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Efficient/Fast way to get permutation of a String in java [duplicate]

What is an elegant way to find all the permutations of a string. E.g. permutation for ba, would be ba and ab, but what about longer string such as abcdefgh? Is there any Java implementation example?

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));
}
}
(via Introduction to Programming in Java)

Use recursion.
Try each of the letters in turn as the first letter and then find all the permutations of the remaining letters using a recursive call.
The base case is when the input is an empty string the only permutation is the empty string.

Here is my solution that is based on the idea of the book "Cracking the Coding Interview" (P54):
/**
* List permutations of a string.
*
* #param s the input string
* #return the list of permutations
*/
public static ArrayList<String> permutation(String s) {
// The result
ArrayList<String> res = new ArrayList<String>();
// If input string's length is 1, return {s}
if (s.length() == 1) {
res.add(s);
} else if (s.length() > 1) {
int lastIndex = s.length() - 1;
// Find out the last character
String last = s.substring(lastIndex);
// Rest of the string
String rest = s.substring(0, lastIndex);
// Perform permutation on the rest string and
// merge with the last character
res = merge(permutation(rest), last);
}
return res;
}
/**
* #param list a result of permutation, e.g. {"ab", "ba"}
* #param c the last character
* #return a merged new list, e.g. {"cab", "acb" ... }
*/
public static ArrayList<String> merge(ArrayList<String> list, String c) {
ArrayList<String> res = new ArrayList<>();
// Loop through all the string in the list
for (String s : list) {
// For each string, insert the last character to all possible positions
// and add them to the new list
for (int i = 0; i <= s.length(); ++i) {
String ps = new StringBuffer(s).insert(i, c).toString();
res.add(ps);
}
}
return res;
}
Running output of string "abcd":
Step 1: Merge [a] and b:
[ba, ab]
Step 2: Merge [ba, ab] and c:
[cba, bca, bac, cab, acb, abc]
Step 3: Merge [cba, bca, bac, cab, acb, abc] and d:
[dcba, cdba, cbda, cbad, dbca, bdca, bcda, bcad, dbac, bdac, badc, bacd, dcab, cdab, cadb, cabd, dacb, adcb, acdb, acbd, dabc, adbc, abdc, abcd]

Of all the solutions given here and in other forums, I liked Mark Byers the most. That description actually made me think and code it myself.
Too bad I cannot voteup his solution as I am newbie.
Anyways here is my implementation of his description
public class PermTest {
public static void main(String[] args) throws Exception {
String str = "abcdef";
StringBuffer strBuf = new StringBuffer(str);
doPerm(strBuf,0);
}
private static void doPerm(StringBuffer str, int index){
if(index == str.length())
System.out.println(str);
else { //recursively solve this by placing all other chars at current first pos
doPerm(str, index+1);
for (int i = index+1; i < str.length(); i++) {//start swapping all other chars with current first char
swap(str,index, i);
doPerm(str, index+1);
swap(str,i, index);//restore back my string buffer
}
}
}
private static void swap(StringBuffer str, int pos1, int pos2){
char t1 = str.charAt(pos1);
str.setCharAt(pos1, str.charAt(pos2));
str.setCharAt(pos2, t1);
}
}
I prefer this solution ahead of the first one in this thread because this solution uses StringBuffer. I wouldn't say my solution doesn't create any temporary string (it actually does in system.out.println where the toString() of StringBuffer is called). But I just feel this is better than the first solution where too many string literals are created. May be some performance guy out there can evalute this in terms of 'memory' (for 'time' it already lags due to that extra 'swap')

A very basic solution in Java is to use recursion + Set ( to avoid repetitions ) if you want to store and return the solution strings :
public static Set<String> generatePerm(String input)
{
Set<String> set = new HashSet<String>();
if (input == "")
return set;
Character a = input.charAt(0);
if (input.length() > 1)
{
input = input.substring(1);
Set<String> permSet = generatePerm(input);
for (String x : permSet)
{
for (int i = 0; i <= x.length(); i++)
{
set.add(x.substring(0, i) + a + x.substring(i));
}
}
}
else
{
set.add(a + "");
}
return set;
}

All the previous contributors have done a great job explaining and providing the code. I thought I should share this approach too because it might help someone too. The solution is based on (heaps' algorithm )
Couple of things:
Notice the last item which is depicted in the excel is just for helping you better visualize the logic. So, the actual values in the last column would be 2,1,0 (if we were to run the code because we are dealing with arrays and arrays start with 0).
The swapping algorithm happens based on even or odd values of current position. It's very self explanatory if you look at where the swap method is getting called.You can see what's going on.
Here is what happens:
public static void main(String[] args) {
String ourword = "abc";
String[] ourArray = ourword.split("");
permute(ourArray, ourArray.length);
}
private static void swap(String[] ourarray, int right, int left) {
String temp = ourarray[right];
ourarray[right] = ourarray[left];
ourarray[left] = temp;
}
public static void permute(String[] ourArray, int currentPosition) {
if (currentPosition == 1) {
System.out.println(Arrays.toString(ourArray));
} else {
for (int i = 0; i < currentPosition; i++) {
// subtract one from the last position (here is where you are
// selecting the the next last item
permute(ourArray, currentPosition - 1);
// if it's odd position
if (currentPosition % 2 == 1) {
swap(ourArray, 0, currentPosition - 1);
} else {
swap(ourArray, i, currentPosition - 1);
}
}
}
}

