Scale Engineer

LFU Cache

Hard
#0447Time: `get`: O(1) `put`: O(C), where C is the cache capacity. The need to scan the entire cache for eviction makes this approach inefficient for large capacities.Space: O(C), where C is the cache capacity, to store the key-value pairs and their metadata in the hash map.13 companies
Patterns
Design
Algorithms
LRU Cache
Data structures
Hash TableLinked ListDoubly-Linked List
Companies
DoorDashKLAPayPalServiceNow

Prompt

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.

 

 

Approaches

3 approaches with complexity analysis and trade-offs.

This approach uses a single hash map for O(1) lookups. Each entry in the map stores not only the value but also metadata: its usage frequency and a timestamp for its last access. When an eviction is necessary, the entire cache is scanned to find the item with the lowest frequency. Any ties are broken by choosing the item that was least recently used (i.e., has the oldest timestamp).

Algorithm

  • Initialize a HashMap<Integer, CacheEntry> to store cache data, where CacheEntry holds the value, frequency, and a lastAccessTime.
  • Maintain a global timer variable, incremented on each operation to provide a sequential timestamp.
  • For get(key):
    1. Retrieve CacheEntry from the map. If it doesn't exist, return -1.
    2. Increment the entry's frequency.
    3. Update its lastAccessTime with the current timer value and increment the timer.
    4. Return the value.
  • For put(key, value):
    1. If the key exists, update its value, increment its frequency, and update its lastAccessTime.
    2. If the key does not exist:
      • Check if the cache is full (map.size() == capacity).
      • If full, iterate through all entries in the map to find the one with the minimum frequency. If there's a tie, choose the one with the minimum lastAccessTime.
      • Remove this LFU/LRU entry from the map.
      • Create a new CacheEntry with frequency 1 and the current timestamp, and insert it into the map.

Walkthrough

The implementation revolves around a HashMap<Integer, CacheEntry> and a global timer. The CacheEntry class is a simple container for the value, frequency, and lastAccessTime.

get(key) Operation: This is straightforward. We look up the key in the map. If it exists, we update its frequency and timestamp, then return the value. This is an O(1) operation.

put(key, value) Operation: If the key already exists, we update its value and metadata, similar to a get operation. If the key is new, we first check if the cache is full. If it is, the eviction process begins. We must perform a linear scan through all the entries in the hash map's values. During this scan, we keep track of the entry with the lowest frequency found so far, using the last access time as a tie-breaker. Once the victim entry is identified, it's removed. Finally, the new entry is inserted with a frequency of 1. This eviction scan makes the put operation O(C), where C is the capacity.

class LFUCache {    class CacheEntry {        int value;        int frequency;        int accessTime;         CacheEntry(int value, int frequency, int accessTime) {            this.value = value;            this.frequency = frequency;            this.accessTime = accessTime;        }    }     private Map<Integer, CacheEntry> cache;    private int capacity;    private int timer;     public LFUCache(int capacity) {        this.cache = new HashMap<>();        this.capacity = capacity;        this.timer = 0;    }     public int get(int key) {        if (!cache.containsKey(key)) {            return -1;        }        CacheEntry entry = cache.get(key);        entry.frequency++;        entry.accessTime = timer++;        return entry.value;    }     public void put(int key, int value) {        if (capacity == 0) {            return;        }         if (cache.containsKey(key)) {            CacheEntry entry = cache.get(key);            entry.value = value;            entry.frequency++;            entry.accessTime = timer++;        } else {            if (cache.size() >= capacity) {                int keyToEvict = -1;                int minFreq = Integer.MAX_VALUE;                int minTime = Integer.MAX_VALUE;                 for (Map.Entry<Integer, CacheEntry> mapEntry : cache.entrySet()) {                    CacheEntry current = mapEntry.getValue();                    if (current.frequency < minFreq) {                        minFreq = current.frequency;                        minTime = current.accessTime;                        keyToEvict = mapEntry.getKey();                    } else if (current.frequency == minFreq && current.accessTime < minTime) {                        minTime = current.accessTime;                        keyToEvict = mapEntry.getKey();                    }                }                cache.remove(keyToEvict);            }            cache.put(key, new CacheEntry(value, 1, timer++));        }    }}

Complexity

Time

`get`: O(1) `put`: O(C), where C is the cache capacity. The need to scan the entire cache for eviction makes this approach inefficient for large capacities.

