In this tutorial, we will learn how to find the permutation of a String in a Java Program. It’s a tricky question and asked mostly in Java interviews.
Algorithm for Permutation of a String in Java
We will first take the first character from the String and permute with the remaining chars. If String = “ABC” First char = A and remaining chars permutations are BC and CB. Now we can insert first char in the available positions in the permutations. BC -> ABC, BAC, BCA CB -> ACB, CAB, CBA We can write a recursive function to return the permutations and then another function to insert the first characters to get the complete list of permutations.
Java Program to Print Permutations of a String
package com.journaldev.java.string;
import java.util.HashSet;
import java.util.Set;
/**
* Java Program to find all permutations of a String
* @author Pankaj
*
*/
public class StringFindAllPermutations {
public static Set<String> permutationFinder(String str) {
Set<String> perm = new HashSet<String>();
//Handling error scenarios
if (str == null) {
return null;
} else if (str.length() == 0) {
perm.add("");
return perm;
}
char initial = str.charAt(0); // first character
String rem = str.substring(1); // Full string without first character
Set<String> words = permutationFinder(rem);
for (String strNew : words) {
for (int i = 0;i<=strNew.length();i++){
perm.add(charInsert(strNew, initial, i));
}
}
return perm;
}
public static String charInsert(String str, char c, int j) {
String begin = str.substring(0, j);
String end = str.substring(j);
return begin + c + end;
}
public static void main(String[] args) {
String s = "AAC";
String s1 = "ABC";
String s2 = "ABCD";
System.out.println("\nPermutations for " + s + " are: \n" + permutationFinder(s));
System.out.println("\nPermutations for " + s1 + " are: \n" + permutationFinder(s1));
System.out.println("\nPermutations for " + s2 + " are: \n" + permutationFinder(s2));
}
}
I have used Set to store the string permutations. So that duplicates are removed automatically.
Output
Permutations for AAC are:
[AAC, ACA, CAA]
Permutations for ABC are:
[ACB, ABC, BCA, CBA, CAB, BAC]
Permutations for ABCD are:
[DABC, CADB, BCAD, DBAC, BACD, ABCD, ABDC, DCBA, ADBC, ADCB, CBDA, CBAD, DACB, ACBD, CDBA, CDAB, DCAB, ACDB, DBCA, BDAC, CABD, BADC, BCDA, BDCA]
That’s all for finding all permutations of a String in Java.
You can download the example program code from our GitHub Repository.
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for (int i = 0; i perm.add(charInsert(strNew, initial, i)); can u explain me what this mean… as it is showing error … And (String strNew : words) conversion is not done… check this program
- ASHISH MISHRA
Hi Ashish, Looks like code got corrupted and some lines are removed when I posted it here, I have updated it and it’s working fine now. Thanks for the input, you can try the updated program and it will work.
- Pankaj
Hi Pankaj, Thank you for this program. Can you please explainend the program once by taking an example ? I m understanding the logic clearly but have some doubt how its traversing. Thanks in Advace
- Altaf Ahmad
run program in Eclipse in debug mode and see how it’s getting calculated. That would solve all your doubts.
- Pankaj
a.b.c=c.b.a write a program to make the above statement in valid .remember this is not class or interface .do it
- Nizamuddin
sorry write a program to make the above statement is valid in program .remember this is not class or interface.
- Nizamuddin
I am not sure what you are trying to convey in the comment, care to explain?
- Pankaj
using a.b.c=c.b.a; statement how to write java program ?
- prasad
Excellent one! I was finding just this one!
- Tahmid
please write the code for input value is :String str=“KriSHnA” output value is String str=“kRIshNa” i…e., if we taken case sensitive string in the output string should be opposite in there case sensitivity.
- srinivasa.v
public static void reversCase(String str) { String newStr = “”; for(int i=0; i<str.length(); i++) { if(str.charAt(i) == str.toUpperCase().charAt(i)) newStr += str.toLowerCase().charAt(i); if(str.charAt(i) == str.toLowerCase().charAt(i)) newStr += str.toUpperCase().charAt(i); } System.out.println(newStr); }
- Praveen
Hi Pankaj, superb man… Thank you for the post. really good to easy understand.
