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HackerRank Abstract Classes - Polymorphism solution in c++ programming

In this HackerRank abstract classes polymorphism problem in c++ programming language, Abstract base classes in C++ can only be used as base classes. Thus, they are allowed to have virtual member functions without definitions.

A cache is a component that stores data so future requests for that data can be served faster. The data stored in a cache might be the results of an earlier computation, or the duplicates of data stored elsewhere. A cache hit occurs when the requested data can be found in a cache, while a cache miss occurs when it cannot. Cache hits are served by reading data from the cache which is faster than recomputing a result or reading from a slower data store. Thus, the more requests that can be served from the cache, the faster the system performs.

One of the popular cache replacement policies is: "least recently used" (LRU). It discards the least recently used items first.

For example, if a cache with a capacity to store 5 keys has the following state(arranged from most recently used key to least recently used key) -

5 3 2 1 4

Now, If the next key comes as 1(which is a cache hit), then the cache state in the same order will be -

1 5 3 2 4

Now, If the next key comes as 6(which is a cache miss), then the cache state in the same order will be -

6 1 5 3 2

You can observe that 4 has been discarded because it was the least recently used key and since the capacity of cache is 5, it could not be retained in the cache any longer.

You have to write a class LRUCache which extends the class Cache and uses the member functions and variables to implement an LRU cache.

HackerRank Abstract Classes - Polymorphism solution in c++ programming


HackerRank Abstract classes polymorphism problem solution in c++ programming.

#include <iostream>
#include <vector>
#include <map>
#include <string>
#include <algorithm>
#include <set>
#include <cassert>
using namespace std;

struct Node{
   Node* next;
   Node* prev;
   int value;
   int key;
   Node(Node* p, Node* n, int k, int val):prev(p),next(n),key(k),value(val){};
   Node(int k, int val):prev(NULL),next(NULL),key(k),value(val){};
};

class Cache{
   
   protected: 
   map<int,Node*> mp; //map the key to the node in the linked list
   int cp;  //capacity
   Node* tail; // double linked list tail pointer
   Node* head; // double linked list head pointer
   virtual void set(int, int) = 0; //set function
   virtual int get(int) = 0; //get function

};
#include <iostream>
#include <list>
#define key first
#define val second
class LRUCache {
    int cp;
    map<int, list<pair<int, int> >::iterator> mp;
    list<pair<int, int> > lru;
public:
    LRUCache(int capacity) : cp(capacity){}
    void set(int key, int val) {
        if(mp.find(key) != mp.end()) {
            mp[key]->key = key;
            mp[key]->val = val;
        }
        else {
            lru.push_front({key, val});
            mp[key] = lru.begin();
            if(lru.size() > cp) {
                mp.erase(lru.back().key);
                lru.pop_back();
            }
        }
    }
    int get(int key) {
        if(mp.find(key) != mp.end()) {
            lru.push_front(*mp[key]);
            lru.erase(mp[key]);
            mp[key] = lru.begin();
            return mp[key]->val;
        }
        else
            return -1;
    }
};
int main() {
   int n, capacity,i;
   cin >> n >> capacity;
   LRUCache l(capacity);
   for(i=0;i<n;i++) {
      string command;
      cin >> command;
      if(command == "get") {
         int key;
         cin >> key;
         cout << l.get(key) << endl;
      } 
      else if(command == "set") {
         int key, value;
         cin >> key >> value;
         l.set(key,value);
      }
   }
   return 0;
}


Second solution

class LRUCache : public Cache {
public:
    LRUCache(size_t capacity)
    {
        cp = capacity;
        tail = 0;
        head = 0;
    }
    
    ~LRUCache() {
        while(tail) {
            auto node = tail;
            tail = tail->prev;
            delete node;
        }
    }
    
    // Set/insert the value off the key, if present, otherwise
    // add the key as the most recently used key. If the cache
    // has reached its capacity, it should replace the least
    // recently used key with a new key.
    void set(int key, int value) {
        Node* node;
        auto it = mp.find(key);
        if(mp.end() == it) {
            if(mp.size() < cp) {
                node = new Node(key, value);
                if(tail) {
                    tail->next = node;
                    node->prev = tail;
                } else {
                    tail = node;
                    head = node;
                }
            } else {
                node = tail;
                mp.erase(tail->value);
            }
            mp.insert(make_pair(key, node));
        } else {
            node = it->second;
        }
        node->key = key;
        node->value = value;
        set_mru(node);
    }
    
    // Get the value (will always be positive) of the key
    // if the key exists in the cache and then make that key
    // as the most recently used key; otherwise, return -1.
    int get(int key) {
        auto it = mp.find(key);
        if(mp.end() == it)
            return -1;

        auto node = it->second;
        set_mru(node);
        return node->value;
    }

private:
    void set_mru(Node* node) {
        if(node->next) {
            node->next->prev = node->prev;
        }
        if(node->prev) {
            node->prev->next = node->next;
        }
        if(tail == node)
            tail = node->prev;
        node->prev = 0;
        node->next = head;
        head = node;
    }
};


Post a Comment

2 Comments

  1. It's great buddy that you provide solutions to these Hackerrank problems. It proves to be very helpful. But it'll be more helpful if you provide little explanation to these solutions.

    ReplyDelete
  2. #include

    class LRUCache{

    list lru;
    int cp;
    map mp;
    public:
    LRUCache(int s){
    cp=s;
    }
    int get(int k){
    int v=-1;
    auto f=mp.find(k);
    if(f!=mp.end()){
    v=f->second;
    lru.remove(k);
    lru.push_front(k);
    }
    return v;
    }
    void set(int k,int v){
    auto f=mp.find(k);
    if(f==mp.end()){
    mp.insert({k,v});
    lru.push_front(k);
    if(lru.size()>cp){
    mp.erase(lru.back());
    lru.pop_back();
    }
    }else{
    lru.remove(k);
    lru.push_front(k);
    mp[k]=v;
    }
    }

    };

    ReplyDelete