# 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)**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).

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