Divide a sorted array in K parts with sum of difference of max and min minimized in each part

Given an ascending sorted array arr[] of size N and an integer K, the task is to partition the given array into K non-empty subarrays such that the sum of differences of the maximum and the minimum of each subarray is minimized.

Examples: 

Input: arr[] = {4, 8, 15, 16, 23, 42}, K = 3 
Output: 12 
Explanation: 
The given array can be split into three sub arrays in the following way: 
{4, 8}, {15, 16, 23}, {42} 
Here, the sum of difference of the minimum and maximum element in each of the subarrays respectively are: 
4 + 8 + 0 = 12.

Input: arr[] = {1, 2, 3, 4, 5}, K = 5 
Output: 0 

Approach: One observation that needs to be made is that we clearly know that the sum of the difference between the maximum and the minimum element of the subarray is minimum only when we choose the adjacent elements as the maximum and the minimum element of the subarray. So: 

• Let's say we have to split the array into K + 1 parts such that the first part will be arr[0 ... i₁-1], the second part will be arr[i₁... i₂-1], and so on.
• So, the sum of the differences between the K parts will be:

sum = arr[i₁-1] - arr[0] + arr[i₂-1] - arr[i₁] + .... + arr[n] - arr[i_K] 
 

• After rearranging the above value, we get:

sum = -arr[0] - (arr[i₁] - arr[i₁ - 1]) - (arr[i₂] - arr[i₂ - 1]) - ... -(arr[i_K] - arr[i_K - 1]) + arr[N - 1] 
 

• Clearly, the value to be computed is formed from the difference between the adjacent elements of the array. If this difference is maximum, then the sum will be minimum.
• Therefore, the idea is to iterate over the array and store the difference between the adjacent elements of the array in another array.
• Now, sort this array in descending order. We know that the maximum values of the difference should be taken to get the minimum difference.
• Therefore, subtract the first K - 1 values from the difference of the first and the last element of the array. This gives the sum of the remaining differences of the K subarrays formed from the array.

Below is the implementation of the above approach:  

// C++ program to find the minimum 
// sum of differences possible for 
// the given array when the array 
// is divided into K subarrays 
#include<bits/stdc++.h> 
using namespace std; 

// Function to find the minimum 
// sum of differences possible for 
// the given array when the array 
// is divided into K subarrays 
int calculate_minimum_split(int n, int a[], int k) 
{ 

    // Array to store the differences 
    // between two adjacent elements 
    int p[n - 1]; 
    
    // Iterating through the array 
    for(int i = 1; i < n; i++) 
    
        // Storing differences to p 
        p[i - 1] = a[i] - a[i - 1]; 
        
    // Sorting p in descending order 
    sort(p, p + n - 1, greater<int>()); 
    
    // Sum of the first k-1 values of p 
    int min_sum = 0; 
    for(int i = 0; i < k - 1; i++) 
        min_sum += p[i]; 
        
    // Computing the result 
    int res = a[n - 1] - a[0] - min_sum; 
    
    return res; 
} 

// Driver code 
int main() 
{ 
    int arr[6] = { 4, 8, 15, 16, 23, 42 }; 
    int k = 3; 
    int n = sizeof(arr) / sizeof(int); 

    cout << calculate_minimum_split(n, arr, k); 
} 

// This code is contributed by ishayadav181 

// Java program to find the minimum
// sum of differences possible for
// the given array when the array
// is divided into K subarrays
import java.util.*;

class GFG{

// Function to find the minimum
// sum of differences possible for
// the given array when the array
// is divided into K subarrays
static int calculate_minimum_split(int n, int a[], 
                                   int k)
{
    
    // Array to store the differences
    // between two adjacent elements
    Integer[] p = new Integer[n - 1];

    // Iterating through the array
    for(int i = 1; i < n; i++)

        // Storing differences to p
        p[i - 1] = a[i] - a[i - 1];

