How to find all permutation of a String in Java

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

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

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

Hi Pankaj, superb man… Thank you for the post. really good to easy understand.

- Anitha Reddy

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

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

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

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

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

Java code which will print all possible permutation of any string .,and string will provided at runtime by user…plz help me…

- Pragya

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

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