460. LFU Cache
Problem Description
In this problem, we need to design a Least Frequently Used (LFU) cache. An LFU cache is a type of cache that removes the least frequently accessed item when it needs space. If there are multiple items with the same lowest frequency, it removes the least recently used among those.
The LFUCache
class requires the following operations:
LFUCache(int capacity)
: Constructor that initializes the cache with a given capacity.int get(int key)
: Retrieves the value linked to thekey
from the cache. If thekey
is not present, it should return-1
.void put(int key, int value)
: Inserts or updates the value of thekey
. If inserting a new item causes the cache to exceed itscapacity
, the least frequently used key should be removed. In case of a tie (when multiple keys have the same frequency), the least recently used key should be removed.
Additionally, each key
in the cache will have a use counter
which increments every time the key
is accessed through get
or put
operations. The goal is to implement get
and put
methods that each run with an average time complexity of O(1)
.
Intuition
The solution to this problem requires us to perform insertion, access, and removal operations with O(1)
complexity. We can satisfy these requirements with a combination of data structures: hashtables and doubly linked lists.
- Hashtable: Allows
O(1)
access to nodes by their keys. - Doubly Linked Lists: Allow
O(1)
insertion and removal of nodes. We use one doubly linked list per frequency count, so each list contains all nodes that have been accessed a certain number of times.
For the LFUCache
, we maintain the following:
- A
map
that maps keys to their corresponding node which contains the value and frequency of access. - A
freq_map
that maps frequencies to a doubly linked list containing all nodes with that frequency. - A variable
min_freq
to keep track of the minimum frequency of all keys in the cache for easy access to the least frequently used item.
When a get
or put
operation is called, we update the corresponding node's frequency, move it to the respective frequency list in the freq_map
, and update min_freq
if necessary. If a put
operation is performed on a full cache, we remove the least frequently used item. When there's a tie, we remove the least recently used item from the list corresponding to the lowest frequency.
By doing this, we ensure constant-time O(1)
complexity for get
and put
operations while maintaining the structure of the LFU cache.
Learn more about Linked List patterns.
Solution Approach
The implementation for the LFU Cache problem involves designing a set of classes to efficiently manage our cache operations. Here we will discuss the algorithms, patterns, and data structures utilized in the solution code provided:
Node Class
A Node
represents an item in the cache, with properties for its key
, value
, frequency (freq)
, prev
(previous node), and next
(next node). This will be used to store each entry and will be inserted into a doubly linked list.
DoublyLinkedList Class
A DoublyLinkedList
is used to keep track of all items with the same frequency. It supports adding a node to the head (add_first
), removing a specific node (remove
), and removing the last node (remove_last
). An empty list can be checked with is_empty
.
LFUCache Class
Data Structures:
self.map
: A dictionary (hashtable) to map keys to their corresponding node (Node
objects).self.freq_map
: A dictionary to map frequencies to their corresponding doubly linked list, where each list contains all nodes with that frequency.self.min_freq
: An integer tracking the current minimum frequency amongst all keys in the cache.
Constructor:
- Initializes the data structures and sets the capacity of the cache.
get Method:
- Retrieves the node from the map using the given
key
. - If the
key
is missing or the capacity is 0, returns-1
. - If the node is found, it calls the
incr_freq
method to update the frequency of the node and returns itsvalue
.
put Method:
- If the capacity is 0, the function ends early as no items can be put into the cache.
- If the
key
is present in the map, it updates thevalue
in the corresponding node and increases the frequency by callingincr_freq
. - If the
key
is not present and the cache is at fullcapacity
, it removes the least frequently used item by (1) accessing the linked list for the current minimum frequency, (2) removing the last node from that list, and (3) removing the node's key from themap
. - A new node is created and added to the
map
and to the frequency map under frequency1
. Themin_freq
is set to1
.
incr_freq Method:
- Removes the node from its current frequency list in
freq_map
. - If the removal of the node results in an empty list and that frequency was the
min_freq
, incrementsmin_freq
. - Increments the node's frequency, inserts it into the linked list in the
freq_map
that corresponds to the new frequency.
add_node Method:
- A helper used to add a node to the linked list in the
freq_map
corresponding to the node’s frequency.
With these components combined, we ensure each operation on the cache is O(1)
on average by efficiently tracking both the least frequently and least recently used items.
