Minimum cost of choosing the array element

Given an array arr[] of N integers and an integer M and the cost of selecting any array element(say x) at any day(say d), is x*d. The task is to minimize the cost of selecting 1, 2, 3, ..., N array where each day at most M elements is allowed to select.
Examples: 
 

Input: arr[] = {6, 19, 3, 4, 4, 2, 6, 7, 8}, M = 2 
Output: 2 5 11 18 30 43 62 83 121 
Explanation: 
For selecting 1, 2, 3, .. , N elements when at most 2 elements are allowed to select each day:
The Cost of selecting 1 element: 
select one smallest element on day 1, then cost is 2*1 = 2 
The Cost of selecting 2 elements: 
select two smallest elements on day 1, then cost is (2+3)*1 = 5
The Cost of selecting 3 elements: 
select 2nd and 3rd smallest elements on day 1, then cost is (3+4)*1 = 7 
select 1st smallest element on day 2, then cost is 2*2 = 4 
So, the total cost is 7 + 4 = 11 
Similarly, we can find the cost for selecting 4, 5, 6, 7, 8 and 9 elements is 18, 30, 43, 62, 83 and 121 respectively.
Input: arr[] = {6, 19, 12, 6, 7, 9}, M = 3 
Output: 6 12 19 34 52 78 
 

Approach: The idea is to use Prefix Sum Array. 
 

1. Sort the given array in increasing order.
2. Store the prefix sum of the sorted array in pref[]. This prefix sum gives the minimum cost of selecting the 1, 2, 3, ... N array elements when atmost one element is allowed to select each day.
3. To find the minimum cost when atmost M element is allowed to select each day, update the prefix array pref[] from index M to N as: 
    

pref[i] = pref[i] + pref[i-M]

1. 
   For Example: 
    

arr[] = {6, 9, 3, 4, 4, 2, 6, 7, 8}
After sorting arr[]:
arr[] = {2, 3, 4, 4, 6, 6, 7, 8, 9}

Prefix array is:
pref[] = {2, 5, 9, 13, 19, 25, 32, 40, 49}
Now at every index i, pref[i] gives the cost 
of selecting i array element when atmost one 
element is allowed to select each day.

Now for M = 3, when at most 3 elements
are allowed to select each day, then 
by update every index(from M to N)
of pref[] as:
pref[i] = pref[i] + pref[i-M] 

the cost of selecting elements 
from (i-M+1)th to ith index on day 1,
the cost of selecting elements 
from (i-M)th to (i-2*M)th index on day 2
...
...
...
the cost of selecting elements 
from (i-n*M)th to 0th index on day N.

1. 
    
2. After the above step, every index(say i) of prefix array pref[] stores the cost selecting i elements when atmost M elements are allowed to select each day.

Below is the implementation of the above approach: 
 

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function that find the minimum cost of 
// selecting array element
void minimumCost(int arr[], int N, int M) {
    
    // Sorting the given array in 
    // increasing order
    sort(arr, arr + N);
    
    // To store the prefix sum of arr[] 
    int pref[N];
    
    pref[0] = arr[0];
    
    for(int i = 1; i < N; i++) {
        pref[i] = arr[i] + pref[i-1];
    }
    
    // Update the pref[] to find the cost
    // selecting array element by selecting
    // at most M element
    for(int i = M; i < N; i++) {
        pref[i] += pref[i-M];
    }
    
    // Print the pref[] for the result
    for(int i = 0; i < N; i++) {
        cout << pref[i] << ' ';
    }
    
}

// Driver Code
int main()
{
    int arr[] = {6, 19, 3, 4, 4, 2, 6, 7, 8};
    int M = 2;
    int N = sizeof(arr)/sizeof(arr[0]);
    
    minimumCost(arr, N, M);
    return 0;
}

// Java program for the above approach
import java.util.*;

class GFG{

// Function that find the minimum cost of 
// selecting array element
static void minimumCost(int arr[], int N, int M) 
{
    
    // Sorting the given array in 
    // increasing order
    Arrays.sort(arr);
    