Let's use input abc as an example.
Start off with just the last element (c) in a set (["c"]), then add the second last element (b) to its front, end and every possible positions in the middle, making it ["bc", "cb"] and then in the same manner it will add the next element from the back (a) to each string in the set making it:
"a" + "bc" = ["abc", "bac", "bca"] and "a" + "cb" = ["acb" ,"cab", "cba"]
Thus entire permutation:
["abc", "bac", "bca","acb" ,"cab", "cba"]
Code:
public class Test
{
static Set<String> permutations;
static Set<String> result = new HashSet<String>();
public static Set<String> permutation(String string) {
permutations = new HashSet<String>();
int n = string.length();
for (int i = n - 1; i >= 0; i--)
{
shuffle(string.charAt(i));
}
return permutations;
}
private static void shuffle(char c) {
if (permutations.size() == 0) {
permutations.add(String.valueOf(c));
} else {
Iterator<String> it = permutations.iterator();
for (int i = 0; i < permutations.size(); i++) {
String temp1;
for (; it.hasNext();) {
temp1 = it.next();
for (int k = 0; k < temp1.length() + 1; k += 1) {
StringBuilder sb = new StringBuilder(temp1);
sb.insert(k, c);
result.add(sb.toString());
}
}
}
permutations = result;
//'result' has to be refreshed so that in next run it doesn't contain stale values.
result = new HashSet<String>();
}
}
public static void main(String[] args) {
Set<String> result = permutation("abc");
System.out.println("\nThere are total of " + result.size() + " permutations:");
Iterator<String> it = result.iterator();
while (it.hasNext()) {
System.out.println(it.next());
}
}
}

This one is without recursion
public static void permute(String s) {
if(null==s || s.isEmpty()) {
return;
}
// List containing words formed in each iteration
List<String> strings = new LinkedList<String>();
strings.add(String.valueOf(s.charAt(0))); // add the first element to the list
// Temp list that holds the set of strings for
// appending the current character to all position in each word in the original list
List<String> tempList = new LinkedList<String>();
for(int i=1; i< s.length(); i++) {
for(int j=0; j<strings.size(); j++) {
tempList.addAll(merge(s.charAt(i), strings.get(j)));
}
strings.removeAll(strings);
strings.addAll(tempList);
tempList.removeAll(tempList);
}
for(int i=0; i<strings.size(); i++) {
System.out.println(strings.get(i));
}
}
/**
* helper method that appends the given character at each position in the given string
* and returns a set of such modified strings
* - set removes duplicates if any(in case a character is repeated)
*/
private static Set<String> merge(Character c, String s) {
if(s==null || s.isEmpty()) {
return null;
}
int len = s.length();
StringBuilder sb = new StringBuilder();
Set<String> list = new HashSet<String>();
for(int i=0; i<= len; i++) {
sb = new StringBuilder();
sb.append(s.substring(0, i) + c + s.substring(i, len));
list.add(sb.toString());
}
return list;
}

Well here is an elegant, non-recursive, O(n!) solution:
public static StringBuilder[] permutations(String s) {
if (s.length() == 0)
return null;
int length = fact(s.length());
StringBuilder[] sb = new StringBuilder[length];
for (int i = 0; i < length; i++) {
sb[i] = new StringBuilder();
}
for (int i = 0; i < s.length(); i++) {
char ch = s.charAt(i);
int times = length / (i + 1);
for (int j = 0; j < times; j++) {
for (int k = 0; k < length / times; k++) {
sb[j * length / times + k].insert(k, ch);
}
}
}
return sb;
}

One of the simple solution could be just keep swapping the characters recursively using two pointers.
public static void main(String[] args)
{
String str="abcdefgh";
perm(str);
}
public static void perm(String str)
{ char[] char_arr=str.toCharArray();
helper(char_arr,0);
}
public static void helper(char[] char_arr, int i)
{
if(i==char_arr.length-1)
{
// print the shuffled string
String str="";
for(int j=0; j<char_arr.length; j++)
{
str=str+char_arr[j];
}
System.out.println(str);
}
else
{
for(int j=i; j<char_arr.length; j++)
{
char tmp = char_arr[i];
char_arr[i] = char_arr[j];
char_arr[j] = tmp;
helper(char_arr,i+1);
char tmp1 = char_arr[i];
char_arr[i] = char_arr[j];
char_arr[j] = tmp1;
}
}
}

python implementation
def getPermutation(s, prefix=''):
if len(s) == 0:
print prefix
for i in range(len(s)):
getPermutation(s[0:i]+s[i+1:len(s)],prefix+s[i] )
getPermutation('abcd','')

This is what I did through basic understanding of Permutations and Recursive function calling. Takes a bit of time but it's done independently.
public class LexicographicPermutations {
public static void main(String[] args) {
// TODO Auto-generated method stub
String s="abc";
List<String>combinations=new ArrayList<String>();
combinations=permutations(s);
Collections.sort(combinations);
System.out.println(combinations);
}
private static List<String> permutations(String s) {
// TODO Auto-generated method stub
List<String>combinations=new ArrayList<String>();
if(s.length()==1){
combinations.add(s);
}
else{
for(int i=0;i<s.length();i++){
List<String>temp=permutations(s.substring(0, i)+s.substring(i+1));
for (String string : temp) {
combinations.add(s.charAt(i)+string);
}
}
}
return combinations;
}}
which generates Output as [abc, acb, bac, bca, cab, cba].
Basic logic behind it is
For each character, consider it as 1st character & find the combinations of remaining characters. e.g. [abc](Combination of abc)->.
a->[bc](a x Combination of (bc))->{abc,acb}
b->[ac](b x Combination of (ac))->{bac,bca}
c->[ab](c x Combination of (ab))->{cab,cba}
And then recursively calling each [bc],[ac] & [ab] independently.

