Given an ascendingly 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 = 3Output:12Explanation:

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 = 5Output: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 … i
_{1}-1], the second part will be arr[i_{1}… i_{2}-1] and so on. - So, the sum of the differences between the K parts will be:

sum = arr[i

_{1}-1] – arr[0] + arr[i_{2}-1] – arr[i_{1}] + …. + arr[n] – arr[i_{K}]

- After rearranging the above value, we get:

sum = -arr[0] – (arr[i

_{1}] – arr[i_{1}– 1]) – (arr[i_{2}] – arr[i_{2}– 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++

`// 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

`// 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

`# 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#

`// 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` |

**Output:**

12

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

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