How to implement the Sum of Subsets problem in Java

Question

Does anyone know how to implement the Sum-of-Subsets problem in Java from this pseudo code?

w = an array of positive integers sorted in non-decreasing order.
W = the target sum value
include = an array or arraylist of the solutions who's weight adds up to W. After the print statement, this array can be deleted so that it can store the next solution.
weight = weight of elements in the include array.
total = weight of the remaining elements not in the include array.

public static void sum_of_subsets(index i, int weight, int total)
{
     if(promising(i))
     {
          if(weight == W)
          {
               System.out.print(include[1] through include[i]);
          }
          else
          {
               include[i + 1] = "yes";     //Include w[i + 1]
               sum_of)subsets(i + 1, weight + w[i + 1], total - w[i + 1]);
               include[i + 1] = "no";      //Do not include w[i + 1]
               sum_of_subsets(i + 1, weight, total - w[i + 1]);
          }
     }
}

public static boolean promising(index i);
{
     return (weight + total >= W) && (weight == W || weight + w[i + 1] <= W);
}

this is really baffling me, so if you could add comments that would be great!!!

Answer 1

Algorithms Coded in Java

First the algorithm removes all numbers that are larger than the sum to begin with.

Then for the largest number smaller than the sum, it checks if there are any numbers in the list that it can add to itself to get the sum. Once we have either found a pair, or the sum is greater than the desired sum, we can break since the list is sorted. We then consider the second largest number and see if we can make a pair with that, and so on.

   /**
    * This will find how many pairs of numbers in the given array sum
    * up to the given number.
    *
    * @param array - array of integers
    * @param sum - The sum
    * @return int - number of pairs.
    */
public static int sumOfSubset(int[] array, int sum)
{
        // This has a complexity of O ( n lg n )
        Arrays.sort(array);

        int pairCount = 0;
        int leftIndex = 0;
        int rightIndex = array.length - 1;

        // The portion below has a complextiy of
        //  O ( n ) in the worst case.
        while (array[rightIndex] > sum + array[0])
        {
            rightIndex--;    
        }

        while (leftIndex < rightIndex)
        {
            if (array[leftIndex] + array[rightIndex] == sum)
            {
                pairCount++;
                leftIndex++;
                rightIndex--;
            }
            else if(array[leftIndex] + array[rightIndex]  < sum)
            {
                leftIndex++;
            }
            else
            {
                rightIndex--;   
            }
        }

        return pairCount;
}

The algorithm above does not return the pairs, but that is trivially to add.