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Minimize cost to split an array into K subsets such that the cost of each element is its product with its position in the subset

  • Last Updated : 13 May, 2021

Given an array arr[] of size N and a positive integer K, the task is to find the minimum possible cost to split the array into K subsets, where the cost of ith element ( 1-based indexing ) of each subset is equal to the product of that element and i.

Examples:

Input: arr[] = { 2, 3, 4, 1 }, K = 3 
Output: 11 
Explanation: 
Split the array arr[] into K(= 3) subsets { { 4, 1 }, { 2 }, { 3 } } 
Total cost of 1st subset = 4 * 1 + 1 * 2 = 6 
Total cost of 2nd subset = 2 * 1 = 2 
Total cost of 3rd subset = 3 * 1 = 3 
Therefore, the total cost of K(= 3) subsets is 6 + 2 + 3 = 11.

Input: arr[] = { 9, 20, 7, 8 }, K=2 
Output: 59 
Explanation: 
Dividing the array arr[] into K(= 3) subsets { { 20, 8 }, { 9, 7 } } 
Total cost of 1st subset = 20 * 1 + 8 * 2 = 36 
Total cost of 2nd subset = 9 * 1 + 7 * 2 = 23 
Therefore, the total cost of K(= 3) subsets is 36 + 23 = 59

Approach: The problem can be solved using Greedy technique. The idea is to divide the array elements such all elements in respective subsets is in decreasing order. Follow the steps below to solve the problem:



  • Sort the given array in descending order.
  • Initialize a variable, say totalCost, to store the minimum cost to split the array into K subsets.
  • Initialize a variable, say X, to store the position of an element in a subset.
  • Iterate over the range [1, N] using variable i. For every ith operation, increment the value of totalCost by ((arr[i]+ …+ arr[i + K]) * X) and update i = i + K, X += 1.
  • Finally, print the value of totalCost.

Below is the implementation of the above approach:

C++




// C++ program to implement
// the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the minimum cost to
// split array into K subsets
int getMinCost(int* arr, int n, int k)
{
    // Sort the array in descending order
    sort(arr, arr + n, greater<int>());
 
    // Stores minimum cost to split
    // the array into K subsets
    int min_cost = 0;
 
    // Stores position of
    // elements of a subset
    int X = 0;
 
    // Iterate over the range [1, N]
    for (int i = 0; i < n; i += k) {
 
        // Calculate the cost to select
        // X-th element of every subset
        for (int j = i; j < i + k && j < n; j++) {
 
            // Update min_cost
            min_cost += arr[j] * (X + 1);
        }
 
        // Update X
        X++;
    }
 
    return min_cost;
}
 
// Driver Code
int main()
{
    int arr[] = { 9, 20, 7, 8 };
 
    int K = 2;
 
    int N = sizeof(arr)
            / sizeof(arr[0]);
 
    // Function call
    cout << getMinCost(arr, N, K) << endl;
}

Java




// Java program to implement
// the above approach
import java.util.*;
class GFG
{
 
// reverses an array
static void reverse(int a[], int n)
{
    int i, k, t;
    for (i = 0; i < n / 2; i++)
    {
        t = a[i];
        a[i] = a[n - i - 1];
        a[n - i - 1] = t;
    }
}
  
// Function to find the minimum cost to
// split array into K subsets
static int getMinCost(int[] arr, int n, int k)
{
   
    // Sort the array in descending order
    Arrays.sort(arr);
    reverse(arr, n);
  
    // Stores minimum cost to split
    // the array into K subsets
    int min_cost = 0;
  
    // Stores position of
    // elements of a subset
    int X = 0;
  
    // Iterate over the range [1, N]
    for (int i = 0; i < n; i += k)
    {
  
        // Calculate the cost to select
        // X-th element of every subset
        for (int j = i; j < i + k && j < n; j++)
        {
  
