Given an integer array (of length n), find and return all the subsets of input array using recursion in Java

Question

Suppose my input array is [15,20,12] The required answer is a 2D array
The Required is output is as followed
[12
20
20 12
15
15 12
15 20
15 20 12
]

Answer 1

Here you go.

public static void main(String[] args) {

    int[] nums= {15, 20, 12};
    int[][] subsets = subsets(nums);
    for (int i = 1; i < subsets.length; i++) {
        System.out.println(Arrays.toString(subsets[i]));
    }
}

public static int[][] subsets(int input[]) {
    List<List<Integer>> list = new ArrayList<>();
    subsetsHelper(list, new ArrayList<>(), input, 0);
    return convertListToArray(list);
}

private static void subsetsHelper(List<List<Integer>> list , List<Integer> resultList, int [] nums, int start){
    list.add(new ArrayList<>(resultList));
    for(int i = start; i < nums.length; i++){
       // add element
        resultList.add(nums[i]);
       // Explore
        subsetsHelper(list, resultList, nums, i + 1);
       // remove
        resultList.remove(resultList.size() - 1);
    }
}

private static int[][] convertListToArray(List<List<Integer>> list) {
    int[][] array = new int[list.size()][];
    for (int i = 0; i < array.length; i++) {
        array[i] = new int[list.get(i).size()];
    }
    for(int i=0; i<list.size(); i++){
        for (int j = 0; j < list.get(i).size(); j++) {
            array[i][j] = list.get(i).get(j);
        }
    }
    return array;

}

1.As each recursion call will represent subset here, add resultList(see recursion code above) to the list of subsets in each call. 2.Iterate over elements of a set. 3.In each iteration Add elements to the list explore(recursion) and make start = i+1 to go through remaining elements of the array. Remove element from the list

Output:

[15]
[15, 20]
[15, 20, 12]
[15, 12]
[20]
[20, 12]
[12]

Answer 2

Not clear if it's homework or practical case. This is how would I solve it using Guava PowerSet:

public static void main(String[] args) {
    Integer input[] = {15,20,12};
    List<Integer> rev = Lists.reverse(Arrays.asList(input));
    Set<Integer> indices = IntStream.range(0, input.length).boxed().collect(ImmutableSet.toImmutableSet());
    Object output[] = Sets.powerSet(indices).stream()
            .filter(indexset -> !indexset.isEmpty())
            .map(indexset -> indexset.stream().map(i -> rev.get(i)).collect(Collectors.collectingAndThen(toList(), Lists::reverse)))
            .map(List::toArray)
            .toArray();
    System.out.println(Arrays.deepToString(output));
}

Answer 3

Disclaimer:

1. This is my original work. No part of the solution has been copied from anywhere.
2. My solution works perfectly for 3 elements. However, this needs to be improved to work for arrays of other sizes. Despite this, I am publishing it so that OP or anyone else can extend this solution to work for an array of any size.
3. This question is close to the power set except for the fact that a power set can not have duplicate elements. If this exception is removed from this question, there are many solutions available eg at 1 , 2 , 3 etc.

---

import java.util.Arrays;

public class Main {
    public static void main(String[] args) {
        int[] arr = { 15, 20, 12 };
        System.out.println(Arrays.deepToString(subsets(arr)));
    }

    public static int[][] subsets(int input[]) {
        int[][] subarrs = new int[(int) Math.pow(2, input.length) - 1][input.length];
        int[] indices = { 0 };
        subsetsHelper(input, subarrs, 0, 0, 0, indices);
        return subarrs;
    }

    private static void subsetsHelper(int input[], int[][] subarrs, int index, int i, int j, int[] indices) {
        if (i == input.length) {
            subarrs[index] = input;
            return;
        }
        int[] subarr = new int[indices.length];
        for (int x = 0; x < subarr.length; x++) {
            subarr[x] = input[indices[x]];
        }
        subarrs[index] = subarr;
        if (j == input.length - 1) {
            subsetsHelper(input, subarrs, index + 1, i + 1, i + 1, new int[] { i + 1 });
        } else {
            subsetsHelper(input, subarrs, index + 1, i, j + 1, new int[] { i, j + 1 });
        }
    }
}

Output:

 [[15], [15, 20], [15, 12], [20], [20, 12], [12], [15, 20, 12]]

Answer 4

public static int[][]returnallsub(int arr[], int si){

    if(si==arr.length)
    {int[][]ret=new int [1][0];
return ret;
    }
    
    
    int [][]rss =returnallsub(arr,si+1);
    int [][]tss=new int[rss.length*2][];
    
    int i=0;
for(;i<rss.length;i++) {
tss[i]=new int [rss[i].length];

}
       int k=0;
    for(;k<rss.length;k++) {
tss[i]=new int [rss[k].length+1];
  i++;
}
    
    
        int j=0;
   for(i=0;i<rss.length;i++) {
  for(j=0;j<rss[i].length;j++){
      
      tss[i][j]=rss[i][j];
  }
}
    
     for(k=i;k<tss.length;k++)  {
      tss[k][0]=arr[si];
  }
    
      int r=i;
      for(i=0;i<rss.length;i++) {
  for(j=0;j<rss[i].length;j++){
      
      tss[r][j+1]=rss[i][j];
      
  }
          r++;
} 
 return tss; 
    
    
}



public static int[][] subsets(int arr[]) {
   
     int start=0;
  return  returnallsub( arr,0);
    
    


}

}