Open In App

Java Program for 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.

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));``    ``}``}`

Output:

`11`

Time Complexity: O (n * log n)

Space Complexity: O(1) as no extra space has been taken.

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).

Below is the code for the above approach.

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 = Arrays.stream(arr).min().getAsInt();``    ``int` `second_min = Integer.MAX_VALUE;``    ``for``(``int` `i = ``0``; i < n ; i ++)``    ``{``        ``// If arr[i] is not equal to first min``        ``if` `(arr[i] != first_min)``            ``second_min = Math.min(arr[i],second_min);``    ``}``    ``int` `first_max = Arrays.stream(arr).max().getAsInt();``    ``int` `second_max = Integer.MIN_VALUE;``    ``for` `(``int` `i = ``0``; i < n ; i ++)``    ``{``        ``// If arr[i] is not equal to first max``        ``if` `(arr[i] != first_max)``            ``second_max = Math.max(arr[i],second_max);``    ``}``    ` `    ``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));``    ``}``}` `// This code is contributed by Aman Kumar`

Output

`11`

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

Please refer complete article on Maximum difference between groups of size two for more details!