            // Update min_cost
            min_cost += arr[j] * (X + 1);
        }
  
        // Update X
        X++;
    }
    return min_cost;
}
   
// Driver code
public static void main(String[] args)
{
    int arr[] = { 9, 20, 7, 8 };
    int K = 2;
    int N = arr.length;
  
    // Function call
    System.out.println( getMinCost(arr, N, K));
}
}
 
// This code is contributed by susmitakundugoaldanga

Python3




# Python program to implement
# the above approach
 
# Function to find the minimum cost to
# split array into K subsets
def getMinCost(arr, n, k):
   
    # Sort the array in descending order
    arr.sort(reverse = True)
 
    # Stores minimum cost to split
    # the array into K subsets
    min_cost = 0;
 
    # Stores position of
    # elements of a subset
    X = 0;
 
    # Iterate over the range [1, N]
    for i in range(0, n, k):
 
        # Calculate the cost to select
        # X-th element of every subset
        for j in range(i, n, 1):
           
            # Update min_cost
            if(j < i + k):
                min_cost += arr[j] * (X + 1);
 
        # Update X
        X += 1;
    return min_cost;
 
# Driver code
if __name__ == '__main__':
    arr = [9, 20, 7, 8];
    K = 2;
    N = len(arr);
 
    # Function call
    print(getMinCost(arr, N, K));
 
# This code is contributed by 29AjayKumar

C#




// C# program to implement
// the above approach
using System;
class GFG
{
 
// reverses an array
static void reverse(int []a, int n)
{
    int i, k, t;
    for (i = 0; i < n / 2; i++)
    {
        t = a[i];
        a[i] = a[n - i - 1];
        a[n - i - 1] = t;
    }
}
  
// Function to find the minimum cost to
// split array into K subsets
static int getMinCost(int[] arr, int n, int k)
{
   
    // Sort the array in descending order
    Array.Sort(arr);
    reverse(arr, n);
  
    // Stores minimum cost to split
    // the array into K subsets
    int min_cost = 0;
  
    // Stores position of
    // elements of a subset
    int X = 0;
  
    // Iterate over the range [1, N]
    for (int i = 0; i < n; i += k)
    {
  
        // Calculate the cost to select
        // X-th element of every subset
        for (int j = i; j < i + k && j < n; j++)
        {
  
            // Update min_cost
            min_cost += arr[j] * (X + 1);
        }
  
        // Update X
        X++;
    }
    return min_cost;
}
   
// Driver code
public static void Main(String[] args)
{
    int []arr = { 9, 20, 7, 8 };
    int K = 2;
    int N = arr.Length;
  
    // Function call
    Console.WriteLine( getMinCost(arr, N, K));
}
}
 
// This code is contributed by shikhasingrajput

Javascript




<script>
 
// JavaScript program to implement
// the above approach
 
// Reverses an array
function reverse(a, n)
{
    var i, k, t;
    for(i = 0; i < n / 2; i++)
    {
        t = a[i];
        a[i] = a[n - i - 1];
        a[n - i - 1] = t;
    }
}
 
// Function to find the minimum cost to
// split array into K subsets
function getMinCost(arr, n, k)
{
     
    // Sort the array in descending order
    arr.sort((a, b) => b - a);
     
    // Stores minimum cost to split
    // the array into K subsets
    var min_cost = 0;
     
    // Stores position of
    // elements of a subset
    var X = 0;
     
    // Iterate over the range [1, N]
    for(var i = 0; i < n; i += k)
    {
         
        // Calculate the cost to select
        // X-th element of every subset
        for(var j = i; j < i + k && j < n; j++)
        {
             
            // Update min_cost
            min_cost += arr[j] * (X + 1);
        }
         
        // Update X
        X++;
    }
    return min_cost;
}
 
// Driver code
var arr = [ 9, 20, 7, 8 ];
var K = 2;
var N = arr.length;
 
// Function call
document.write(getMinCost(arr, N, K));
 
// This code is contributed by rdtank
 
</script>
Output: 
59

 

Time Complexity: O(N * log(N))
Auxiliary Space: O(N)

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