Maximum difference between groups of size two

Difficulty Level : Easy
Last Updated : 04 May, 2021

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 <bits/stdc++.h>
#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[0] + arr[1];
    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[0]);
    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[0] + arr[1];
    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

<?php
// PHP program to find minimum
// difference between groups of
// highest and lowest sums.
function CalculateMax($arr, $n)
{
    // Sorting the whole array.
    sort($arr);
     
    $min_sum = $arr[0] +
               $arr[1];
    $max_sum = $arr[$n - 1] +
               $arr[$n - 2];
 
    return abs($max_sum -
               $min_sum);
}
 
// Driver code
$arr = array (6, 7, 1, 11 );
$n = sizeof($arr);
echo CalculateMax($arr, $n), "\n" ;
 
// This code is contributed by ajit
?>

Javascript

<script>
 
    // Javascript program to
    // find minimum difference
    // between groups of highest and lowest
    // sums.
     
    function CalculateMax(arr, n)
    {
        // Sorting the whole array.
        arr.sort(function(a, b){return a - b});
 
        let min_sum = arr[0] + arr[1];
        let max_sum = arr[n-1] + arr[n-2];
 
        return (Math.abs(max_sum - min_sum));
    }
     
    let arr = [ 6, 7, 1, 11 ];
    let n = arr.length;
    document.write(CalculateMax(arr, n));
         
</script>

Output:

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).

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