Space

O(C), where C is the cache capacity, to store the key-value pairs and their metadata in the hash map.

Trade-offs

Pros

  • Relatively simple to conceptualize and implement.

  • The get operation is efficient with O(1) time complexity.

Cons

  • The put operation has a time complexity of O(C) where C is the capacity, which fails to meet the problem's O(1) requirement.

  • Performance degrades significantly as the cache capacity increases.

Solutions

public class LFU_Cache { public class LFUCache { // Save the key, value HashMap < Integer , Integer > vals ; // Save the key to the value of the number of visits HashMap < Integer , Integer > counts ; // 频率和一个里面所有key都是当前频率的list之间的映射 HashMap < Integer , LinkedHashSet < Integer >> lists ; int capacity ; // Initialize the frequency of data occurrences int min = - 1 ; public LFUCache ( int cap ) { capacity = cap ; vals = new HashMap <>(); counts = new HashMap <>(); lists = new HashMap <>(); } public int get ( int key ) { if (! vals . containsKey ( key )) return - 1 ; int count = counts . get ( key ); counts . put ( key , count + 1 ); lists . get ( count ). remove ( key ); // Determine whether min should add 1 if ( count == min && lists . get ( count ). size () == 0 ) { min ++; } if (! lists . containsKey ( count + 1 )) { lists . put ( count + 1 , new LinkedHashSet <>()); } lists . get ( count + 1 ). add ( key ); return vals . get ( key ); } public void put ( int key , int value ) { if ( capacity <= 0 ) return ; if ( vals . containsKey ( key )) { vals . put ( key , value ); get ( key ); return ; } if ( vals . size () >= capacity ) { int minFreKey = lists . get ( min ). iterator (). next (); lists . get ( min ). remove ( minFreKey ); vals . remove ( minFreKey ); counts . remove ( minFreKey ); } vals . put ( key , value ); counts . put ( key , 1 ); min = 1 ; if (! lists . containsKey ( 1 )) { lists . put ( 1 , new LinkedHashSet <>()); } lists . get ( 1 ). add ( key ); } } } ############ class LFUCache { private final Map < Integer , Node > map ; private final Map < Integer , DoublyLinkedList > freqMap ; private final int capacity ; private int minFreq ; public LFUCache ( int capacity ) { this . capacity = capacity ; map = new HashMap <>( capacity , 1 ); freqMap = new HashMap <>(); } public int get ( int key ) { if ( capacity == 0 ) { return - 1 ; } if (! map . containsKey ( key )) { return - 1 ; } Node node = map . get ( key ); incrFreq ( node ); return node . value ; } public void put ( int key , int value ) { if ( capacity == 0 ) { return ; } if ( map . containsKey ( key )) { Node node = map . get ( key ); node . value = value ; incrFreq ( node ); return ; } if ( map . size () == capacity ) { DoublyLinkedList list = freqMap . get ( minFreq ); map . remove ( list . removeLast (). key ); } Node node = new Node ( key , value ); addNode ( node ); map . put ( key , node ); minFreq = 1 ; } private void incrFreq ( Node node ) { int freq = node . freq ; DoublyLinkedList list = freqMap . get ( freq ); list . remove ( node ); if ( list . isEmpty ()) { freqMap . remove ( freq ); if ( freq == minFreq ) { minFreq ++; } } node . freq ++; addNode ( node ); } private void addNode ( Node node ) { int freq = node . freq ; DoublyLinkedList list = freqMap . getOrDefault ( freq , new DoublyLinkedList ()); list . addFirst ( node ); freqMap . put ( freq , list ); } private static class Node { int key ; int value ; int freq ; Node prev ; Node next ; Node ( int key , int value ) { this . key = key ; this . value = value ; this . freq = 1 ; } } private static class DoublyLinkedList { private final Node head ; private final Node tail ; public DoublyLinkedList () { head = new Node (- 1 , - 1 ); tail = new Node (- 1 , - 1 ); head . next = tail ; tail . prev = head ; } public void addFirst ( Node node ) { node . prev = head ; node . next = head . next ; head . next . prev = node ; head . next = node ; } public Node remove ( Node node ) { node . next . prev = node . prev ; node . prev . next = node . next ; node . next = null ; node . prev = null ; return node ; } public Node removeLast () { return remove ( tail . prev ); } public boolean isEmpty () { return head . next == tail ; } } }

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