Split array into K Subarrays to minimize sum of difference between min and max

Given a sorted array arr[] of size N and integer K, the task is to split the array into K non-empty subarrays such that the sum of the difference between the maximum element and the minimum element of each subarray is minimized. 

Note: Every element of the array must be included in one subarray and each subarray should be non-empty.

Examples:

Input: arr[] = {5, 9, 16, 17, 24, 43}, K = 3 
Output: 12
Explanation: {5, 9}, {16, 17, 24}, {43} are the three subarrays because 
(9-5) + (24-16) + (43-43) = 12 is minimum of all possible subarrays.

Input: arr[] = [5, 7, 8, 8, 11, 14, 22}, K = 3
Output: 13
Explanation: {5, 7}, {8, 8}, {11, 14, 22}. So, (7 - 5) + (8 - 8) + (22 - 11) = 13  
The given array of length 5 cannot be split into subsets of 4. 

Approach: The problem can be solved based on the following idea:

We have to choose K-1 indexes from the array to form K subarrays.

If N is the length of the array and suppose we have chosen K-1 indices (i.e.  a < b < . . . < c) then the required sum of K-subarrays will be 
(arr[a] - arr[0]) + (arr[b] - arr[a+1]) + . . . + (arr[c][/c] - arr[b+1]) + (arr[N-1] - arr[c language="+1"][/c]).

On rearranging: (arr[N-1] - arr[0]) +(arr[a] - arr[a+1]) + (arr[b] - arr[b+1]) + . . . + (arr[c][/c] - arr[c language="+1"][/c])).

So, initially answer is  sum = arr[N-1] - arr[0] and to minimize the answer add K - 1 minimum pairs to sum.

Follow the below steps to implement the idea:

• Create an array and store the difference between adjacent elements.
• Sort the difference array.
• Pick the first K elements from the array.
• Add these values to the difference between arr[N-1] and arr[0].

Below is the implementation of the above approach:  

// C++ code to implement the approach

#include <bits/stdc++.h>
using namespace std;

// Function to find  minimum sum of
// difference between the maximum
// and the minimum element of each subarray
int minimumSum(int arr[], int n, int k)
{
    // Calculate the difference
    // between the adjacent elements
    // and store it in diff array
    int diff[n - 1];

    for (int i = 0; i < n - 1; i++)
        diff[i] = (arr[i] - arr[i + 1]);

    sort(diff, diff + n - 1);

    int sum = arr[n - 1] - arr[0];

    for (int i = 0; i < k - 1; i++)
        sum += diff[i];

    // Return the required sum
    return sum;
}

// Driver code
int main()
{
    int arr[] = { 5, 9, 16, 17, 24, 43 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int K = 3;

    // Function call
    cout << minimumSum(arr, N, K);
    return 0;
}

// Java code to implement the approach

import java.io.*;
import java.util.*;

class GFG {

    // Function to find  minimum sum of
    // difference between the maximum
    // and the minimum element of each subarray
    static int minimumSum(int arr[], int n, int k)
    {
        // Calculate the difference
        // between the adjacent elements
        // and store it in diff array
        int[] diff = new int[n - 1];

        for (int i = 0; i < n - 1; i++)
            diff[i] = (arr[i] - arr[i + 1]);

        Arrays.sort(diff);

        int sum = arr[n - 1] - arr[0];

        for (int i = 0; i < k - 1; i++)
            sum += diff[i];

        // Return the required sum
        return sum;
    }

    public static void main(String[] args)
    {
        int arr[] = { 5, 9, 16, 17, 24, 43 };
        int N = arr.length;
        int K = 3;

        // Function call
        System.out.print(minimumSum(arr, N, K));
    }
}

// This code is contributed by lokeshmvs21.

# Python3 code to implement the approach

# Function to find  minimum sum of
# difference between the maximum
# and the minimum element of each subarray
def minimumSum(arr, n, k) :
    
    # Calculate the difference
    # between the adjacent elements
    # and store it in diff array
    diff = [0] * (n - 1)

    for i in range(n-1):
        diff[i] = (arr[i] - arr[i + 1])

    diff.sort()

    sum = arr[n - 1] - arr[0]

    for i in range(k-1):
        sum += diff[i]

    # Return the required sum
    return sum

# Driver code
if __name__ == "__main__":
    
    arr = [ 5, 9, 16, 17, 24, 43 ]
    N = len(arr)
    K = 3

    # Function call
    print(minimumSum(arr, N, K))

    # This code is contributed by sanjoy_62.

// C# code to implement the approach
using System;

public class GFG {
    // Function to find  minimum sum of
    // difference between the maximum
    // and the minimum element of each subarray
    static int minimumSum(int[] arr, int n, int k)
    {
        // Calculate the difference
        // between the adjacent elements
        // and store it in diff array
        int[] diff = new int[n - 1];

        for (int i = 0; i < n - 1; i++)
            diff[i] = (arr[i] - arr[i + 1]);

        Array.Sort(diff);

        int sum = arr[n - 1] - arr[0];

        for (int i = 0; i < k - 1; i++)
            sum += diff[i];

        // Return the required sum
        return sum;
    }

    static public void Main()
    {
        int[] arr = { 5, 9, 16, 17, 24, 43 };
        int N = arr.Length;
        int K = 3;

        // Function call
        Console.WriteLine(minimumSum(arr, N, K));
    }
}

// This code is contributed by Rohit Pradhan

<script>
        // JavaScript code for the above approach

        // Function to find  minimum sum of
        // difference between the maximum
        // and the minimum element of each subarray
        function minimumSum(arr, n, k)
        {
        
            // Calculate the difference
            // between the adjacent elements
            // and store it in diff array
            let diff = new Array(n - 1);

            for (let i = 0; i < n - 1; i++)
                diff[i] = (arr[i] - arr[i + 1]);

            diff.sort(function (a, b) { return a - b })

            let sum = arr[n - 1] - arr[0];

            for (let i = 0; i < k - 1; i++)
                sum += diff[i];

            // Return the required sum
            return sum;
        }

        // Driver code
        let arr = [5, 9, 16, 17, 24, 43];
        let N = arr.length;
        let K = 3;

        // Function call
        document.write(minimumSum(arr, N, K));

 // This code is contributed by Potta Lokesh

    </script>

12

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