Use recursion.
when the input is an empty string the only permutation is an empty string.Try for each of the letters in the string by making it as the first letter and then find all the permutations of the remaining letters using a recursive call.
import java.util.ArrayList;
import java.util.List;
class Permutation {
private static List<String> permutation(String prefix, String str) {
List<String> permutations = new ArrayList<>();
int n = str.length();
if (n == 0) {
permutations.add(prefix);
} else {
for (int i = 0; i < n; i++) {
permutations.addAll(permutation(prefix + str.charAt(i), str.substring(i + 1, n) + str.substring(0, i)));
}
}
return permutations;
}
public static void main(String[] args) {
List<String> perms = permutation("", "abcd");
String[] array = new String[perms.size()];
for (int i = 0; i < perms.size(); i++) {
array[i] = perms.get(i);
}
int x = array.length;
for (final String anArray : array) {
System.out.println(anArray);
}
}
}

this worked for me..
import java.util.Arrays;
public class StringPermutations{
public static void main(String args[]) {
String inputString = "ABC";
permute(inputString.toCharArray(), 0, inputString.length()-1);
}
public static void permute(char[] ary, int startIndex, int endIndex) {
if(startIndex == endIndex){
System.out.println(String.valueOf(ary));
}else{
for(int i=startIndex;i<=endIndex;i++) {
swap(ary, startIndex, i );
permute(ary, startIndex+1, endIndex);
swap(ary, startIndex, i );
}
}
}
public static void swap(char[] ary, int x, int y) {
char temp = ary[x];
ary[x] = ary[y];
ary[y] = temp;
}
}

Java implementation without recursion
public Set<String> permutate(String s){
Queue<String> permutations = new LinkedList<String>();
Set<String> v = new HashSet<String>();
permutations.add(s);
while(permutations.size()!=0){
String str = permutations.poll();
if(!v.contains(str)){
v.add(str);
for(int i = 0;i<str.length();i++){
String c = String.valueOf(str.charAt(i));
permutations.add(str.substring(i+1) + c + str.substring(0,i));
}
}
}
return v;
}

Let me try to tackle this problem with Kotlin:
fun <T> List<T>.permutations(): List<List<T>> {
//escape case
if (this.isEmpty()) return emptyList()
if (this.size == 1) return listOf(this)
if (this.size == 2) return listOf(listOf(this.first(), this.last()), listOf(this.last(), this.first()))
//recursive case
return this.flatMap { lastItem ->
this.minus(lastItem).permutations().map { it.plus(lastItem) }
}
}
Core concept: Break down long list into smaller list + recursion
Long answer with example list [1, 2, 3, 4]:
Even for a list of 4 it already kinda get's confusing trying to list all the possible permutations in your head, and what we need to do is exactly to avoid that. It is easy for us to understand how to make all permutations of list of size 0, 1, and 2, so all we need to do is break them down to any of those sizes and combine them back up correctly. Imagine a jackpot machine: this algorithm will start spinning from the right to the left, and write down
return empty/list of 1 when list size is 0 or 1
handle when list size is 2 (e.g. [3, 4]), and generate the 2 permutations ([3, 4] & [4, 3])
For each item, mark that as the last in the last, and find all the permutations for the rest of the item in the list. (e.g. put [4] on the table, and throw [1, 2, 3] into permutation again)
Now with all permutation it's children, put itself back to the end of the list (e.g.: [1, 2, 3][,4], [1, 3, 2][,4], [2, 3, 1][, 4], ...)

import java.io.IOException;
import java.util.ArrayList;
import java.util.Scanner;
public class hello {
public static void main(String[] args) throws IOException {
hello h = new hello();
h.printcomp();
}
int fact=1;
public void factrec(int a,int k){
if(a>=k)
{fact=fact*k;
k++;
factrec(a,k);
}
else
{System.out.println("The string will have "+fact+" permutations");
}
}
public void printcomp(){
String str;
int k;
Scanner in = new Scanner(System.in);
System.out.println("enter the string whose permutations has to b found");
str=in.next();
k=str.length();
factrec(k,1);
String[] arr =new String[fact];
char[] array = str.toCharArray();
while(p<fact)
printcomprec(k,array,arr);
// if incase u need array containing all the permutation use this
//for(int d=0;d<fact;d++)
//System.out.println(arr[d]);
}
int y=1;
int p = 0;
int g=1;
int z = 0;
public void printcomprec(int k,char array[],String arr[]){
for (int l = 0; l < k; l++) {
for (int b=0;b<k-1;b++){
for (int i=1; i<k-g; i++) {
char temp;
String stri = "";
temp = array[i];
array[i] = array[i + g];
array[i + g] = temp;
for (int j = 0; j < k; j++)
stri += array[j];
arr[z] = stri;
System.out.println(arr[z] + " " + p++);
z++;
}
}
char temp;
temp=array[0];
array[0]=array[y];
array[y]=temp;
if (y >= k-1)
y=y-(k-1);
else
y++;
}
if (g >= k-1)
g=1;
else
g++;
}
}

