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

• Difficulty Level : Basic
• Last Updated : 07 May, 2021

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

Example:

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Input: arr[] = {7, 9, 5, 10}
Output:
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

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 the 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);` `    ``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`

## Javascript

 ``
Output:
`1`

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

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