Find all Permutations of a String in java

Lets say you have String as ABC.
So we take out A from ABC
First character =A and RemainingString = BC
As we are applying recursion here, we will find permutations of BC.
Take out B from BC.
First character= B and RemainingString = C
As we are applying recursion here, we will find permutations of C.
When we take out C, our String size becomes 0 and that is our base case here.
First character = C and RemainingString = “”
We insert C to differences indexes of Permutations  of RemainingString(“”), so we get permutation of C as C.
We insert B to different indexes of Permutations of remainingString(C), so we get BC and CB.
C: BC, CB
Now we insert A to different indexes in BC and CB.
BC : ABC , BAC , BCA
CB : ACB, CAB, CBA
so thats how we got all permutations of ABC.

package org.arpit.java2blog;

import java.util.HashSet;
import java.util.Iterator;
import java.util.Set;

public class PermutationOfStringJava {

 public static void main(String[] args) {

     Set set=permutationOfString("ABC");
     System.out.println("Permutations of String ABC are:");
     for (Iterator iterator = set.iterator(); iterator.hasNext();) {
   String string = (String) iterator.next();
   System.out.println(string); 
  }
 }

 public static Set permutationOfString(String str)
 {
  Set permutationSet=new HashSet();
 
  if(str.length()==0)
  {
   permutationSet.add("");
            return permutationSet;
  }
 
  // take out first character of String
  char c=str.charAt(0);
  // Remaining String
  String rem=str.substring(1);
 
 
  Set permutatedSetForRemainingString=permutationOfString(rem);
  for (String permutedString: permutatedSetForRemainingString) {
   for (int j = 0; j <= permutedString.length(); j++) {
    String permutation=insertFirstCharAtDiffPlaces(permutedString,c,j);
    permutationSet.add(permutation);
   }
   
  }
  return permutationSet;
 
 }
 public static String insertFirstCharAtDiffPlaces(String perm,char firstChar,int index)
 {
  // Inserting firstCharacter of orig String at difference places based on index
  return perm.substring(0,index)+firstChar+perm.substring(index);
 }

}

When you run above program, you will get following out:


Permutations of String ABC are:
ACB
ABC
BCA
CBA
CAB
BAC

Ref : https://java2blog.com/find-all-permutations-of-string-in-java/