- Anitha Reddy
I didn’t understand the logic can you please explain in simple form. I understand below logic permutation(prefix + str.charAt(i), str.substring(0, i) + str.substring(i+1, n)); its really very hard from me to understand it if possible can you please explain
- Vijay
Hi Pankaj, Thats a good code you have written there. However, I just have one suggestion which will improve the performance of this code. In you method: public static String charInsert(String str, char c, int j) { String begin = str.substring(0, j); String end = str.substring(j); return begin + c + end; } Your return statement is actually creating 2 more new strings, since the “+” operator creates a new string rather than appending to the existing string. I would rather have a StringBuffer do this thing, since it gives a better performance and is built for cases where you would want to change ur string constantly before a final version. Hope that helps! Thanks!
- Bhavya
Thanks for the suggestion, however StringBuilder will give even better performance. :)
- Pankaj
Yes. I did mean StringBuilder. I tend to get confused between the two very often.
- Bhavya
Its really very easy to understand!! Thanks!!
- rajni
Hi Pankaj, Thank you for the post. Its very clear and easy. What is the ime complexity and of the program?
- Rashmi
There are n! permutations it will take O(n!) time.
- Rudra
I am student of class 11, living in Kolkata. And I am very much interested in computer programming especially in Java. So this particular problem has been given to us as a project. I tried to solve it on my own. However, I failed. I asked a classmate for some help, but she failed to figure out the logic too. We both thought a lot. We ended up with a solution which works only if we know the word before hand. It was a very idiotic one as we had to write n number of for loops if we had to find out the permutation of a word with n number of alphabets. We rejected it. Then we thought about using the Mathematical portion. We thought of creating an array which would store all the letter of the word. And then another which would store all the permutations. We could figure out the number of cells needed by calculating the factorial of the number of words. We have even figured out how to cancel the printing of the words that have already been printed. Then suddenly we discovered that all permutations are even and half are reverse of the permutation. So, if we need to find out the permutation of ABCD. Instead of figuring all 24(4!) possibilities, we can just figure out one half and then we can reverse them. Which are… ABCD ABDC ACBD ACDB ADBC ADCB BACD BADC BCAD BDAC CABD CBAD None of them are common none of them are reverse of each other. Now we can’t figure out any logic to figure out these… That’s where we need some help. Now you’d say to just use your program. But there’s a small problem. There are some functions and other logics you have use which we haven’t yet been taught. So obviously we can’t use them. So if you can just give a solution using arrays and string and scanner and reverse function and all low level functions it would be much help :)
- Anwita Ghosh Dastidar
package Strings; public class PermitationString { public static void main(String[] args) { String s = “ABC”; permutation(s); } public static void permutation(String str) { permutation(“”, str); } private static void permutation(String prefix, String str) { int n = str.length(); if (n == 0) System.out.println(prefix); else { for (int i = 0; i < n; i++) permutation(prefix + str.charAt(i), str.substring(0, i) + str.substring(i+1, n)); } } }
- Siddu
Wow! nice logic thanks for posting. Mozib
- Mozib
String permutation in java simple and easy way. please correct me if my logic is wrong.
package com.java.programming; import java.util.ArrayList; import java.util.LinkedHashMap; import java.util.List; import java.util.Map; import java.util.Map.Entry; /** * @author Brijesh * */ public class StringPermutatio { public static void main(String[] args) { String input = "ABCD"; System.out.println(genPermutation(input).toString()); } public static List genPermutation(String input) { Map firstMap = new LinkedHashMap(); Map secondMap = new LinkedHashMap(); List output = new ArrayList(); char[] chr = input.toCharArray(); for (int i = 0; i < chr.length; i++) { firstMap.put(Character.valueOf(chr[i]), String.valueOf(chr[i])); secondMap.put(Character.valueOf(chr[i]), String.valueOf(chr[i])); } for (Entry chr1 : firstMap.entrySet()) { for (Entry chr2 : secondMap.entrySet()) { if (chr1.getValue().equals(chr2.getValue())) { output.add(String.valueOf(chr1.getValue())); } else { output.add(String.valueOf(chr1.getValue()) + String.valueOf(chr2.getValue())); } } } return output; } }- Brijesh
Hey Brijesh/Team This logic in not OK… I think it will output only 2 Characters String. If i put my name as SHESH then it will give Output as SE, HS, H, HE, ES … etc. BUT actual output should be 5 character String like HHSES, HHSSE, SEHHS, ESSHH, SSEHH etc… If i am wrong plz let me know.
- Shesh narayan
I run this code for a abc. I got only output cba. Please help me
- Ashok
Hi Pankaj , While referring to your above code , I found a bug. I think you are creating new HashSet inside permutationFinder() method , which is wrong. The set should have been passed as a parameter to the method , then only it would give all permutations. Otherwise , it would give for only last iteration.
