Split a given array into K subarrays minimizing the difference between their maximum and minimum

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

Examples: 

Input: arr[] = {1, 3, 3, 7}, K = 4 
Output: 0 
Explanation: 
The given array can be split into 4 subarrays as {1}, {3}, {3}, and {7}. 
The difference between minimum and maximum of each subarray is: 
1. {1}, difference = 1 - 1 = 0 
2. {3}, difference = 3 - 3 = 0 
3. {3}, difference = 3 - 3 = 0 
4. {7}, difference = 7 - 7 = 0 
Therefore, the sum all the difference is 0 which is minimized.

Input: arr[] = {4, 8, 15, 16, 23, 42}, K = 3 
Output: 12 
Explanation: 
The given array can be split into 3 subarrays as {4, 8, 15, 16}, {23}, and {42}. 
The difference between minimum and maximum of each subarray is: 
1. {4, 8, 15, 16}, difference = 16 - 4 = 12 
2. {23}, difference = 23 - 23 = 0 
3. {42}, difference = 42 - 42 = 0 
Therefore, the sum all the difference is 12 which is minimized. 

Approach: To split the given array into K subarrays with the given conditions, the idea is to split at indexes(say i) where the difference between elements arr[i+1] and arr[i] is largest. Below are the steps to implement this approach: 

1. Store the difference between consecutive pairs of elements in the given array arr[] into another array(say temp[]).
2. Sort the array temp[] in increasing order.
3. Initialise the total difference(say diff) as the difference of the first and last element of the given array arr[].
4. Add the first K - 1 values from the array temp[] to the above difference.
5. The value stored in diff is the minimum sum of the difference of maximum and minimum element of the K subarray.

Below is the implementation of the above approach: 

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

// Function to find the subarray
int find(int a[], int n, int k)
{
    vector<int> v;

    // Add the difference to vectors
    for (int i = 1; i < n; ++i) {
        v.push_back(a[i - 1] - a[i]);
    }

    // Sort vector to find minimum k
    sort(v.begin(), v.end());

    // Initialize result
    int res = a[n - 1] - a[0];

    // Adding first k-1 values
    for (int i = 0; i < k - 1; ++i) {
        res += v[i];
    }

// Return the minimized sum
    return res;
}

// Driver Code
int main()
{
// Given array arr[]
    int arr[] = { 4, 8, 15, 16, 23, 42 };

    int N = sizeof(arr) / sizeof(int);

// Given K
int K = 3;

// Function Call
    cout << find(arr, N, K) << endl;

    return 0;
}

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

class GFG{

// Function to find the subarray
static int find(int a[], int n, int k)
{
    Vector<Integer> v = new Vector<Integer>();

    // Add the difference to vectors
    for(int i = 1; i < n; ++i)
    {
       v.add(a[i - 1] - a[i]);
    }

    // Sort vector to find minimum k
    Collections.sort(v);

    // Initialize result
    int res = a[n - 1] - a[0];

    // Adding first k-1 values
    for(int i = 0; i < k - 1; ++i)
    {
       res += v.get(i);
    }
    
    // Return the minimized sum
    return res;
}

// Driver Code
public static void main(String[] args)
{
    
    // Given array arr[]
    int arr[] = { 4, 8, 15, 16, 23, 42 };

    int N = arr.length;
    
    // Given K
    int K = 3;
    
    // Function Call
    System.out.print(find(arr, N, K) + "\n");
}
}

// This code is contributed by Amit Katiyar

# Python3 program for the above approach

# Function to find the subarray
def find(a, n, k):
    
    v = []
    
    # Add the difference to vectors
    for i in range(1, n):
        v.append(a[i - 1] - a[i])
        
    # Sort vector to find minimum k
    v.sort()
    
    # Initialize result
    res = a[n - 1] - a[0]
    
    # Adding first k-1 values
    for i in range(k - 1):
        res += v[i]
    
    # Return the minimized sum
    return res
    
# Driver code
arr = [ 4, 8, 15, 16, 23, 42 ]

# Length of array
N = len(arr)

K = 3

# Function Call
print(find(arr, N, K))

# This code is contributed by sanjoy_62

// C# program for the above approach
using System;
using System.Collections.Generic;

class GFG{

// Function to find the subarray
static int find(int []a, int n, int k)
{
    List<int> v = new List<int>();

    // Add the difference to vectors
    for(int i = 1; i < n; ++i)
    {
       v.Add(a[i - 1] - a[i]);
    }

    // Sort vector to find minimum k
    v.Sort();

    // Initialize result
    int res = a[n - 1] - a[0];

    // Adding first k-1 values
    for(int i = 0; i < k - 1; ++i)
    {
       res += v[i];
    }
    
    // Return the minimized sum
    return res;
}

// Driver Code
public static void Main(String[] args)
{
    
    // Given array []arr
    int []arr = { 4, 8, 15, 16, 23, 42 };
    int N = arr.Length;
    
    // Given K
    int K = 3;
    
    // Function Call
    Console.Write(find(arr, N, K) + "\n");
}
}

// This code is contributed by Amit Katiyar

<script>
// javascript program for the above approach

    // Function to find the subarray
    function find(a , n , k) {
        var v = [];

        // Add the difference to vectors
        for (i = 1; i < n; ++i) {
            v.push(a[i - 1] - a[i]);
        }

        // Sort vector to find minimum k
        v.sort((a,b)=>a-b);

        // Initialize result
        var res = a[n - 1] - a[0];

        // Adding first k-1 values
        for (i = 0; i < k - 1; ++i) {
            res += v[i];
        }

        // Return the minimized sum
        return res;
    }

    // Driver Code
    

        // Given array arr
        var arr = [ 4, 8, 15, 16, 23, 42 ];

        var N = arr.length;

        // Given K
        var K = 3;

        // Function Call
        document.write(find(arr, N, K) + "\n");

// This code contributed by aashish1995
</script>

12

Time Complexity: O(N), where N is the number of elements in the array. 
Auxiliary Space: O(N), where N is the number of elements in the array.