/** Returns an array list containing all
* permutations of the characters in s. */
public static ArrayList<String> permute(String s) {
ArrayList<String> perms = new ArrayList<>();
int slen = s.length();
if (slen > 0) {
// Add the first character from s to the perms array list.
perms.add(Character.toString(s.charAt(0)));
// Repeat for all additional characters in s.
for (int i = 1; i < slen; ++i) {
// Get the next character from s.
char c = s.charAt(i);
// For each of the strings currently in perms do the following:
int size = perms.size();
for (int j = 0; j < size; ++j) {
// 1. remove the string
String p = perms.remove(0);
int plen = p.length();
// 2. Add plen + 1 new strings to perms. Each new string
// consists of the removed string with the character c
// inserted into it at a unique location.
for (int k = 0; k <= plen; ++k) {
perms.add(p.substring(0, k) + c + p.substring(k));
}
}
}
}
return perms;
}

Here is a straightforward minimalist recursive solution in Java:
public static ArrayList<String> permutations(String s) {
ArrayList<String> out = new ArrayList<String>();
if (s.length() == 1) {
out.add(s);
return out;
}
char first = s.charAt(0);
String rest = s.substring(1);
for (String permutation : permutations(rest)) {
out.addAll(insertAtAllPositions(first, permutation));
}
return out;
}
public static ArrayList<String> insertAtAllPositions(char ch, String s) {
ArrayList<String> out = new ArrayList<String>();
for (int i = 0; i <= s.length(); ++i) {
String inserted = s.substring(0, i) + ch + s.substring(i);
out.add(inserted);
}
return out;
}

We can use factorial to find how many strings started with particular letter.
Example: take the input abcd. (3!) == 6 strings will start with every letter of abcd.
static public int facts(int x){
int sum = 1;
for (int i = 1; i < x; i++) {
sum *= (i+1);
}
return sum;
}
public static void permutation(String str) {
char[] str2 = str.toCharArray();
int n = str2.length;
int permutation = 0;
if (n == 1) {
System.out.println(str2[0]);
} else if (n == 2) {
System.out.println(str2[0] + "" + str2[1]);
System.out.println(str2[1] + "" + str2[0]);
} else {
for (int i = 0; i < n; i++) {
if (true) {
char[] str3 = str.toCharArray();
char temp = str3[i];
str3[i] = str3[0];
str3[0] = temp;
str2 = str3;
}
for (int j = 1, count = 0; count < facts(n-1); j++, count++) {
if (j != n-1) {
char temp1 = str2[j+1];
str2[j+1] = str2[j];
str2[j] = temp1;
} else {
char temp1 = str2[n-1];
str2[n-1] = str2[1];
str2[1] = temp1;
j = 1;
} // end of else block
permutation++;
System.out.print("permutation " + permutation + " is -> ");
for (int k = 0; k < n; k++) {
System.out.print(str2[k]);
} // end of loop k
System.out.println();
} // end of loop j
} // end of loop i
}
}

//insert each character into an arraylist
static ArrayList al = new ArrayList();
private static void findPermutation (String str){
for (int k = 0; k < str.length(); k++) {
addOneChar(str.charAt(k));
}
}
//insert one char into ArrayList
private static void addOneChar(char ch){
String lastPerStr;
String tempStr;
ArrayList locAl = new ArrayList();
for (int i = 0; i < al.size(); i ++ ){
lastPerStr = al.get(i).toString();
//System.out.println("lastPerStr: " + lastPerStr);
for (int j = 0; j <= lastPerStr.length(); j++) {
tempStr = lastPerStr.substring(0,j) + ch +
lastPerStr.substring(j, lastPerStr.length());
locAl.add(tempStr);
//System.out.println("tempStr: " + tempStr);
}
}
if(al.isEmpty()){
al.add(ch);
} else {
al.clear();
al = locAl;
}
}
private static void printArrayList(ArrayList al){
for (int i = 0; i < al.size(); i++) {
System.out.print(al.get(i) + " ");
}
}

//Rotate and create words beginning with all letter possible and push to stack 1
//Read from stack1 and for each word create words with other letters at the next location by rotation and so on
/* eg : man
1. push1 - man, anm, nma
2. pop1 - nma , push2 - nam,nma
pop1 - anm , push2 - amn,anm
pop1 - man , push2 - mna,man
*/
public class StringPermute {
static String str;
static String word;
static int top1 = -1;
static int top2 = -1;
static String[] stringArray1;
static String[] stringArray2;
static int strlength = 0;
public static void main(String[] args) throws IOException {
System.out.println("Enter String : ");
InputStreamReader isr = new InputStreamReader(System.in);
BufferedReader bfr = new BufferedReader(isr);
str = bfr.readLine();
word = str;
strlength = str.length();
int n = 1;
for (int i = 1; i <= strlength; i++) {
n = n * i;
}
stringArray1 = new String[n];
stringArray2 = new String[n];
push(word, 1);
doPermute();
display();
}
public static void push(String word, int x) {
if (x == 1)
stringArray1[++top1] = word;
else
stringArray2[++top2] = word;
}
public static String pop(int x) {
if (x == 1)
return stringArray1[top1--];
else
return stringArray2[top2--];
}
public static void doPermute() {
for (int j = strlength; j >= 2; j--)
popper(j);
}
public static void popper(int length) {
// pop from stack1 , rotate each word n times and push to stack 2
if (top1 > -1) {
while (top1 > -1) {
word = pop(1);
for (int j = 0; j < length; j++) {
rotate(length);
push(word, 2);
}
}
}
// pop from stack2 , rotate each word n times w.r.t position and push to
// stack 1
else {
while (top2 > -1) {
word = pop(2);
for (int j = 0; j < length; j++) {
rotate(length);
push(word, 1);
}
}
}
}
public static void rotate(int position) {
char[] charstring = new char[100];
for (int j = 0; j < word.length(); j++)
charstring[j] = word.charAt(j);
int startpos = strlength - position;
char temp = charstring[startpos];
for (int i = startpos; i < strlength - 1; i++) {
charstring[i] = charstring[i + 1];
}
charstring[strlength - 1] = temp;
word = new String(charstring).trim();
}
public static void display() {
int top;
if (top1 > -1) {
while (top1 > -1)
System.out.println(stringArray1[top1--]);
} else {
while (top2 > -1)
System.out.println(stringArray2[top2--]);
}
}
}

