ARTS #126

Algorithm

本周选择的算法题是：LFU Cache。

规则

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=[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

class Node:
    """
    双向链表的节点抽象
    """

    def __init__(self, key, value):
        self.key = key
        self.value = value
        self.frequency = 1
        self.previous = self.next = None

class LinkedList:
    """
    双向链表，用于记录同一个频率下的所有节点
    """

    def __init__(self):
        self.dummy = Node(None, None)
        self.dummy.next = self.dummy.previous = self.dummy
        self.size = 0

    def __len__(self):
        return self.size
    
    def push(self, node):
        node.next = self.dummy.next
        node.previous = self.dummy
        node.next.previous = node
        self.dummy.next = node
        self.size += 1

    def pop(self, node = None):
        if self.size == 0: return None

        if node is None:
            node = self.dummy.previous

        node.previous.next = node.next
        node.next.previous = node.previous
        self.size -= 1

        return node

class LFUCache:

    def __init__(self, capacity: int):
        self.capacity = capacity
        self.size = self.min_frequency = 0
        self.nodes = {}

        self.linkedLists = defaultdict(LinkedList)

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

        node = self.nodes.get(key)
        self._update(node)

        return node.value

    def put(self, key: int, value: int) -> None:
        if self.capacity == 0: return None
        node = self.nodes.get(key)

        if node is None:
            if self.size == self.capacity:
                node = self.linkedLists[self.min_frequency].pop()
                del self.nodes[node.key]
                self.size -= 1

            node = Node(key, value)
            self.nodes[key] = node
            self.size += 1
            self.min_frequency = 1
            self.linkedLists[1].push(node)
        else:
            self._update(node)
            node.value = value
    
    def _update(self, node):
        linkedList = self.linkedLists[node.frequency]
        linkedList.pop(node)

        if self.min_frequency == node.frequency and not self.linkedLists[self.min_frequency]:
            self.min_frequency += 1
        
        node.frequency += 1
        self.linkedLists[node.frequency].push(node)

Review

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Tip

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Algorithm

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