Sample case
[5,3,-2]
possible combinations sums: 5, 3, -2, 8, 6, 1, ..
answer - [8,6,5]

Example :
[1,2,3,1,1000]

Combinations:

Duplicates of earlier ones are striked out, for example (3,1) matches the earlier (1,3).

(), (1), (2), (3), (1), (1000), (1,2), (1,3), (1,1), (1,1000), (2,3), (2,1), (2,1000), (3,1), (3,1000), (1,1000), (1,2,3), (1,2,1), (1,2,1000), (1,3,1), (1,3,1000), (1,1,1000), (2,3,1), (2,3,1000), (2,1,1000), (3,1,1000), (1,2,3,1), (1,2,3,1000), (1,2,1,1000), (1,3,1,1000), (2,3,1,1000), (1,2,3,1,1000)

And the corresponding sums:

0, 1, 2, 3, 1, 1000, 3, 4, 2, 1001, 5, 3, 1002, 4, 1003, 1001, 6, 4, 1003, 5, 1004, 1002, 6, 1005, 1003, 1004, 7, 1006, 1004, 1005, 1006, 1007

Getting top k=3, sums = 1007, 1006, 1005

So output is [1007, 1006, 1005].

Constraints:

Array size n = 1 to 105
Array elements -109 to 109
k ranges from 1 to 2000

import java.util.*;
class HelloWorld {
	static void subsetSums(int arr[], int n,TreeSet treeadd)
	{
		// There are total 2^n subsets
		int total = 1 << n;

			// Consider all numbers from 0 to 2^n - 1
			for (int i = 0; i < total; i++) {
				int sum = 0;
				// Consider binary representation of
				// current i to decide which elements
				// to pick.
				for (int j = 0; j < n; j++)
					if ((i & (1 << j)) != 0)
						sum += arr[j];

				treeadd.add(sum);
	}
	// Driver code
	public static void main(String args[])
	{
		TreeSet<Integer> treeadd = new TreeSet<Integer>();
		int arr[] = new int[] { 1,2,3,1,1000 };
		int n = arr.length;
		int k=3;
		subsetSums(arr, n,treeadd);
		Iterator<Integer>iterator = treeadd.descendingIterator();
		while (iterator.hasNext() && k>0) {
				System.out.println(iterator.next());
				k--;
			}
	}
}