- vaibhav dave
Hi Pankaj, Thanks for the post. I think the set which is declared inside method findPermutation should have been passed to method. Coz , everytime a new instance is created…which will not provide expected result.
- vaibhav dave
Set words = permutationFinder(rem); Why this line is used
- Manoj
public static void permute(String string){ permute(“”, string); } private static void permute(String permutation, String remainingCharacters){ if(remainingCharacters==null){ return; } else if(remainingCharacters.length()==0){//print permutation for the first character System.out.println(permutation); } for(int i =0; i<remainingCharacters.length();i++){ char firstChar=remainingCharacters.charAt(i);//first character to permute permute(permutation+firstChar,remainingCharacters.substring(0,i)+remainingCharacters.substring(i+1));//add char to the permutation and permute the remaining characters } }
- Siyabonga Mdletshe
Could you please explain the above procedure
- Aditya
is this works no where you are assigning the result
- Nayakam Vishnuvardhan
This is for first set of values abc , aca debugged output. if input if “abc” a bc ab c abc here how the the second word is changed to c. ac b acb please explain how is that changed.
- Nayakam Vishnuvardhan
Java code which will print all possible permutation of any string .,and string will provided at runtime by user…plz help me…
- Pragya
Just extend the program to use Scanner class to get input from the user.
- Pankaj
Why are we using Set for Permutations , the output for any string with repeating characters would be Combinations and not Permutations. We should be using List instead of Set. Please let me know if I am wrong.
- Aman
just to remove duplication.
- Avi
thanks for code that’s a huge help for me
- Baljinder kaur
Code to accept User input and display permutations using the recursive method. public class PermutationExample { public static void main(String args[]) throws Exception { Scanner input = new Scanner(System.in); System.out.print(“Enter String: “); String chars = input.next(); showPermutations(””, chars); } public static void showPermutations(String str, String chars) { if (chars.length() <= 1) System.out.println(str + chars); else for (int i = 0; i < chars.length(); i++) { try { String newString = chars.substring(0, i) + chars.substring(i + 1); showPermutations(str + chars.charAt(i), newString); } catch (Exception e) { e.printStackTrace(); } } } }
- Santy
Hi Pankaj Thanks for your program! When I ran the flow with some input, it throws StringIndexOutOfBoundException. While debugging the flow, I could not find any kind of string data getting stored into the variable “words” in the statement Set words = permutationFinder(); Also, lets take the input as ABC. In the first iteration, it takes the rem as BC, and further the rem goes to empty string. So, it has been observed that the flow can never reach the for loop. Please check and let me know if I am going wrong in my understanding anywhere. The same kind of code I found at crunchify.com with method names different. Thank you
- Datt
public class GFG { static void printPermutn(String str, String ans) { if (str.length() == 0) { System.out.print(ans + " "); // str.length() is ‘0’ } for (int i = 0; i < str.length(); i++) { // str.length is not ‘0’ … Can you Explain this Please. char ch = str.charAt(i); String ros = str.substring(0, i) + str.substring(i + 1); printPermutn(ros, ch+ans); } } public static void main(String[] args) { String s = “abc”; printPermutn(s, “”); } }
- Hemanth
I stole your code to write a method that checks whether any permutation ↴ of an input string is a palindrome. Thanks! import java.util.*; class Movies { public static boolean hasPalindromePermutation(String input){ String[] strArray = permute(input).toArray(new String[0]); for(String cur : strArray){ if(cur.equals(isPalindrome(cur))){ return true; } } return false; } public static Set permute(String str){ Set perm = new HashSet(); //Handling error scenarios if (str == null) { return null; } else if (str.length() == 0) { perm.add(“”); return perm; } char initial = str.charAt(0); // first character String rem = str.substring(1); // Full string without first character Set words = permute(rem); for (String strNew : words) { // System.out.println(strNew); for (int i = 0; i<=strNew.length(); i++){ perm.add(swapper(strNew, initial, i)); } } return perm; } public static String swapper(String str, char c, int j) { String begin = str.substring(0, j); String end = str.substring(j); return begin + c + end; } public static String isPalindrome(String str){ char[] arr = str.toCharArray(); int low = 0; int high = arr.length-1; char temp; while(low<high){ temp = arr[low]; arr[low] = arr[high]; arr[high] = temp; low++; high–; } String result = new String(arr); return result; } public static void main(String[ ] args) {; System.out.println(hasPalindromePermutation(“civic”)); } }
- Beto
