Added Shop In Candy Store

problems/shop-in-candy-store/Readme.md

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+# Shop in Candy Store
+
+In a candy store there are N different types of candies available  and the prices of all the N different types of candies are provided to you.
+You are now provided with an attractive offer.
+You can buy a single candy from the store and get atmost K other candies ( all are different types ) for free.
+Now you have to answer two questions. Firstly, you have to tell what is the minimum amount of money you have to spend to buy all the N different candies. Secondly, you have to tell what is the maximum amount of money you have to spend to buy all the N different candies.
+In both the cases you must utilize the offer i.e. you buy one candy and get K other candies for free.
+
+
+# Input  
+The first line of the input contains T the number of test cases. Each test case consists of two lines. The first line of each test case contains the values of N and K as described above.  Then in the next line N integers follow denoting the price of each of the N different candies.
+
+
+# Output
+For each test case output a single line containing 2 space separated integers , the first denoting the minimum amount of money required to be spent and the second denoting the maximum amount of money to be spent.
+Remember to output the answer of each test case in a new line.
+
+# Constraints      
+1 <= T <= 50
+1 <= N <= 1000
+ 0 <= K <= N-1
+1 <= Ai <= 100
+
+# Example:
+Input    
+ 1
+ 4  2
+ 3 2 1 4
+
+Output
+ 3 7
+
+# Explanation
+As according to the offer if you but one candy you can take atmost two more for free.
+So in the first case you buy the candy which costs 1 and take candies worth 3 and 4 for free, also you buy candy worth 2 as well.
+So min cost = 1+2 =3.
+In the second case I buy the candy which costs 4 and take candies worth 1 and 2 for free, also I  buy candy worth 3 as well.
+So max cost = 3+4 =7.
+
+# Topics : 
+Greedy || Sorting

solutions/cpp/ShopInCandyStore.hpp

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+#include <bits/stdc++.h>
+using namespace std;
+int main() 
+{
+	int t; cin>>t;
+	while(t--)
+	{
+		int n,k; cin>>n>>k; long int a[n],min=0,max=0,c=0;
+		for(int i=0;i<n;i++) cin>>a[i];
+			sort(a,a+n);
+		int x=(n/(k+1)); if((n%(k+1))>0) {x++;}
+
+		for(int i=0;i<x;i++) {min+=a[i];}
+			for(int i=n-1;c<x;i--) {max+=a[i];c++;}
+				cout<<min<<" "<<max<<endl;
+		}
+		return 0;
+	}