Another simple way is to loop through the string, pick the character that is not used yet and put it to a buffer, continue the loop till the buffer size equals to the string length. I like this back tracking solution better because:
Easy to understand
Easy to avoid duplication
The output is sorted
Here is the java code:
List<String> permute(String str) {
if (str == null) {
return null;
}
char[] chars = str.toCharArray();
boolean[] used = new boolean[chars.length];
List<String> res = new ArrayList<String>();
StringBuilder sb = new StringBuilder();
Arrays.sort(chars);
helper(chars, used, sb, res);
return res;
}
void helper(char[] chars, boolean[] used, StringBuilder sb, List<String> res) {
if (sb.length() == chars.length) {
res.add(sb.toString());
return;
}
for (int i = 0; i < chars.length; i++) {
// avoid duplicates
if (i > 0 && chars[i] == chars[i - 1] && !used[i - 1]) {
continue;
}
// pick the character that has not used yet
if (!used[i]) {
used[i] = true;
sb.append(chars[i]);
helper(chars, used, sb, res);
// back tracking
sb.deleteCharAt(sb.length() - 1);
used[i] = false;
}
}
}
Input str: 1231
Output list: {1123, 1132, 1213, 1231, 1312, 1321, 2113, 2131, 2311, 3112, 3121, 3211}
Noticed that the output is sorted, and there is no duplicate result.

Recursion is not necessary, even you can calculate any permutation directly, this solution uses generics to permute any array.
Here is a good information about this algorihtm.
For C# developers here is more useful implementation.
public static void main(String[] args) {
String word = "12345";
Character[] array = ArrayUtils.toObject(word.toCharArray());
long[] factorials = Permutation.getFactorials(array.length + 1);
for (long i = 0; i < factorials[array.length]; i++) {
Character[] permutation = Permutation.<Character>getPermutation(i, array, factorials);
printPermutation(permutation);
}
}
private static void printPermutation(Character[] permutation) {
for (int i = 0; i < permutation.length; i++) {
System.out.print(permutation[i]);
}
System.out.println();
}
This algorithm has O(N) time and space complexity to calculate each permutation.
public class Permutation {
public static <T> T[] getPermutation(long permutationNumber, T[] array, long[] factorials) {
int[] sequence = generateSequence(permutationNumber, array.length - 1, factorials);
T[] permutation = generatePermutation(array, sequence);
return permutation;
}
public static <T> T[] generatePermutation(T[] array, int[] sequence) {
T[] clone = array.clone();
for (int i = 0; i < clone.length - 1; i++) {
swap(clone, i, i + sequence[i]);
}
return clone;
}
private static int[] generateSequence(long permutationNumber, int size, long[] factorials) {
int[] sequence = new int[size];
for (int j = 0; j < sequence.length; j++) {
long factorial = factorials[sequence.length - j];
sequence[j] = (int) (permutationNumber / factorial);
permutationNumber = (int) (permutationNumber % factorial);
}
return sequence;
}
private static <T> void swap(T[] array, int i, int j) {
T t = array[i];
array[i] = array[j];
array[j] = t;
}
public static long[] getFactorials(int length) {
long[] factorials = new long[length];
long factor = 1;
for (int i = 0; i < length; i++) {
factor *= i <= 1 ? 1 : i;
factorials[i] = factor;
}
return factorials;
}
}

My implementation based on Mark Byers's description above:
static Set<String> permutations(String str){
if (str.isEmpty()){
return Collections.singleton(str);
}else{
Set <String> set = new HashSet<>();
for (int i=0; i<str.length(); i++)
for (String s : permutations(str.substring(0, i) + str.substring(i+1)))
set.add(str.charAt(i) + s);
return set;
}
}

Permutation of String:
public static void main(String args[]) {
permu(0,"ABCD");
}
static void permu(int fixed,String s) {
char[] chr=s.toCharArray();
if(fixed==s.length())
System.out.println(s);
for(int i=fixed;i<s.length();i++) {
char c=chr[i];
chr[i]=chr[fixed];
chr[fixed]=c;
permu(fixed+1,new String(chr));
}
}

Here is another simpler method of doing Permutation of a string.
public class Solution4 {
public static void main(String[] args) {
String a = "Protijayi";
per(a, 0);
}
static void per(String a , int start ) {
//bse case;
if(a.length() == start) {System.out.println(a);}
char[] ca = a.toCharArray();
//swap
for (int i = start; i < ca.length; i++) {
char t = ca[i];
ca[i] = ca[start];
ca[start] = t;
per(new String(ca),start+1);
}
}//per
}

A java implementation to print all the permutations of a given string considering duplicate characters and prints only unique characters is as follow:
import java.util.Set;
import java.util.HashSet;
public class PrintAllPermutations2
{
public static void main(String[] args)
{
String str = "AAC";
PrintAllPermutations2 permutation = new PrintAllPermutations2();
Set<String> uniqueStrings = new HashSet<>();
permutation.permute("", str, uniqueStrings);
}
void permute(String prefixString, String s, Set<String> set)
{
int n = s.length();
if(n == 0)
{
if(!set.contains(prefixString))
{
System.out.println(prefixString);
set.add(prefixString);
}
}
else
{
for(int i=0; i<n; i++)
{
permute(prefixString + s.charAt(i), s.substring(0,i) + s.substring(i+1,n), set);
}
}
}
}