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Start EvaluatorExample Walkthrough
Let's say we have an LFU Cache with a capacity of 2, and we perform the following operations:
-
put(1, 10)
: The cache is empty, so this will create a new node with key1
and value10
. The frequency is set to1
. Our frequency map has1 -> Node(1)
andmin_freq
is now1
. -
put(2, 20)
: A new node with key2
and value20
is created. Similar to the first step, this node's frequency is also1
and is added to the same frequency list. The frequency map now is1 -> Node(1) <-> Node(2)
. -
get(1)
: This operation will get the node with key1
, which has a value of10
. The frequency of this node will be incremented to2
. Since it's the only node with this frequency, it will be added to a new frequency list. The frequency map updates to1 -> Node(2)
and2 -> Node(1)
.min_freq
remains1
. -
put(3, 30)
: We need to add a new node with key3
and value30
. Since the cache has reached its capacity, we must remove an item. We look atmin_freq
which is1
and remove the least recently used item from that frequency's list, which isNode(2)
. We then addNode(3)
with a frequency of1
to the list. The frequency map is now1 -> Node(3)
and2 -> Node(1)
. -
get(3)
: The node with key3
is accessed, increasing its frequency to2
. Now both nodes with keys1
and3
have the same frequency. The lists are updated:1 ->
becomes empty and2 -> Node(1) <-> Node(3)
.min_freq
is updated to2
because there are no nodes with a frequency of1
after this operation. -
get(1)
: The node with key1
is accessed again, so its frequency increases to3
. We update the lists such that2 -> Node(3)
and a new list3 -> Node(1)
is created.min_freq
remains2
.
The sequence of operations above demonstrates how the LFU Cache responds to different commands. The cache maintains the frequency of each node with the help of the freq_map
, and updates the min_freq
accordingly to ensure that the least frequently and least recently used node is always easily identifiable and removable in constant time.
Solution Implementation
1from collections import defaultdict
2
3class Node:
4 def __init__(self, key: int, value: int) -> None:
5 self.key = key
6 self.value = value
7 self.freq = 1
8 self.prev = None
9 self.next = None
10
11
12class DoublyLinkedList:
13 def __init__(self) -> None:
14 self.head = Node(-1, -1) # Sentinel head node
15 self.tail = Node(-1, -1) # Sentinel tail node
16 self.head.next = self.tail
17 self.tail.prev = self.head
18
19 def add_first(self, node: Node) -> None:
20 """Add a node right after the head."""
21 node.prev = self.head
22 node.next = self.head.next
23 self.head.next.prev = node
24 self.head.next = node
25
26 def remove(self, node: Node) -> Node:
27 """Remove an existing node from the list."""
28 node.prev.next = node.next
29 node.next.prev = node.prev
30 # Clear the removed node's pointers
31 node.next, node.prev = None, None
32 return node
33
34 def remove_last(self) -> Node:
35 """Remove the last node and return it."""
36 return self.remove(self.tail.prev)
37
38 def is_empty(self) -> bool:
39 """Check if the list is empty."""
40 return self.head.next == self.tail
41
42
43class LFUCache:
44 def __init__(self, capacity: int):
45 self.capacity = capacity
46 self.min_freq = 0
47 self.key_node_map = defaultdict(Node)
48 self.freq_node_list_map = defaultdict(DoublyLinkedList)
49
50 def get(self, key: int) -> int:
51 """Get the value of the key if the key exists in the cache."""
52 if self.capacity == 0 or key not in self.key_node_map:
53 return -1
54
55 node = self.key_node_map[key]
56 self._increase_freq(node)
57 return node.value
58
59 def put(self, key: int, value: int) -> None:
60 """Set or insert the value if the key is not already present."""
61 if self.capacity == 0:
62 return
63
64 if key in self.key_node_map:
65 node = self.key_node_map[key]
66 node.value = value
67 self._increase_freq(node)
68 return
69
70 if len(self.key_node_map) == self.capacity:
71 least_freq_list = self.freq_node_list_map[self.min_freq]
72 node_to_remove = least_freq_list.remove_last()
73 del self.key_node_map[node_to_remove.key]
74
75 new_node = Node(key, value)
76 self._add_node(new_node)
77 self.key_node_map[key] = new_node
78 self.min_freq = 1 # Reset min_freq to 1 as a new node is added
79
80 def _increase_freq(self, node: Node) -> None:
81 """Increase the frequency count of a node."""
82 freq = node.freq
83 node_list = self.freq_node_list_map[freq]
84 node_list.remove(node)
85 if node_list.is_empty():
86 del self.freq_node_list_map[freq]
87 if freq == self.min_freq:
88 self.min_freq += 1
89
90 node.freq += 1
91 self._add_node(node)
92
93 def _add_node(self, node: Node) -> None:
94 """Add a node to the list that corresponds to its frequency count."""
