# Split array into two subarrays such that difference of their maximum is minimum

Given an array arr[] consisting of integers, the task is to split the given array into two sub-arrays such that difference between their maximum elements is minimum.
Example:

Input: arr[] = {7, 9, 5, 10}
Output: 1
Explanation:
The subarrays are {5, 10} and {7, 9} with the difference between their maximums = 10 – 9 = 1.

Input: arr[] = {6, 6, 6}
Output: 0

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

Approach:
We can observe that we need to split the array into two subarrays such that:

• If the maximum element occurs more than once in the array, it needs to be present in both the subarrays at least once.
• Otherwise, the largest and the second-largest elements should be present in different subarrays.

This ensures that the difference between the maximum elements of the two subarrays is maximized.
Hence, we need to sort the array, and then the difference between largest 2 elements, i.e. arr[n – 1] and arr[n – 2], is the required answer.

Below is the implementation of the above approach:

## C++

 // C++ Program to split a given // array such that the difference // between their maximums is minimized.    #include using namespace std;    int findMinDif(int arr[], int N) {     // Sort the array     sort(arr, arr + N);        // Return the difference     // between two highest     // elements     return (arr[N - 1] - arr[N - 2]); }    // Driver Program int main() {        int arr[] = { 7, 9, 5, 10 };     int N = sizeof(arr) / sizeof(arr[0]);        cout << findMinDif(arr, N);     return 0; }

## Java

 // Java Program to split a given array  // such that the difference between  // their maximums is minimized. import java.util.*;    class GFG{    static int findMinDif(int arr[], int N) {            // Sort the array     Arrays.sort(arr);            // Return the difference between      // two highest elements     return (arr[N - 1] - arr[N - 2]); }    // Driver code public static void main(String[] args) {     int arr[] = { 7, 9, 5, 10 };     int N = arr.length;        System.out.println(findMinDif(arr, N)); } }    // This code is contributed by offbeat

## Python3

 # Python3 Program to split a given  # array such that the difference  # between their maximums is minimized.  def findMinDif(arr, N):            # Sort the array      arr.sort()        # Return the difference      # between two highest      # elements      return (arr[N - 1] - arr[N - 2])    # Driver Program  arr = [ 7, 9, 5, 10 ] N = len(arr) print(findMinDif(arr, N))    # This code is contributed by yatinagg

## C#

 // C# Program to split a given array  // such that the difference between  // their maximums is minimized. using System; class GFG{    static int findMinDif(int []arr, int N) {            // Sort the array     Array.Sort(arr);            // Return the difference between      // two highest elements     return (arr[N - 1] - arr[N - 2]); }    // Driver code public static void Main() {     int []arr = { 7, 9, 5, 10 };     int N = arr.Length;        Console.Write(findMinDif(arr, N)); } }    // This code is contributed by Code_Mech

Output:

1

Time complexity: O(N*log(N)), N is the number of elements of the array.

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Improved By : yatinagg, offbeat, Code_Mech

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