String permutaions using Es6
Using reduce() method
const permutations = str => {
if (str.length <= 2)
return str.length === 2 ? [str, str[1] + str[0]] : [str];
return str
.split('')
.reduce(
(acc, letter, index) =>
acc.concat(permutations(str.slice(0, index) + str.slice(index + 1)).map(val => letter + val)),
[]
);
};
console.log(permutations('STR'));

In case anyone wants to generate the permutations to do something with them, instead of just printing them via a void method:
static List<int[]> permutations(int n) {
class Perm {
private final List<int[]> permutations = new ArrayList<>();
private void perm(int[] array, int step) {
if (step == 1) permutations.add(array.clone());
else for (int i = 0; i < step; i++) {
perm(array, step - 1);
int j = (step % 2 == 0) ? i : 0;
swap(array, step - 1, j);
}
}
private void swap(int[] array, int i, int j) {
int buffer = array[i];
array[i] = array[j];
array[j] = buffer;
}
}
int[] nVector = new int[n];
for (int i = 0; i < n; i++) nVector [i] = i;
Perm perm = new Perm();
perm.perm(nVector, n);
return perm.permutations;
}

How to add space(indent) before text except first line when print out. How to move this code to other class just call from main,

package com.Chall;
import java.util.ArrayList;
import java.util.List;
public class Main {
public static void main(String[] args) {
for (String part : getParts("Indent space in every line ecept first printed line", 3, 5)) {
System.out.println(part);
}
}
private static List<String> getParts(String string, int partitionSize, int maxLine) {
List<String> parts = new ArrayList<>();
int len = string.length();
int count = 0;
for (int i = 0; i < len; i += partitionSize) {
parts.add(string.substring(i, Math.min(len, i + partitionSize)));
count++;
if (maxLine == count) {
break;
}
}
return parts;
}
}

for (3,5) you get:
Ind
ent
sp
ace
in
for (7,5) you get:
Indent
space i
n every
line e
cept fi
With this code :
public static void main(String[] args) {
printList(getParts("Indent space in every line ecept first printed line", 3, 5));
}
private static void printList(List<String> parts) {
for (String part : parts) {
System.out.println(part);
}
}
private static List<String> getParts(String string, int partitionSize, int maxLine) {
List<String> parts = new ArrayList<>();
int len = string.length();
int count = 0;
String indent = ""; //<------
for (int i = 0; i < len; i += partitionSize) {
parts.add(indent + string.substring(i, Math.min(len, i + partitionSize)).trim());
count++;
indent = " "; //<-------------
if (maxLine == count) {
break;
}
}
return parts;
}
use trim() to remove space, and then add one

Alternate display of 2 strings in Java

I have a java program where the following is what I wanted to achieve:
first input: ABC
second input: xyz
output: AxByCz
and my Java program is as follows:
import java.io.*;
class DisplayStringAlternately
{
public static void main(String[] arguments)
{
String firstC[], secondC[];
firstC = new String[] {"A","B","C"};
secondC = new String[] {"x","y","z"};
displayStringAlternately(firstC, secondC);
}
public static void displayStringAlternately (String[] firstString, String[] secondString)
{
int combinedLengthOfStrings = firstString.length + secondString.length;
for(int counter = 1, i = 0; i < combinedLengthOfStrings; counter++, i++)
{
if(counter % 2 == 0)
{
System.out.print(secondString[i]);
}
else
{
System.out.print(firstString[i]);
}
}
}
}
however I encounter the following runtime error:
Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 3
AyC at DisplayStringAlternately.displayStringAlternately(DisplayStringAlternately.java:23)
at DisplayStringAlternately.main(DisplayStringAlternately.java:12)
Java Result: 1
What mistake is in my Java program?

If both arrays have same length for loop should continue while i < anyArray.length.
Also you don't need any counter to determine from which array you should print first. Just hardcode that first element will be printed from firstString and next one from secondString.
So your displayStringAlternately method can look like
public static void displayStringAlternately(String[] firstString,
String[] secondString) {
for (int i = 0; i < firstString.length; i++) {
System.out.print(firstString[i]);
System.out.print(secondString[i]);
}
}
Anyway your code throws ArrayIndexOutOfBoundsException because each time you decide from which array print element you are incrementing i, so effectively you are jumping through arrays this way
i=0 i=2
{"A","B","C"};
{"x","y","z"};
i=1 i=3
^^^-here is the problem
so as you see your code tries to access element from second array which is not inside of it (it is out of its bounds).

As you commented, If both arrays length is same, you can simply do
firstC = new String[] {"A","B","C"};
secondC = new String[] {"x","y","z"};
Then
for(int i = 0; i < firstC.length; i++) {
System.out.print(firstC[i]);
System.out.print(secondC[i]);
}

Using the combined length of the Strings is wrong, since, for example, secondString[i] would cause an exception when i >= secondString.length.