    // Sorting p in descending order
    Arrays.sort(p, new Comparator<Integer>() 
    {
        public int compare(Integer a, Integer b)
        {
            return b - a;
        }
    });

    // Sum of the first k-1 values of p
    int min_sum = 0;
    for(int i = 0; i < k - 1; i++)
        min_sum += p[i];

    // Computing the result
    int res = a[n - 1] - a[0] - min_sum;

    return res;
}

// Driver code
public static void main(String[] args)
{
    int arr[] = { 4, 8, 15, 16, 23, 42 };
    int k = 3;
    int n = arr.length;

    System.out.println(calculate_minimum_split(
                       n, arr, k));
}
}

// This code is contributed by offbeat

# Python3 program to find the minimum
# sum of differences possible for
# the given array when the array
# is divided into K subarrays

# Function to find the minimum
# sum of differences possible for
# the given array when the array
# is divided into K subarrays
def calculate_minimum_split(a, k):

    # Array to store the differences 
    # between two adjacent elements
    p =[] 
    n = len(a)

    # Iterating through the array
    for i in range(1, n):
        
        # Appending differences to p
        p.append(a[i]-a[i-1])

    # Sorting p in descending order
    p.sort(reverse = True)
 
    # Sum of the first k-1 values of p
    min_sum = sum(p[:k-1])
 
    # Computing the result
    res = a[n-1]-a[0]-min_sum 
    
    return res

if __name__ == "__main__":
    arr = [4, 8, 15, 16, 23, 42] 
    K = 3

    print(calculate_minimum_split(arr, K))

// C# program to find the minimum  
// sum of differences possible for  
// the given array when the array  
// is divided into K subarrays  
using System;

class GFG{
    
// Function to find the minimum  
// sum of differences possible for  
// the given array when the array  
// is divided into K subarrays  
static int calculate_minimum_split(int n, int[] a,
                                   int k)  
{  
    
    // Array to store the differences  
    // between two adjacent elements  
    int[] p = new int[n - 1];  
      
    // Iterating through the array  
    for(int i = 1; i < n; i++)  
      
        // Storing differences to p  
        p[i - 1] = a[i] - a[i - 1];  
          
    // Sorting p in descending order  
    Array.Sort(p); 
    Array.Reverse(p); 
      
    // Sum of the first k-1 values of p  
    int min_sum = 0;  
    for(int i = 0; i < k - 1; i++)  
        min_sum += p[i];  
          
    // Computing the result  
    int res = a[n - 1] - a[0] - min_sum;  
      
    return res;  
}  

// Driver code
static void Main()
{
    int[] arr = { 4, 8, 15, 16, 23, 42 };  
    int k = 3;  
    int n = arr.Length;  

    Console.Write(calculate_minimum_split(
                  n, arr, k));
}
}

// This code is contributed by divyeshrabadiya07

<script>
// Javascript program to find the minimum 
// sum of differences possible for 
// the given array when the array 
// is divided into K subarrays 


// Function to find the minimum
// sum of differences possible for
// the given array when the array
// is divided into K subarrays
function calculate_minimum_split(n, a, k)
{
     
    // Array to store the differences 
    // between two adjacent elements 
    let p = Array.from({length: n-1}, (_, i) => 0);
       
    // Iterating through the array 
    for(let i = 1; i < n; i++) 
       
        // Storing differences to p 
        p[i - 1] = a[i] - a[i - 1]; 
           
    // Sorting p in descending order 
    p.sort((a, b) => a - b);
    p.reverse();
       
    // Sum of the first k-1 values of p 
    let min_sum = 0; 
    for(let i = 0; i < k - 1; i++) 
        min_sum += p[i]; 
           
    // Computing the result 
    let res = a[n - 1] - a[0] - min_sum; 
       
    return res; 
}

// Driver Code
    
       let arr = [ 4, 8, 15, 16, 23, 42 ];
    let k = 3;
    let n = arr.length;
 
    document.write(calculate_minimum_split(
                       n, arr, k));
  
           
</script>

12

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

Auxiliary Space: O(N)