Question

C++ program c that reads a list of costs for a few items. Program should calculate the best price for the “buy-two-get-one-free” deal.

The first line (= 5 in the example) indicates the number of items for you to buy. The second line presents the costs of five items. Note that if you buy two items with your money, you can get one item free. For example, let’s say that you buy the first two items (= 2 and 7) with your money. Then, you can choose one item for free. Let’s say that you select the item with the cost 6 for free. After that, there are two items left (= 5 and 5). Since you have to buy the two items anyway, you have to buy them with your money. In this case, you have to pay total 19 (= 2, 7, 5, and 5).

Sample Input 0

5
2 7 5 6 5

Sample Output 0

18

Sample Input 1

3
10 30 20

Sample Output 1

30

Answer 1

IDE Used: DEV C++

Explanation of algorithm used: First, calculate the number of sets including 3 items. For this, we divide the total items by 3. The dividend in integer value will be the total number of free items. Free items will be those items with the highest costs. For example, let the total number of items be 8 then 8/3 will be 2 so 2 free items. Then arrange the items in ascending order of their cost. Calculate the sum of cost without the two maximum costs i.e the cost of 2 free items. This sum of the cost will be the best cost of buy2 get 1 free offer.

Code

#include <iostream>
using namespace std;

int main() {
  
   //number of items to buys
   int nof_items;
  
   //number of free items
   int free_items;
  
//to store the final best cost
   int best_cost;
  
   //counter variables i and j can also be used
   int ct1, ct2;
  
   //to store temporary cost while swapping
   int temp_cost;
  
   //array to store the cost of all items
   int price_array[999];
  
   //getting the number of items from the user
   cout<<"Enter the number of items you want to buy: ";
   cin>>nof_items;
  
   cout<<"\nEnter the cost of each item\n";
   for(ct1=0;ct1<nof_items;ct1++ )
   {
       cin>>price_array[ct1];
   }
  
   //calculate the numeber of free items
   //which is equal number of sets of 3 items
   free_items = nof_items/3;
  
   //find the best cost for buy 2 get 1 offer
   //for this we will arrange the cost and array in ascending order
   for(ct1=0;ct1<nof_items;ct1++)
   {
       for(ct2=0;ct2<nof_items-ct1-1;ct2++)
       {
           if(price_array[ct2]>price_array[ct2+1])
           {
               //swapping the values
               temp_cost = price_array[ct2];
               price_array[ct2] = price_array[ct2+1];
               price_array[ct2+1] = temp_cost;
           }
       }
   }
  
   //then we will find the sum of array excluding the free items
   //that is removing the highest cost items
   for(ct1=0;ct1<nof_items-free_items; ct1++)
   {
       best_cost = best_cost+price_array[ct1];
   }
  
   //printing the sum of best price
   printf("Best price of buy2 get 1 deal is %d", best_cost);
   return 0;
}

Code Snippets

Sample Outputs