    // To store the prefix sum of arr[] 
    int []pref = new int[N];
    pref[0] = arr[0];
    
    for(int i = 1; i < N; i++)
    {
        pref[i] = arr[i] + pref[i - 1];
    }
    
    // Update the pref[] to find the cost
    // selecting array element by selecting
    // at most M element
    for(int i = M; i < N; i++)
    {
        pref[i] += pref[i - M];
    }
    
    // Print the pref[] for the result
    for(int i = 0; i < N; i++) 
    {
        System.out.print(pref[i] + " ");
    }
}

// Driver Code
public static void main(String[] args)
{
    int arr[] = { 6, 19, 3, 4, 4, 2, 6, 7, 8 };
    int M = 2;
    int N = arr.length;
    
    minimumCost(arr, N, M);
}
}

// This code is contributed by sapnasingh4991

# Python3 program for the above approach 

# Function that find the minimum cost 
# of selecting array element 
def minimumCost(arr, N, M):
    
    # Sorting the given array in 
    # increasing order 
    arr.sort()
    
    # To store the prefix sum of arr[] 
    pref = []
    
    pref.append(arr[0])
    
    for i in range(1, N):
        pref.append(arr[i] + pref[i - 1])
    
    # Update the pref[] to find the cost 
    # selecting array element by selecting 
    # at most M element 
    for i in range(M, N):
        pref[i] += pref[i - M]
    
    # Print the pref[] for the result 
    for i in range(N):
        print(pref[i], end = ' ')

# Driver Code 
arr = [ 6, 19, 3, 4, 4, 2, 6, 7, 8 ]
M = 2
N = len(arr)

minimumCost(arr, N, M);

# This code is contributed by yatinagg

// C# program for the above approach
using System;

class GFG{

// Function that find the minimum cost  
// of selecting array element
static void minimumCost(int []arr, int N,
                                   int M) 
{
    
    // Sorting the given array  
    // in increasing order
    Array.Sort(arr);
    
    // To store the prefix sum of []arr 
    int []pref = new int[N];
    pref[0] = arr[0];
    
    for(int i = 1; i < N; i++)
    {
       pref[i] = arr[i] + pref[i - 1];
    }
    
    // Update the pref[] to find the cost
    // selecting array element by selecting
    // at most M element
    for(int i = M; i < N; i++)
    {
       pref[i] += pref[i - M];
    }
    
    // Print the pref[] for the result
    for(int i = 0; i < N; i++) 
    {
       Console.Write(pref[i] + " ");
    }
}

// Driver Code
public static void Main(String[] args)
{
    int []arr = { 6, 19, 3, 4, 4,  
                  2, 6, 7, 8 };
    int M = 2;
    int N = arr.Length;
    
    minimumCost(arr, N, M);
}
}

// This code is contributed by Amit Katiyar

<script>

// Javascript program for the above approach

// Function that find the minimum cost of 
// selecting array element
function minimumCost(arr, N, M) 
{
      
    // Sorting the given array in 
    // increasing order
    arr.sort((a, b) => a - b);
      
    // To store the prefix sum of arr[] 
    let pref = Array.from({length: N}, (_, i) => 0);
    pref[0] = arr[0];
      
    for(let i = 1; i < N; i++)
    {
        pref[i] = arr[i] + pref[i - 1];
    }
      
    // Update the pref[] to find the cost
    // selecting array element by selecting
    // at most M element
    for(let i = M; i < N; i++)
    {
        pref[i] += pref[i - M];
    }
      
    // Print the pref[] for the result
    for(let i = 0; i < N; i++) 
    {
       document.write(pref[i] + " ");
    }
}

// Driver Code
    
    let arr = [ 6, 19, 3, 4, 4, 2, 6, 7, 8 ];
    let M = 2;
    let N = arr.length;
      
    minimumCost(arr, N, M);
      
</script>

2 5 11 18 30 43 62 83 121

Time Complexity: O(N*log N), where N is the number of element in the array.

Space Complexity: O(N) as pref array has been created. Here, N is the number of element in the array.