# Maximum difference between groups of size two

• Difficulty Level : Easy
• Last Updated : 13 Jun, 2022

Given an array of even number of elements, form groups of 2 using these array elements such that the difference between the group with highest sum and the one with lowest sum is maximum.
Note: An element can be a part of one group only and it has to be a part of at least 1 group.
Examples:

```Input : arr[] = {1, 4, 9, 6}
Output : 10
Groups formed will be (1, 4) and (6, 9),
the difference between highest sum group
(6, 9) i.e 15 and lowest sum group (1, 4)
i.e 5 is 10.

Input : arr[] = {6, 7, 1, 11}
Output : 11
Groups formed will be (1, 6) and (7, 11),
the difference between highest sum group
(7, 11) i.e 18 and lowest sum group (1, 6)
i.e 7 is 11.```

Simple Approach: We can solve this problem by making all possible combinations and checking each set of combination differences between the group with the highest sum and with the lowest sum. A total of n*(n-1)/2 such groups would be formed (nC2).
Time Complexity: O(n^3), because it will take O(n^2) to generate groups and to check against each group n iterations will be needed thus overall it takes O(n^3) time.
Efficient Approach: We can use the greedy approach. Sort the whole array and our result is sum of last two elements minus sum of first two elements.

## C++

 `// CPP program to find minimum difference``// between groups of highest and lowest``// sums.``#include ``#define ll long long int``using` `namespace` `std;` `ll CalculateMax(ll arr[], ``int` `n)``{``    ``// Sorting the whole array.``    ``sort(arr, arr + n);``   ` `    ``int` `min_sum = arr + arr;``    ``int` `max_sum = arr[n-1] + arr[n-2];` `    ``return` `abs``(max_sum - min_sum);``}` `// Driver code``int` `main()``{``    ``ll arr[] = { 6, 7, 1, 11 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);``    ``cout << CalculateMax(arr, n) << endl;``    ``return` `0;``}`

## Java

 `// Java program to find minimum difference``// between groups of highest and lowest``// sums.``import` `java.util.Arrays;``import` `java.io.*;` `class` `GFG {``static` `int`  `CalculateMax(``int`  `arr[], ``int` `n)``{``    ``// Sorting the whole array.``    ``Arrays.sort(arr);``    ` `    ``int` `min_sum = arr[``0``] + arr[``1``];``    ``int` `max_sum = arr[n-``1``] + arr[n-``2``];` `    ``return` `(Math.abs(max_sum - min_sum));``}` `// Driver code``    ` `    ``public` `static` `void` `main (String[] args) {` `    ``int` `arr[] = { ``6``, ``7``, ``1``, ``11` `};``    ``int` `n = arr.length;``    ``System.out.println (CalculateMax(arr, n));``    ``}``}`

## Python3

 `# Python 3 program to find minimum difference``# between groups of highest and lowest``def` `CalculateMax(arr, n):` `    ``# Sorting the whole array.``    ``arr.sort()``    ``min_sum ``=` `arr[``0``] ``+` `arr[``1``]``    ``max_sum ``=` `arr[n ``-` `1``] ``+` `arr[n ``-` `2``]``    ``return` `abs``(max_sum ``-` `min_sum)` `# Driver code``arr ``=` `[``6``, ``7``, ``1``, ``11``]``n ``=` `len``(arr)``print``(CalculateMax(arr, n))` `# This code is contributed``# by Shrikant13`

## C#

 `// C# program to find minimum difference``// between groups of highest and lowest``// sums.``using` `System;` `public` `class` `GFG{` `static` `int` `CalculateMax(``int` `[]arr, ``int` `n)``{``    ``// Sorting the whole array.``    ``Array.Sort(arr);``    ` `    ``int` `min_sum = arr + arr;``    ``int` `max_sum = arr[n-1] + arr[n-2];` `    ``return` `(Math.Abs(max_sum - min_sum));``}` `// Driver code``    ` `    ``static` `public` `void` `Main (){``    ``int` `[]arr = { 6, 7, 1, 11 };``    ``int` `n = arr.Length;``    ``Console.WriteLine(CalculateMax(arr, n));``    ``}``//This code is contributed by Sachin.   ``}`

## PHP

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## Javascript

 ``

Output

`11`

Output:

`11`

Time Complexity: O (n * log n)
Further Optimization :
Instead of sorting, we can find a maximum two and minimum of two in linear time and reduce the time complexity to O(n).

