Any idea of approach to this question?

Question

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

Expected Time Complexity : O(nlogn)

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.

Answer 1

Define a function get_min() . This will take A,k
n := size of A
sort the list A
res := 0, i:= 0
while n is non-zero, do
res := res + A[i]
n := n-k
i := i + 1
return res
Define a function get_max() . This will take A, k
n := size of A
sort the list A
res := 0, idx := 0
i:= n-1
while i>=idx, do
res := res + A[i]
idx := idx + k
i := i - 1
return res
From the main method call these two functions to get the results
get_min(A, k)
get_max(A, k)

if you want another approach you can visit here
https://www.google.com/amp/s/www.geeksforgeeks.org/find-minimum-maximum-amount-buy-n-candies/amp/

Answer 2

Start by sorting prices of candy in ascending order.

The key lies in figuring out the equation which tells us how many 'pairs' of puchased candy and free candy we can get, based on N and K. This way, you do not even need to use a loop for allocating candies.

purchase_c = N%(K+1) is not 0
purchase_c += N/(K+1)

You now have the number of candies you need to purchase. Proceed to use the following logic on the sorted list of prices:

Max logic: Buy most expensive candies.
Min logic: Buy cheapest candies.

https://code.sololearn.com/cSc1TCxYQ9fe/?ref=app