Try the below working code with high performance
public static void main(String[] arguments)
{
String firstC[], secondC[];
firstC = new String[] {"A","B","C"};
secondC = new String[] {"x","y","z"};
StringBuilder builder = new StringBuilder();
for (int i = 0; i < firstC.length; i++) {
builder.append(firstC[i]);
builder.append(secondC[i]);
}
System.out.println(builder.toString());
}

public class concad {
public void main(String[] args) {
String s1 = "RAMESH";
String s2 = "SURESH";
int i;
int j;
for (i = 0; i < s1.length(); i++) {
System.out.print(s1.charAt(i));
for (j = i; j <= i; j++) {
if (j == i) {
System.out.print(s2.charAt(j));
}
}
}
}
}

I have taken two strings as mentioned.Then pass one counter variable in inner for-loop with second string,Then for every even position pass with code "counter%2".Check this out if any concern then comment below.
public class AlternatePosition {
public static void main(String[] arguments) {
String abc = "abcd";
String def = "efgh";
displayStringAlternately(abc, def);
}
public static void displayStringAlternately(String firstString, String secondString) {
for (int i = 0; i < firstString.length(); i++) {
for (int counter = 1, j = 0; j < secondString.length(); counter++, j++) {
if (counter % 2 == 0) {
System.out.print(secondString.charAt(i));
break;
} else {
System.out.print(firstString.charAt(i));
}
}
}
}
}

How to remove duplicate character from a string in java?

In my program, the user enters a string, and it first finds the largest mode of characters in the string. Next, my program is supposed to remove all duplicates of a character in a string, (user input: aabc, program prints: abc) which I'm not entirely certain on how to do. I can get it to remove duplicates from some strings, but not all. For example, when the user puts "aabc" it will print "abc", but if the user puts "aabbhh", it will print "abbhh." Also, before I added the removeDup method to my program, it would only print the maxMode once, but after I added the removeDup method, it began to print the maxMode twice. How do I keep it from printing it twice?
Note: I cannot convert the strings to an array.
import java.util.Scanner;
public class JavaApplication3 {
static class MyStrings {
String s;
void setMyStrings(String str) {
s = str;
}
int getMode() {
int i;
int j;
int count = 0;
int maxMode = 0, maxCount = 1;
for (i = 0; i< s.length(); i++) {
maxCount = count;
count = 0;
for (j = s.length()-1; j >= 0; j--) {
if (s.charAt(j) == s.charAt(i))
count++;
if (count > maxCount){
maxCount = count;
maxMode = i;
}
}
}
System.out.println(s.charAt(maxMode)+" = largest mode");
return maxMode;
}
String removeDup() {
getMode();
int i;
int j;
String rdup = "";
for (i = 0; i< s.length(); i++) {
int count = 1;
for (j = 0; j < rdup.length(); j++) {
if (s.charAt(i) == s.charAt(j)){
count++;
}
}
if (count == 1){
rdup += s.charAt(i);
}
}
System.out.print(rdup);
System.out.println();
return rdup;
}
}
public static void main (String[] args) {
Scanner in = new Scanner(System.in);
MyStrings setS = new MyStrings();
String s;
System.out.print("Enter string:");
s = in.nextLine();
setS.setMyStrings(s);
setS.getMode();
setS.removeDup();
}
}

Try this method...should work fine!
String removeDup()
{
getMode();
int i;
int j;
String rdup = "";
for (i = 0; i< s.length(); i++) {
int count = 1;
for (j = i+1; j < s.length(); j++) {
if (s.charAt(i) == s.charAt(j)) {
count++;
}
}
if (count == 1){
rdup += s.charAt(i);
}
}
// System.out.print(rdup);
System.out.println();
return rdup;
}

Welcome to StackOverflow!
You're calling getMode() both outside and inside of removeDup(), which is why it's printing it twice.
In order to remove all duplicates, you'll have to call removeDup() over and over until all the duplicates are gone from your string. Right now you're only calling it once.
How might you do that? Think about how you're detecting duplicates, and use that as the end condition for a while loop or similar.
Happy coding!

Shouldn't this be an easier way? Also, i'm still learning.
import java.util.*;
public class First {
public static void main(String arg[])
{
Scanner sc= new Scanner(System.in);
StringBuilder s=new StringBuilder(sc.nextLine());
//String s=new String();
for(int i=0;i<s.length();i++){
String a=s.substring(i, i+1);
while(s.indexOf(a)!=s.lastIndexOf(a)){s.deleteCharAt(s.lastIndexOf(a));}
}
System.out.println(s.toString());
}
}

You can do this:
public static void main(String[] args) {
String str = new String("PINEAPPLE");
Set <Character> letters = new <Character>HashSet();
for (int i = 0; i < str.length(); i++) {
letters.add(str.charAt(i));
}
System.out.println(letters);
}

I think an optimized version which supports ASCII codes can be like this:
public static void main(String[] args) {
System.out.println(removeDups("*PqQpa abbBBaaAAzzK zUyz112235KKIIppP!!QpP^^*Www5W38".toCharArray()));
}
public static String removeDups(char []input){
long ocr1=0l,ocr2=0l,ocr3=0;
int index=0;
for(int i=0;i<input.length;i++){
int val=input[i]-(char)0;
long ocr=val<126?val<63?ocr1:ocr2:ocr3;
if((ocr& (1l<<val))==0){//not duplicate
input[index]=input[i];
index++;
}
if(val<63)
ocr1|=(1l<<val);
else if(val<126)
ocr2|=(1l<<val);
else
ocr3|=(1l<<val);
}
return new String(input,0,index);
}
please keep in mind that each of orc(s) represent a mapping of a range of ASCII characters and each java long variable can grow as big as (2^63) and since we have 128 characters in ASCII so we need three ocr(s) which basically maps the occurrences of the character to a long number.
ocr1: (char)0 to (char)62
ocr2: (char)63 to (char)125
ocr3: (char)126 to (char)128
Now if a duplicate was found the
(ocr& (1l<<val))
will be greater than zero and we skip that char and finally we can create a new string with the size of index which shows last non duplicate items index.
You can define more orc(s) and support other character-sets if you want.