95 freq = node.freq
96 node_list = self.freq_node_list_map[freq]
97 node_list.add_first(node)
98
99# Example of how to use LFUCache:
100# cache = LFUCache(capacity)
101# value = cache.get(key)
102# cache.put(key, value)
103
1import java.util.HashMap;
2import java.util.Map;
3
4class LFUCache {
5
6 private final Map<Integer, Node> cache; // The map to store the key-node pairs
7 private final Map<Integer, DoublyLinkedList> frequencyMap; // The map to store the frequency-node list pairs
8 private final int capacity; // The capacity of the LFU cache
9 private int minimumFrequency; // The current minimum frequency of the nodes in the cache
10
11 public LFUCache(int capacity) {
12 this.capacity = capacity;
13 cache = new HashMap<>();
14 frequencyMap = new HashMap<>();
15 }
16
17 public int get(int key) {
18 if (capacity == 0) {
19 return -1;
20 }
21 if (!cache.containsKey(key)) {
22 return -1;
23 }
24
25 // get the node and increment its frequency
26 Node node = cache.get(key);
27 increaseFrequency(node);
28 return node.value;
29 }
30
31 public void put(int key, int value) {
32 if (capacity == 0) {
33 return;
34 }
35 if (cache.containsKey(key)) {
36 // update the value and frequency of the existing node
37 Node node = cache.get(key);
38 node.value = value;
39 increaseFrequency(node);
40 return;
41 }
42 if (cache.size() == capacity) {
43 // evict the least frequently used node, choosing the least recently used one if there's a tie
44 DoublyLinkedList leastFreqList = frequencyMap.get(minimumFrequency);
45 Node toRemove = leastFreqList.removeLast();
46 cache.remove(toRemove.key);
47 }
48 // create a new node and add it to the cache and frequency map
49 Node newNode = new Node(key, value);
50 addNodeToFrequencyList(newNode);
51 cache.put(key, newNode);
52 minimumFrequency = 1; // reset the minimum frequency
53 }
54
55 private void increaseFrequency(Node node) {
56 int currentFrequency = node.frequency;
57 DoublyLinkedList currentList = frequencyMap.get(currentFrequency);
58 currentList.remove(node);
59 if (currentList.isEmpty()) {
60 frequencyMap.remove(currentFrequency);
61 if (currentFrequency == minimumFrequency) {
62 minimumFrequency++; // increment the minimum frequency if necessary
63 }
64 }
65 node.frequency++;
66 addNodeToFrequencyList(node);
67 }
68
69 private void addNodeToFrequencyList(Node node) {
70 // add the node to the new frequency's list or create a new list if it does not exist
71 int newFrequency = node.frequency;
72 DoublyLinkedList newList = frequencyMap.getOrDefault(newFrequency, new DoublyLinkedList());
73 newList.addFirst(node);
74 frequencyMap.put(newFrequency, newList);
75 }
76
77 private static class Node {
78 int key;
79 int value;
80 int frequency; // the frequency of accesses to this node
81 Node prev;
82 Node next;
83
84 Node(int key, int value) {
85 this.key = key;
86 this.value = value;
87 frequency = 1; // initially, all nodes have frequency 1
88 }
89 }
90
91 private static class DoublyLinkedList {
92 // dummy head and tail make it easier to add/remove nodes
93 private final Node head;
94 private final Node tail;
95
96 public DoublyLinkedList() {
97 head = new Node(-1, -1);
98 tail = new Node(-1, -1);
99 head.next = tail;
100 tail.prev = head;
101 }
102
103 public void addFirst(Node node) {
104 // add a new node right after the dummy head
105 node.prev = head;
106 node.next = head.next;
107 head.next.prev = node;
108 head.next = node;
109 }
110
111 public Node remove(Node node) {
112 // remove a node from its current position
113 node.prev.next = node.next;
114 node.next.prev = node.prev;
115 node.next = null;
116 node.prev = null;
117 return node;
118 }
119
120 public Node removeLast() {
121 // remove the last node before the dummy tail (the least recently used one)
122 if (isEmpty()) {
123 return null; // list is empty, nothing to remove
124 }
125 return remove(tail.prev);
126 }
127
128 public boolean isEmpty() {
129 // check if the list is empty by seeing if the dummy head is directly connected to the dummy tail
130 return head.next == tail;
131 }
132 }
133}
134
1#include <unordered_map>
2using namespace std;
3
4// Node class representing a node in the doubly linked list
5class Node {
6public:
7 int key; // Key of the node
8 int value; // Value of the node
9 int freq; // Frequency of accesses
10 Node* prev; // Pointer to the previous node in the list
11 Node* next; // Pointer to the next node in the list
12