Hi Pankaj Thanks for your program! When I ran the flow with some input, it throws StringIndexOutOfBoundException. While debugging the flow, I could not find any kind of string data getting stored into the variable “words” in the statement Set words = permutationFinder(); Also, lets take the input as ABC. In the first iteration, it takes the rem as BC, and further the rem goes to empty string. So, it has been observed that the flow can never reach the for loop. Please check and let me know if I am going wrong in my understanding anywhere. The same kind of code I found at crunchify.com with method names different. Thank you
- Datt
public class GFG { static void printPermutn(String str, String ans) { if (str.length() == 0) { System.out.print(ans + " "); // str.length() is ‘0’ } for (int i = 0; i < str.length(); i++) { // str.length is not ‘0’ … Can you Explain this Please. char ch = str.charAt(i); String ros = str.substring(0, i) + str.substring(i + 1); printPermutn(ros, ch+ans); } } public static void main(String[] args) { String s = “abc”; printPermutn(s, “”); } }
- Hemanth
I stole your code to write a method that checks whether any permutation ↴ of an input string is a palindrome. Thanks! import java.util.*; class Movies { public static boolean hasPalindromePermutation(String input){ String[] strArray = permute(input).toArray(new String[0]); for(String cur : strArray){ if(cur.equals(isPalindrome(cur))){ return true; } } return false; } public static Set permute(String str){ Set perm = new HashSet(); //Handling error scenarios if (str == null) { return null; } else if (str.length() == 0) { perm.add(“”); return perm; } char initial = str.charAt(0); // first character String rem = str.substring(1); // Full string without first character Set words = permute(rem); for (String strNew : words) { // System.out.println(strNew); for (int i = 0; i<=strNew.length(); i++){ perm.add(swapper(strNew, initial, i)); } } return perm; } public static String swapper(String str, char c, int j) { String begin = str.substring(0, j); String end = str.substring(j); return begin + c + end; } public static String isPalindrome(String str){ char[] arr = str.toCharArray(); int low = 0; int high = arr.length-1; char temp; while(low<high){ temp = arr[low]; arr[low] = arr[high]; arr[high] = temp; low++; high–; } String result = new String(arr); return result; } public static void main(String[ ] args) {; System.out.println(hasPalindromePermutation(“civic”)); } }
- Beto
Hi Pankaj Thanks for your program! When I ran the flow with some input, it throws StringIndexOutOfBoundException. While debugging the flow, I could not find any kind of string data getting stored into the variable “words” in the statement Set words = permutationFinder(); Also, lets take the input as ABC. In the first iteration, it takes the rem as BC, and further the rem goes to empty string. So, it has been observed that the flow can never reach the for loop. Please check and let me know if I am going wrong in my understanding anywhere. The same kind of code I found at crunchify.com with method names different. Thank you
- Datt
public class GFG { static void printPermutn(String str, String ans) { if (str.length() == 0) { System.out.print(ans + " "); // str.length() is ‘0’ } for (int i = 0; i < str.length(); i++) { // str.length is not ‘0’ … Can you Explain this Please. char ch = str.charAt(i); String ros = str.substring(0, i) + str.substring(i + 1); printPermutn(ros, ch+ans); } } public static void main(String[] args) { String s = “abc”; printPermutn(s, “”); } }
- Hemanth
I stole your code to write a method that checks whether any permutation ↴ of an input string is a palindrome. Thanks! import java.util.*; class Movies { public static boolean hasPalindromePermutation(String input){ String[] strArray = permute(input).toArray(new String[0]); for(String cur : strArray){ if(cur.equals(isPalindrome(cur))){ return true; } } return false; } public static Set permute(String str){ Set perm = new HashSet(); //Handling error scenarios if (str == null) { return null; } else if (str.length() == 0) { perm.add(“”); return perm; } char initial = str.charAt(0); // first character String rem = str.substring(1); // Full string without first character Set words = permute(rem); for (String strNew : words) { // System.out.println(strNew); for (int i = 0; i<=strNew.length(); i++){ perm.add(swapper(strNew, initial, i)); } } return perm; } public static String swapper(String str, char c, int j) { String begin = str.substring(0, j); String end = str.substring(j); return begin + c + end; } public static String isPalindrome(String str){ char[] arr = str.toCharArray(); int low = 0; int high = arr.length-1; char temp; while(low<high){ temp = arr[low]; arr[low] = arr[high]; arr[high] = temp; low++; high–; } String result = new String(arr); return result; } public static void main(String[ ] args) {; System.out.println(hasPalindromePermutation(“civic”)); } }
- Beto