Maximum difference between groups of size two

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. 

Implementation:

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

11

Time Complexity: O (n * log n)

Auxiliary Space: O(1)

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).  

Implementation:

C++

// CPP program to find minimum difference
// between groups of highest and lowest
// sums.
#include <bits/stdc++.h>
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[0]);
    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

<script>

    // Javascript program to
    // find minimum difference
    // between groups of highest and lowest
    // sums.

    function CalculateMax(arr, n)
    {
    let first_min = Number.MAX_VALUE;
    let second_min = Number.MAX_VALUE;
    for(let 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];
    }

    let first_max = Number.MIN_VALUE;
    let second_max = Number.MIN_VALUE;
    for (let 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);
    }

    let arr = [ 6, 7, 1, 11 ];
    let n = arr.length;
    document.write(CalculateMax(arr, n));

</script>

Output

11

Complexity Analysis:

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