13 // Constructor initializes a node with a key and a value
14 Node(int key, int value) : key(key), value(value), freq(1), prev(nullptr), next(nullptr) {}
15};
16
17// DoublyLinkedList class representing a doubly linked list
18class DoublyLinkedList {
19public:
20 Node* head; // Pointer to the dummy head node
21 Node* tail; // Pointer to the dummy tail node
22
23 // Constructor initializes the dummy head and tail and links them
24 DoublyLinkedList() {
25 head = new Node(-1, -1);
26 tail = new Node(-1, -1);
27 head->next = tail;
28 tail->prev = head;
29 }
30
31 // Adds a node right after the head of the list
32 void add_first(Node* node) {
33 node->prev = head;
34 node->next = head->next;
35 head->next->prev = node;
36 head->next = node;
37 }
38
39 // Removes the given node from the list and returns it
40 Node* remove(Node* node) {
41 node->next->prev = node->prev;
42 node->prev->next = node->next;
43 node->next = nullptr;
44 node->prev = nullptr;
45 return node;
46 }
47
48 // Removes the last node before the tail and returns it
49 Node* remove_last() {
50 return remove(tail->prev);
51 }
52
53 // Checks if the list is empty, i.e., only contains dummy head and tail
54 bool is_empty() {
55 return head->next == tail;
56 }
57};
58
59// LFUCache class representing the Least Frequently Used (LFU) Cache
60class LFUCache {
61public:
62 LFUCache(int capacity) : capacity(capacity), min_freq(0) {}
63
64 int get(int key) {
65 if (capacity == 0 || key_to_node_map.find(key) == key_to_node_map.end()) {
66 return -1;
67 }
68 Node* node = key_to_node_map[key];
69 increase_frequency(node);
70 return node->value;
71 }
72
73 void put(int key, int value) {
74 if (capacity == 0) return;
75
76 if (key_to_node_map.find(key) != key_to_node_map.end()) {
77 Node* node = key_to_node_map[key];
78 node->value = value;
79 increase_frequency(node);
80 } else {
81 if (key_to_node_map.size() == capacity) {
82 DoublyLinkedList* list = freq_to_list_map[min_freq];
83 Node* removed_node = list->remove_last();
84 key_to_node_map.erase(removed_node->key);
85 }
86 Node* new_node = new Node(key, value);
87 add_node(new_node);
88 key_to_node_map[key] = new_node;
89 min_freq = 1;
90 }
91 }
92
93private:
94 int capacity; // Capacity of the cache
95 int min_freq; // Minimum frequency of nodes in the cache
96 unordered_map<int, Node*> key_to_node_map; // Maps keys to their corresponding nodes
97 unordered_map<int, DoublyLinkedList*> freq_to_list_map; // Maps frequencies to lists of nodes with that frequency
98
99 // Increases the frequency of a node
100 void increase_frequency(Node* node) {
101 int freq = node->freq;
102 DoublyLinkedList* list = freq_to_list_map[freq];
103 list->remove(node);
104 if (list->is_empty()) {
105 freq_to_list_map.erase(freq);
106 if (freq == min_freq) min_freq++;
107 }
108 node->freq++;
109 add_node(node);
110 }
111
112 // Adds a node to the list corresponding to its frequency
113 void add_node(Node* node) {
114 int freq = node->freq;
115 if (freq_to_list_map.find(freq) == freq_to_list_map.end()) {
116 freq_to_list_map[freq] = new DoublyLinkedList();
117 }
118 DoublyLinkedList* list = freq_to_list_map[freq];
119 list->add_first(node);
120 }
121};
122
123// Usage:
124// LFUCache* cache = new LFUCache(capacity);
125// int value = cache->get(key);
126// cache->put(key, value);
127
1// Representation of a node in the doubly linked list
2class Node {
3 public key: number;
4 public value: number;
5 public freq: number;
6 public prev: Node | null;
7 public next: Node | null;
8
9 constructor(key: number, value: number) {
10 this.key = key;
11 this.value = value;
12 this.freq = 1; // Initial frequency is set to 1
13 this.prev = null;
14 this.next = null;
15 }
16}
17
18// Representation of a doubly linked list
19class DoublyLinkedList {
20 public head: Node;
21 public tail: Node;
22
23 constructor() {
24 this.head = new Node(-1, -1); // Dummy head
25 this.tail = new Node(-1, -1); // Dummy tail
26 this.head.next = this.tail;
27 this.tail.prev = this.head;
28 }
29
30 // Adds a node right after the list's head
31 addFirst(node: Node): void {
32 node.prev = this.head;
33 node.next = this.head.next;
34 this.head.next!.prev = node;
35 this.head.next = node;
36 }
37
38 // Removes a node from the list
39 remove(node: Node): Node {
40 node.prev!.next = node.next;
41 node.next!.prev = node.prev;
42 return node;
43 }
44
45 // Removes and returns the last node before the list's tail
46 removeLast(): Node {
