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. There is also an attractive offer by candy store. We can buy a single candy from the store and get at-most K other candies (all are different types) for free.

- Find minimum amount of money we have to spend to buy all the N different candies.
- Find maximum amount of money we have to spend to buy all the N different candies.

In both the cases we must utilize the offer and get maximum possible candies back. If k or more candies are available, we must take k candies for every candy purchase. If less than k candies are available, we must take all candies for a candy purchase.

Examples:

Input : price[] = {3, 2, 1, 4} k = 2 Output : Min = 3, Max = 7 Since k is 2, if we buy one candy we can take atmost two more for free. So in the first case we 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 we buy the candy which costs 4 and take candies worth 1 and 2 for free, also We buy candy worth 3 as well. So max cost = 3 + 4 = 7.

One important thing to note is, we must use the offer and get maximum candies back for every candy purchase. So if we want to minimize the money, we must buy candies of minimum cost and get candies of maximum costs for free. To maximize the money, we must do reverse. Below is algorithm based on this.

First Sort the price array.For finding minimum amount :Start purchasing candies from starting and reduce k free candies from last with every single purchase.For finding maximum amount :Start purchasing candies from the end and reduce k free candies from starting in every single purchase.

## C++

// C++ implementation to find the minimum // and maximum amount #include<bits/stdc++.h> using namespace std; // Function to find the minimum amount // to buy all candies int findMinimum(int arr[], int n, int k) { int res = 0; for (int i=0; i<n ; i++) { // Buy current candy res += arr[i]; // And take k candies for free // from the last n = n-k; } return res; } // Function to find the maximum amount // to buy all candies int findMaximum(int arr[], int n, int k) { int res = 0, index = 0; for (int i=n-1; i>=index; i--) { // Buy candy with maximum amount res += arr[i]; // And get k candies for free from // the starting index += k; } return res; } // Driver code int main() { int arr[] = {3, 2, 1, 4}; int n = sizeof(arr) / sizeof(arr[0]); int k = 2; sort(arr, arr+n); cout << findMinimum(arr, n, k)<<" " << findMaximum(arr, n, k)<<endl; return 0; }

## Java

// Java implementation to find the // minimum and maximum amount import java.util.*; class GFG { // Function to find the minimum // amount to buy all candies static int findMinimum(int arr[], int n, int k) { int res = 0; for (int i = 0; i < n; i++) { // Buy current candy res += arr[i]; // And take k candies for free // from the last n = n - k; } return res; } // Function to find the maximum // amount to buy all candies static int findMaximum(int arr[], int n, int k) { int res = 0, index = 0; for (int i = n - 1; i >= index; i--) { // Buy candy with maximum amount res += arr[i]; // And get k candies for free from // the starting index += k; } return res; } // Driver code public static void main(String[] args) { int arr[] = { 3, 2, 1, 4 }; int n = arr.length; int k = 2; Arrays.sort(arr); System.out.println(findMinimum(arr, n, k) + " " + findMaximum(arr, n, k)); } } // This code is contributed by prerna saini

## Python3

# Python implementation # to find the minimum # and maximum amount # Function to find # the minimum amount # to buy all candies def findMinimum(arr,n,k): res = 0 i=0 while(n): # Buy current candy res += arr[i] # And take k # candies for free # from the last n = n-k i+=1 return res # Function to find # the maximum amount # to buy all candies def findMaximum(arr, n, k): res = 0 index = 0 i=n-1 while(i>=index): # Buy candy with # maximum amount res += arr[i] # And get k candies # for free from # the starting index += k i -= 1 return res # Driver code arr = [3, 2, 1, 4] n = len(arr) k = 2 arr.sort() print(findMinimum(arr, n, k)," ", findMaximum(arr, n, k)) # This code is contributed # by Anant Agarwal.

## C#

// C# implementation to find the // minimum and maximum amount using System; public class GFG { // Function to find the minimum // amount to buy all candies static int findMinimum(int []arr, int n, int k) { int res = 0; for (int i = 0; i < n; i++) { // Buy current candy res += arr[i]; // And take k candies for // free from the last n = n - k; } return res; } // Function to find the maximum // amount to buy all candies static int findMaximum(int []arr, int n, int k) { int res = 0, index = 0; for (int i = n - 1; i >= index; i--) { // Buy candy with maximum // amount res += arr[i]; // And get k candies for free // from the starting index += k; } return res; } // Driver code public static void Main() { int []arr = { 3, 2, 1, 4 }; int n = arr.Length; int k = 2; Array.Sort(arr); Console.WriteLine( findMinimum(arr, n, k) + " " + findMaximum(arr, n, k)); } } // This code is contributed by Sam007.

## PHP

<?php // PHP implementation to find the minimum // and maximum amount // Function to find the minimum amount // to buy all candies function findMinimum($arr, $n,$k) { $res = 0; for ($i = 0; $i < $n ; $i++) { // Buy current candy $res += $arr[$i]; // And take k candies for free // from the last $n = $n - $k; } return $res; } // Function to find the maximum amount // to buy all candies function findMaximum($arr, $n, $k) { $res = 0; $index = 0; for ($i = $n - 1; $i >= $index; $i--) { // Buy candy with maximum amount $res += $arr[$i]; // And get k candies // for free from // the starting $index += $k; } return $res; } // Driver Code $arr = array(3, 2, 1, 4); $n = sizeof($arr); $k = 2; sort($arr); sort($arr,$n); echo findMinimum($arr, $n, $k)," " ,findMaximum($arr, $n, $k); return 0; // This code is contributed by nitin mittal. ?>

Output:

3 7

Time Complexity : O(n log n)

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