All subsets of a String in java using recursion

Question

I wrote this code for printing all sets of a String. It also prints the empty subset. Is this the right approach

public class AllSubSet {

    public static void allSet(String s, String t) {
        System.out.print(t + " ");
        if (s.length() > 0) {
            for (int i = 0; i < s.length(); i++) {
                allSet(s.substring(i + 1, s.length()), t + s.charAt(i));
            }
        }
    }

    public static void main(String args[]) {
        String a = "ABC";
        String t = "";
        allSet(a, t);
    }
}

Answer 1

If I have understood correctly, you're aiming for all subset of a String. Through your code for String = "ABC"; //Result is -- > A AB ABC AC B BC C

However, for

String = "ABB";
// Result is --> A AB ABB AB B BB B (You see AB twice as well as letter B).

So to make it more clear for unique subsets, added a set implementation.

TreeSet (extra sorted feature over hash), as below : (using same recursion)

public class Test {

 public static void allSet(String s, String t, Set<String> set) {
     // System.out.print(t + " ");   will print later.
      if(!"".equalsIgnoreCase(t))
      set.add(t);
      if (s.length() > 0) {
          for (int i = 0; i < s.length(); i++) {
              allSet(s.substring(i + 1, s.length()), t + s.charAt(i),set);
          }
      }
  }
public static void main(String args[]) {
    String a = "ABB";
    String t = "";
    Set<String> set = new TreeSet<String>();
    allSet(a, t, set);
    System.out.println(set.toString());
   }



// Now, result --> [A, AB, ABB, B, BB]