Maximum difference between groups of size two
Last Updated :
20 Mar, 2023
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++
#include <bits/stdc++.h>
#define ll long long int
using namespace std;
ll CalculateMax(ll arr[], int n)
{
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);
}
int main()
{
ll arr[] = { 6, 7, 1, 11 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << CalculateMax(arr, n) << endl;
return 0;
}
|
Java
import java.util.Arrays;
import java.io.*;
class GFG {
static int CalculateMax( int arr[], int n)
{
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));
}
public static void main (String[] args) {
int arr[] = { 6 , 7 , 1 , 11 };
int n = arr.length;
System.out.println (CalculateMax(arr, n));
}
}
|
Python3
def CalculateMax(arr, n):
arr.sort()
min_sum = arr[ 0 ] + arr[ 1 ]
max_sum = arr[n - 1 ] + arr[n - 2 ]
return abs (max_sum - min_sum)
arr = [ 6 , 7 , 1 , 11 ]
n = len (arr)
print (CalculateMax(arr, n))
|
C#
using System;
public class GFG{
static int CalculateMax( int []arr, int n)
{
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));
}
static public void Main (){
int []arr = { 6, 7, 1, 11 };
int n = arr.Length;
Console.WriteLine(CalculateMax(arr, n));
}
}
|
PHP
<?php
function CalculateMax( $arr , $n )
{
sort( $arr );
$min_sum = $arr [0] +
$arr [1];
$max_sum = $arr [ $n - 1] +
$arr [ $n - 2];
return abs ( $max_sum -
$min_sum );
}
$arr = array (6, 7, 1, 11 );
$n = sizeof( $arr );
echo CalculateMax( $arr , $n ), "\n" ;
?>
|
Javascript
<script>
function CalculateMax(arr, n)
{
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>
|
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++
#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 (arr[i] < first_min)
{
second_min = first_min;
first_min = arr[i];
}
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 (arr[i] > first_max)
{
second_max = first_max;
first_max = arr[i];
}
else if (arr[i] > second_max && arr[i] != first_max)
second_max = arr[i];
}
return abs (first_max+second_max-first_min-second_min);
}
int main()
{
int arr[] = { 6, 7, 1, 11 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << CalculateMax(arr, n) << endl;
return 0;
}
|
Java
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 (arr[i] < first_min)
{
second_min = first_min;
first_min = arr[i];
}
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 (arr[i] > first_max)
{
second_max = first_max;
first_max = arr[i];
}
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);
}
public static void main (String[] args) {
int arr[] = { 6 , 7 , 1 , 11 };
int n = arr.length;
System.out.println (CalculateMax(arr, n));
}
}
|
Python3
def CalculateMax(arr, n):
first_min = 99999
second_min = 99999
for i in range (n):
if arr[i] < first_min:
second_min = first_min
first_min = arr[i]
elif arr[i] < second_min & arr[i] ! = first_min:
second_min = arr[i]
first_max = - 99999
second_max = - 99999
for i in range (n):
if arr[i] > first_max:
second_max = first_max
first_max = arr[i]
elif arr[i] > second_max & arr[i] ! = first_max:
second_max = arr[i]
return abs (first_max + second_max - first_min - second_min)
arr = [ 6 , 7 , 1 , 11 ]
n = len (arr)
print (CalculateMax(arr, n))
|
C#
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 (arr[i] < first_min)
{
second_min = first_min;
first_min = arr[i];
}
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 (arr[i] > first_max)
{
second_max = first_max;
first_max = arr[i];
}
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);
}
static public void Main (){
int []arr = { 6, 7, 1, 11 };
int n = arr.Length;
Console.WriteLine(CalculateMax(arr, n));
}
}
|
Javascript
<script>
function CalculateMax(arr, n)
{
let first_min = Number.MAX_VALUE;
let second_min = Number.MAX_VALUE;
for (let i = 0; i < n ; i ++)
{
if (arr[i] < first_min)
{
second_min = first_min;
first_min = arr[i];
}
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 (arr[i] > first_max)
{
second_max = first_max;
first_max = arr[i];
}
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>
|
Complexity Analysis:
- Time Complexity: O(N)
- Auxiliary Space: O(1)
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