C# || How To Design A Least Frequently Used (LFU) Cache Using C#

June 01, 2023 admin No comments

The following is a module with functions which demonstrates how to design a least frequently used (LFU) cache using C#.

1. LFU Cache – Problem Statement

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

2. LFU Cache – Solution

The following is a solution which demonstrates how to design a least frequently used (LFU) cache.





		
		
			2. LFU Cache - Solution
			
C#
			
			// ============================================================================
//    Author: Kenneth Perkins
//    Date:   Jun 1, 2022
//    Taken From: http://programmingnotes.org/
//    File:  Solution.cs
//    Description: Demonstrates how to design a least frequently used cache
// ============================================================================
/**
 * Your LFUCache object will be instantiated and called as such:
 * LFUCache obj = new LFUCache(capacity);
 * int param_1 = obj.Get(key);
 * obj.Put(key,value);
 */
public class LFUCache {
    // key: original key, value: frequency and original value.
    private Dictionary<int, Pair<int, Pair<int, int>>> cache;
    // key: frequency, value: All keys that have the same frequency.
    private Dictionary<int, List<Pair<int, int>>> frequencies;
    private int minf;
    private int capacity;

    public LFUCache(int capacity) {
        cache = new Dictionary<int, Pair<int, Pair<int, int>>>();
        frequencies = new Dictionary<int, List<Pair<int, int>>>();
        minf = 0;
        this.capacity = capacity;
    }

    public void Insert(int key, int frequency, int value) {
        if (!frequencies.ContainsKey(frequency)) {
            frequencies[frequency] = new List<Pair<int, int>>();
        }
        frequencies[frequency].Add(new Pair<int, int>(key, value));
        cache[key] = new Pair<int, Pair<int, int>>(frequency, frequencies[frequency].Last());
    }

    public int Get(int key) {
        if (!cache.ContainsKey(key)) {
            return -1;
        }
        var pair = cache[key];
        int f = pair.Key;
        var kv = pair.Value;
        frequencies[f].Remove(kv);
        if (frequencies[f].Count == 0 && minf == f) {
            ++minf;
        }
        Insert(key, f + 1, kv.Value);
        return kv.Value;
    }

    public void Put(int key, int value) {
        if (capacity <= 0) {
            return;
        }
        if (cache.ContainsKey(key)) {
            var it = cache[key];
            it.Value.Value = value;
            Get(key);
            return;
        }
        if (capacity == cache.Count) {
            cache.Remove(frequencies[minf].First().Key);
            frequencies[minf].RemoveAt(0);
        }
        minf = 1;
        Insert(key, 1, value);
    }

    private class Pair<TKey, TValue> {
        public TKey Key;
        public TValue Value;

        public Pair(TKey key, TValue value) {
            Key = key;
            Value = value;
        }
    }
}// http://programmingnotes.org/
			
				
					
				
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						// ============================================================================
//    Author: Kenneth Perkins
//    Date:   Jun 1, 2022
//    Taken From: http://programmingnotes.org/
//    File:  Solution.cs
//    Description: Demonstrates how to design a least frequently used cache
// ============================================================================
/**
 * Your LFUCache object will be instantiated and called as such:
 * LFUCache obj = new LFUCache(capacity);
 * int param_1 = obj.Get(key);
 * obj.Put(key,value);
 */
public class LFUCache {
    // key: original key, value: frequency and original value.
    private Dictionary<int, Pair<int, Pair<int, int>>> cache;
    // key: frequency, value: All keys that have the same frequency.
    private Dictionary<int, List<Pair<int, int>>> frequencies;
    private int minf;
    private int capacity;
 
    public LFUCache(int capacity) {
        cache = new Dictionary<int, Pair<int, Pair<int, int>>>();
        frequencies = new Dictionary<int, List<Pair<int, int>>>();
        minf = 0;
        this.capacity = capacity;
    }
 
    public void Insert(int key, int frequency, int value) {
        if (!frequencies.ContainsKey(frequency)) {
            frequencies[frequency] = new List<Pair<int, int>>();
        }
        frequencies[frequency].Add(new Pair<int, int>(key, value));
        cache[key] = new Pair<int, Pair<int, int>>(frequency, frequencies[frequency].Last());
    }
 
    public int Get(int key) {
        if (!cache.ContainsKey(key)) {
            return -1;
        }
        var pair = cache[key];
        int f = pair.Key;
        var kv = pair.Value;
        frequencies[f].Remove(kv);
        if (frequencies[f].Count == 0 && minf == f) {
            ++minf;
        }
        Insert(key, f + 1, kv.Value);
        return kv.Value;
    }
 
    public void Put(int key, int value) {
        if (capacity <= 0) {
            return;
        }
        if (cache.ContainsKey(key)) {
            var it = cache[key];
            it.Value.Value = value;
            Get(key);
            return;
        }
        if (capacity == cache.Count) {
            cache.Remove(frequencies[minf].First().Key);
            frequencies[minf].RemoveAt(0);
        }
        minf = 1;
        Insert(key, 1, value);
    }
 
    private class Pair<TKey, TValue> {
        public TKey Key;
        public TValue Value;
 
        public Pair(TKey key, TValue value) {
            Key = key;
            Value = value;
        }
    }
}// http://programmingnotes.org/

QUICK NOTES:
The highlighted lines are sections of interest to look out for.

The code is heavily commented, so no further insight is necessary. If you have any questions, feel free to leave a comment below.

Once compiled, you should get this as your output for the example cases:

[null,null,null,1,null,-1,3,null,-1,3,4]

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Posted in: C#, Console Tags: c-sharp, leetcode

C# || How To Design A Least Frequently Used (LFU) Cache Using C#

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