# 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 – 12 = 0
2. {23}, difference = 23 – 23 = 0
3. {42}, difference = 42 – 42 = 0
Therefore, the sum all the difference is 12 which is minimized.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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. Intialise the total difference(say diff) as the difference of 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 K subarray.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach ` `#include ` `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; ` ` `  `    ``// 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

 `// 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 v = ``new` `Vector(); ` ` `  `    ``// 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

 `# 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#

 `// 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; ` ` `  `    ``// 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 `

Output:

```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. My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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Improved By : amit143katiyar, sanjoy_62

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