I want to print all permutations of a given string in Java. To do this I create one auxiliary array boolean used[] to check if I have used some character or not.

private static void permute(String s) {
        int n = s.length();
        boolean[] used = new boolean[n];
        char[] in = s.toCharArray();
        StringBuffer output = new StringBuffer();
        doPermute(in, output, used, n, 0);

//helper method
private static void doPermute(char[] in, StringBuffer output,
            boolean[] used, int n, int level) {
        // TODO Auto-generated method stub
        if(n == level){
        for (int i = 0; i < in.length; i++) {
            used[i] = true;
            doPermute(in, output, used, n, level+1);
            used[i] = false;

//main() method
public static void main(String[] args) {
        String s = "dog";

My main question: is there more elegant/better solution? Can I avoid using this auxiliary array (I tried to but it didn't work)?

  • \$\begingroup\$ Is it ok that the string aa has two identical permutations? \$\endgroup\$ Aug 20, 2014 at 12:31
  • \$\begingroup\$ I assume that there are no repetitions in the characters. This can easily be fixed, is I use Set \$\endgroup\$ Aug 20, 2014 at 12:33

1 Answer 1


I'd bet you're right with used being somehow strange. The standard algorithm I recall doesn't need it. As I tried to improve your code, I ended up with the standard algorithm, so it's pointless to paste it here (someone else will surely do this better).

So some unsolicited general advice instead(*):

When you start with a main, call a static method, which uses another static method, then you're programming purely procedurally. I always recommend to stop it ASAP, e.g., by something like

public static void main(String[] args) {
    new MyClass().go(args);

Then you need no more static and can start putting some fields into your first object.

The other thing is printing instead of doing some reusable work.

(*) I consider this more important the good algorithmization as it took me long to grasp.

  • 1
    \$\begingroup\$ As I tried to improve your code, I ended up with the standard algorithm, so it's pointless to paste it here I would guess that the OP is not entirely aware of what "the standard algorithm" is in this case. \$\endgroup\$ Aug 20, 2014 at 12:41
  • \$\begingroup\$ @SimonAndréForsberg Sure, but that's what Google masters. And someone else can show step-wise improvements. \$\endgroup\$
    – maaartinus
    Aug 20, 2014 at 12:44
  • \$\begingroup\$ Thank you very much for the feedback! Yes, unfortunately I am not aware of the standard algorithm - is it this one: stackoverflow.com/questions/2920315/permutation-of-array ? \$\endgroup\$ Aug 20, 2014 at 12:45
  • 1
    \$\begingroup\$ @user3371223 Yes, what I've meant was the first code block in this answer. \$\endgroup\$
    – maaartinus
    Aug 20, 2014 at 12:48

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