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# Minimum difference between groups of size two

• Difficulty Level : Medium
• Last Updated : 21 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 the highest sum and the one with the lowest sum is minimum.
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[] = {2, 6, 4, 3}
Output : 1
Groups formed will be (2, 6) and (4, 3),
the difference between highest sum group
(2, 6) i.e 8 and lowest sum group (3, 4)
i.e 7 is 1.

Input : arr[] = {11, 4, 3, 5, 7, 1}
Output : 3
Groups formed will be (1, 11), (4, 5) and
(3, 7), the difference between highest
sum group (1, 11) i.e 12 and lowest sum
group (4, 5) i.e 9 is 3.```

Simple Approach: A simple approach would be to try against all combinations of array elements and check against each set of combination difference between the group with the highest sum and the one with the lowest sum. A total of n*(n-1)/2 such groups would be formed (nC2).
Time Complexity: O(n^3) To generate groups n^2 iterations will be needed and to check against each group n iterations will be needed and hence n^3 iterations will be needed in the worst case.
Efficient Approach: Efficient approach would be to use the greedy approach. Sort the whole array and generate groups by selecting one element from the start of the array and one from the end.

## C++

 `// CPP program to find minimum difference``// between groups of highest and lowest``// sums.``#include ``#define ll long long int``using` `namespace` `std;` `ll calculate(ll a[], ll n)``{``    ``// Sorting the whole array.``    ``sort(a, a + n);` `    ``// Generating sum groups.``    ``vector s;``    ``for` `(``int` `i = 0, j = n - 1; i < j; i++, j--)``       ``s.push_back(a[i] + a[j]);` `    ``ll mini = *min_element(s.begin(), s.end());``    ``ll maxi = *max_element(s.begin(), s.end());` `    ``return` `abs``(maxi - mini);``}` `int` `main()``{``    ``ll a[] = { 2, 6, 4, 3 };``    ``int` `n = ``sizeof``(a) / (``sizeof``(a));``    ``cout << calculate(a, n) << endl;``    ``return` `0;``}`

## Java

 `// Java program to find minimum``// difference between groups of``// highest and lowest sums.``import` `java.util.Arrays;``import` `java.util.Collections;``import` `java.util.Vector;`  `class` `GFG {` `static` `long` `calculate(``long` `a[], ``int` `n)``{``    ``// Sorting the whole array.``    ``Arrays.sort(a);``    ``int` `i,j;``    ` `    ``// Generating sum groups.``    ``Vector s = ``new` `Vector<>();``    ``for` `(i = ``0``, j = n - ``1``; i < j; i++, j--)``        ``s.add((a[i] + a[j]));``        ` `    ``long` `mini = Collections.min(s);``    ``long` `maxi = Collections.max(s);``    ``return` `Math.abs(maxi - mini);``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``long` `a[] = { ``2``, ``6``, ``4``, ``3` `};``    ``int` `n = a.length;``    ``System.out.println(calculate(a, n));``    ``}``}``// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 program to find minimum``# difference between groups of``# highest and lowest sums.``def` `calculate(a, n):``    ` `    ``# Sorting the whole array.``    ``a.sort();` `    ``# Generating sum groups.``    ``s ``=` `[];``    ``i ``=` `0``;``    ``j ``=` `n ``-` `1``;``    ``while``(i < j):``        ``s.append((a[i] ``+` `a[j]));``        ``i ``+``=` `1``;``        ``j ``-``=` `1``;` `    ``mini ``=` `min``(s);``    ``maxi ``=` `max``(s);` `    ``return` `abs``(maxi ``-` `mini);` `# Driver Code``a ``=` `[ ``2``, ``6``, ``4``, ``3` `];``n ``=` `len``(a);``print``(calculate(a, n));` `# This is contributed by mits`

## C#

 `// C# program to find minimum``// difference between groups of``// highest and lowest sums.``using` `System;``using` `System.Linq;``using` `System.Collections.Generic;` `class` `GFG``{` `    ``static` `long` `calculate(``long` `[]a, ``int` `n)``    ``{``        ``// Sorting the whole array.``        ``Array.Sort(a);``        ``int` `i, j;` `        ``// Generating sum groups.``        ``List<``long``> s = ``new` `List<``long``>();``        ``for` `(i = 0, j = n - 1; i < j; i++, j--)``            ``s.Add((a[i] + a[j]));` `        ``long` `mini = s.Min();``        ``long` `maxi = s.Max();``        ``return` `Math.Abs(maxi - mini);``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``long` `[]a = { 2, 6, 4, 3 };``        ``int` `n = a.Length;``        ``Console.WriteLine(calculate(a, n));``    ``}``}` `//This code is contributed by Rajput-Ji`

## PHP

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## Javascript

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Output:
`1`

Time Complexity: O (n * log n) 