## C++

 `// CPP program to find minimum difference``// between groups of highest and lowest``// sums.``#include ``using` `namespace` `std;` `int` `CalculateMax(``int` `arr[], ``int` `n)``{``    ` `    ``int` `first_min = INT_MAX;``    ``int` `second_min = INT_MAX;``    ``for``(``int` `i = 0; i < n ; i ++)``    ``{``        ``/* If current element is smaller than first``        ``then update both first and second */``        ``if` `(arr[i] < first_min)``        ``{``            ``second_min = first_min;``            ``first_min = arr[i];``        ``}`` ` `        ``/* If arr[i] is in between first and second``        ``then update second */``        ``else` `if` `(arr[i] < second_min && arr[i] != first_min)``            ``second_min = arr[i];``    ``}``    ` `    ``int` `first_max = INT_MIN;``    ``int` `second_max = INT_MIN;``    ``for` `(``int` `i = 0; i < n ; i ++)``    ``{``        ``/* If current element is smaller than first``        ``then update both first and second */``        ``if` `(arr[i] > first_max)``        ``{``            ``second_max = first_max;``            ``first_max = arr[i];``        ``}`` ` `        ``/* If arr[i] is in between first and second``        ``then update second */``        ``else` `if` `(arr[i] > second_max && arr[i] != first_max)``            ``second_max = arr[i];``    ``}``    ` `    ``return` `abs``(first_max+second_max-first_min-second_min);``    ` `}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 6, 7, 1, 11 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);``    ``cout << CalculateMax(arr, n) << endl;``    ``return` `0;``}` `// This code is contributed by Pushpesh Raj`

## Java

 `// Java program to find minimum difference``// between groups of highest and lowest``// sums.``import` `java.util.Arrays;``import` `java.io.*;` `class` `GFG {``static` `int` `CalculateMax(``int` `arr[], ``int` `n)``{``    ``int` `first_min = Integer.MAX_VALUE;``    ``int` `second_min = Integer.MAX_VALUE;``    ``for``(``int` `i = ``0``; i < n ; i ++)``    ``{``        ``/* If current element is smaller than first``        ``then update both first and second */``        ``if` `(arr[i] < first_min)``        ``{``            ``second_min = first_min;``            ``first_min = arr[i];``        ``}`` ` `        ``/* If arr[i] is in between first and second``        ``then update second */``        ``else` `if` `(arr[i] < second_min && arr[i] != first_min)``            ``second_min = arr[i];``    ``}``    ` `    ``int` `first_max = Integer.MIN_VALUE;``    ``int` `second_max = Integer.MIN_VALUE;``    ``for` `(``int` `i = ``0``; i < n ; i ++)``    ``{``        ``/* If current element is smaller than first``        ``then update both first and second */``        ``if` `(arr[i] > first_max)``        ``{``            ``second_max = first_max;``            ``first_max = arr[i];``        ``}`` ` `        ``/* If arr[i] is in between first and second``        ``then update second */``        ``else` `if` `(arr[i] > second_max && arr[i] != first_max)``            ``second_max = arr[i];``    ``}``    ` `    ``return` `Math.abs(first_max+second_max-first_min-second_min);``}` `// Driver code``    ` `    ``public` `static` `void` `main (String[] args) {` `    ``int` `arr[] = { ``6``, ``7``, ``1``, ``11` `};``    ``int` `n = arr.length;``    ``System.out.println (CalculateMax(arr, n));``    ``}``}`

## Python3

 `# Python 3 program to find minimum difference``# between groups of highest and lowest``def` `CalculateMax(arr, n):` `    ``# maxint constant``    ``first_min ``=` `99999``    ``second_min ``=` `99999` `    ``for` `i ``in` `range``(n):``        ``if` `arr[i] < first_min:``            ``second_min ``=` `first_min``            ``first_min ``=` `arr[i]``        ``# If arr[i] is in between first and second``        ``# then update second` `        ``elif` `arr[i] < second_min & arr[i] !``=` `first_min:``            ``second_min ``=` `arr[i]` `    ``# maxint constant``    ``first_max ``=` `-``99999``    ``second_max ``=` `-``99999` `    ``for` `i ``in` `range``(n):``        ``if` `arr[i] > first_max:``            ``second_max ``=` `first_max``            ``first_max ``=` `arr[i]``        ``# If arr[i] is in between first and second``        ``# then update second` `        ``elif` `arr[i] > second_max & arr[i] !``=` `first_max:``            ``second_max ``=` `arr[i]` `    ``return` `abs``(first_max``+``second_max``-``first_min``-``second_min)`  `# Driver code``arr ``=` `[``6``, ``7``, ``1``, ``11``]``n ``=` `len``(arr)``print``(CalculateMax(arr, n))` `# This code is contributed Aarti_Rathi`

## C#

 `// C# program to find minimum difference``// between groups of highest and lowest``// sums.``using` `System;` `public` `class` `GFG{` `static` `int` `CalculateMax(``int` `[]arr, ``int` `n)``{``    ``int` `first_min = ``int``.MaxValue;``    ``int` `second_min = ``int``.MaxValue;``    ``for``(``int` `i = 0; i < n ; i ++)``    ``{``        ``/* If current element is smaller than first``        ``then update both first and second */``        ``if` `(arr[i] < first_min)``        ``{``            ``second_min = first_min;``            ``first_min = arr[i];``        ``}`` ` `        ``/* If arr[i] is in between first and second``        ``then update second */``        ``else` `if` `(arr[i] < second_min && arr[i] != first_min)``            ``second_min = arr[i];``    ``}``    ` `    ``int` `first_max = ``int``.MinValue;``    ``int` `second_max = ``int``.MinValue;``    ``for` `(``int` `i = 0; i < n ; i ++)``    ``{``        ``/* If current element is smaller than first``        ``then update both first and second */``        ``if` `(arr[i] > first_max)``        ``{``            ``second_max = first_max;``            ``first_max = arr[i];``        ``}`` ` `        ``/* If arr[i] is in between first and second``        ``then update second */``        ``else` `if` `(arr[i] > second_max && arr[i] != first_max)``            ``second_max = arr[i];``    ``}``    ` `    ``return` `Math.Abs(first_max+second_max-first_min-second_min);``}` `// Driver code``    ` `    ``static` `public` `void` `Main (){``    ``int` `[]arr = { 6, 7, 1, 11 };``    ``int` `n = arr.Length;``    ``Console.WriteLine(CalculateMax(arr, n));``    ``}``}`

## Javascript

 ``

Output

`11`

Time Complexity: O(N)
Auxiliary Space: O(1)

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