47 const nodeToRemove = this.tail.prev!;
48 return this.remove(nodeToRemove);
49 }
50
51 // Checks if the list is empty (only contains dummy head and tail)
52 isEmpty(): boolean {
53 return this.head.next === this.tail;
54 }
55}
56
57// Representation of the Least Frequently Used (LFU) Cache
58let capacity: number; // Capacity of the cache
59let minFreq: number; // Minimum frequency of nodes in the cache
60let keyToNodeMap = new Map<number, Node>(); // Maps keys to their corresponding nodes
61let freqToListMap = new Map<number, DoublyLinkedList>(); // Maps frequencies to lists of nodes with that frequency
62
63// Increases the frequency of a node, moving it to the appropriate frequency list
64function increaseFrequency(node: Node): void {
65 let freq = node.freq;
66 let list = freqToListMap.get(freq)!;
67 list.remove(node);
68 if (list.isEmpty()) {
69 freqToListMap.delete(freq);
70 if (freq === minFreq) {
71 minFreq++;
72 }
73 }
74 node.freq++;
75 addNode(node);
76}
77
78// Adds a node to the list corresponding to its frequency
79function addNode(node: Node): void {
80 let freq = node.freq;
81 if (!freqToListMap.has(freq)) {
82 freqToListMap.set(freq, new DoublyLinkedList());
83 }
84 let list = freqToListMap.get(freq)!;
85 list.addFirst(node);
86}
87
88// Creates a new LFU Cache with a given capacity
89function createCache(newCapacity: number): void {
90 capacity = newCapacity;
91 minFreq = 0;
92 keyToNodeMap.clear();
93 freqToListMap.clear();
94}
95
96// Retrieves the value for a key from the cache, or -1 if key is not present
97function get(key: number): number {
98 if (capacity === 0 || !keyToNodeMap.has(key)) {
99 return -1;
100 }
101
102 let node = keyToNodeMap.get(key)!;
103 increaseFrequency(node);
104 return node.value;
105}
106
107// Inserts a key-value pair into the cache, or updates the value if the key already exists
108function put(key: number, value: number): void {
109 if (capacity === 0) return;
110
111 if (keyToNodeMap.has(key)) {
112 let node = keyToNodeMap.get(key)!;
113 node.value = value;
114 increaseFrequency(node);
115 } else {
116 if (keyToNodeMap.size >= capacity) {
117 let list = freqToListMap.get(minFreq)!;
118 let removedNode = list.removeLast();
119 keyToNodeMap.delete(removedNode.key);
120 }
121 let newNode = new Node(key, value);
122 addNode(newNode);
123 keyToNodeMap.set(key, newNode);
124 minFreq = 1;
125 }
126}
127
Time and Space Complexity
Time Complexity
get() method:
- The time complexity of the
get
method isO(1)
. - Accessing the node from the hashmap takes constant time, and the
incr_freq
method is designed to execute in constant time by performing a fixed number of pointer reference changes.
put() method:
- The time complexity of the
put
method isO(1)
for most of the operations which include checking the length of the hashmap, accessing the node from the hashmap, updating the value, and incrementing the frequency throughincr_freq
. - However, when the cache is full, we need to remove the least frequently used item, which again is done in constant time because the DoublyLinkedList allows us to remove the last element efficiently and the hashmap provides instant access to any node.
incr_freq() method:
- The
incr_freq
method has a time complexity ofO(1)
because it involves removing a node from one frequency list and adding it to another, which entails only changing a few pointers around. - If the removed frequency list is empty, it pops the list from the frequency map in constant time and potentially increases the minimum frequency.
add_node() method:
- The
add_node
method's time complexity is alsoO(1)
as it simply adds the node to the front of the frequency list, which again involves a few pointer changes.
Space Complexity
LFUCache class:
- The space complexity of the LFUCache class is
O(capacity)
, wherecapacity
is the maximum number of elements that the cache can hold. - Since each key and each frequency list takes up space, and the list grows as more unique items with distinct frequencies are added, the space complexity scales with the number of items, which is limited by capacity.
- The use of two hashmaps (
map
andfreq_map
) ensures that the space used stays within a limit that is proportional tocapacity
.
Learn more about how to find time and space complexity quickly using problem constraints.
Which of the following is a good use case for backtracking?
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