# 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

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 **i ^{th}** 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 = 3Output:11Explanation:

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=2Output:59Explanation:

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**i**operation, increment the value of^{th}**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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