_map.pop(freq) if freq == self.min_freq: self.min_freq += 1 node.freq += 1 self.add_node(node) def add_node(self, node: Node) -> None: freq = node.freq ls = self.freq_map[freq] ls.add_first(node) self.freq_map[freq] = ls # Your LFUCache object will be instantiated and called as such: # obj = LFUCache(capacity) # param_1 = obj.get(key) # obj.put(key,value)(code-box)

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