Can use HashSet as well as normal for loops:
public class RemoveDupliBuffer
{
public static String checkDuplicateNoHash(String myStr)
{
if(myStr == null)
return null;
if(myStr.length() <= 1)
return myStr;
char[] myStrChar = myStr.toCharArray();
HashSet myHash = new HashSet(myStrChar.length);
myStr = "";
for(int i=0; i < myStrChar.length ; i++)
{
if(! myHash.add(myStrChar[i]))
{
}else{
myStr += myStrChar[i];
}
}
return myStr;
}
public static String checkDuplicateNo(String myStr)
{
// null check
if (myStr == null)
return null;
if (myStr.length() <= 1)
return myStr;
char[] myChar = myStr.toCharArray();
myStr = "";
int tail = 0;
int j = 0;
for (int i = 0; i < myChar.length; i++)
{
for (j = 0; j < tail; j++)
{
if (myChar[i] == myChar[j])
{
break;
}
}
if (j == tail)
{
myStr += myChar[i];
tail++;
}
}
return myStr;
}
public static void main(String[] args) {
String myStr = "This is your String";
myStr = checkDuplicateNo(myStr);
System.out.println(myStr);
}

Try this simple answer- works well for simple character string accepted as user input:
import java.util.Scanner;
public class string_duplicate_char {
String final_string = "";
public void inputString() {
//accept string input from user
Scanner user_input = new Scanner(System.in);
System.out.println("Enter a String to remove duplicate Characters : \t");
String input = user_input.next();
user_input.close();
//convert string to char array
char[] StringArray = input.toCharArray();
int StringArray_length = StringArray.length;
if (StringArray_length < 2) {
System.out.println("\nThe string with no duplicates is: "
+ StringArray[1] + "\n");
} else {
//iterate over all elements in the array
for (int i = 0; i < StringArray_length; i++) {
for (int j = i + 1; j < StringArray_length; j++) {
if (StringArray[i] == StringArray[j]) {
int temp = j;//set duplicate element index
//delete the duplicate element by copying the adjacent elements by one place
for (int k = temp; k < StringArray_length - 1; k++) {
StringArray[k] = StringArray[k + 1];
}
j++;
StringArray_length--;//reduce char array length
}
}
}
}
System.out.println("\nThe string with no duplicates is: \t");
//print the resultant string with no duplicates
for (int x = 0; x < StringArray_length; x++) {
String temp= new StringBuilder().append(StringArray[x]).toString();
final_string=final_string+temp;
}
System.out.println(final_string);
}
public static void main(String args[]) {
string_duplicate_char object = new string_duplicate_char();
object.inputString();
}
}

Another easy solution to clip the duplicate elements in a string using HashSet and ArrayList :
import java.util.ArrayList;
import java.util.HashSet;
import java.util.Iterator;
import java.util.Scanner;
public class sample_work {
public static void main(String args[]) {
String input = "";
System.out.println("Enter string to remove duplicates: \t");
Scanner in = new Scanner(System.in);
input = in.next();
in.close();
ArrayList<Character> String_array = new ArrayList<Character>();
for (char element : input.toCharArray()) {
String_array.add(element);
}
HashSet<Character> charset = new HashSet<Character>();
int array_len = String_array.size();
System.out.println("\nLength of array = " + array_len);
if (String_array != null && array_len > 0) {
Iterator<Character> itr = String_array.iterator();
while (itr.hasNext()) {
Character c = (Character) itr.next();
if (charset.add(c)) {
} else {
itr.remove();
array_len--;
}
}
}
System.out.println("\nThe new string with no duplicates: \t");
for (int i = 0; i < array_len; i++) {
System.out.println(String_array.get(i).toString());
}
}
}

your can use this simple code and understand how to remove duplicates values from string.I think this is the simplest way to understand this problem.
class RemoveDup
{
static int l;
public String dup(String str)
{
l=str.length();
System.out.println("length"+l);
char[] c=str.toCharArray();
for(int i=0;i<l;i++)
{
for(int j=0;j<l;j++)
{
if(i!=j)
{
if(c[i]==c[j])
{
l--;
for(int k=j;k<l;k++)
{
c[k]=c[k+1];
}
j--;
}
}
}
}
System.out.println("after concatination lenght:"+l);
StringBuilder sd=new StringBuilder();
for(int i=0;i<l;i++)
{
sd.append(c[i]);
}
str=sd.toString();
return str;
}
public static void main(String[] ar)
{
RemoveDup obj=new RemoveDup();
Scanner sc=new Scanner(System.in);
String st,t;
System.out.println("enter name:");
st=sc.nextLine();
sc.close();
t=obj.dup(st);
System.out.println(t);
}
}

/*
* To change this license header, choose License Headers in Project Properties.
* To change this template file, choose Tools | Templates
* and open the template in the editor.
*/
package javaapplication26;
import java.util.*;
/**
*
* #author THENNARASU
*/
public class JavaApplication26 {
public static void main(String[] args) {
int i,j,k=0,count=0,m;
char a[]=new char[10];
char b[]=new char[10];
Scanner ob=new Scanner(System.in);
String str;
str=ob.next();
a=str.toCharArray();
int c=str.length();
for(j=0;j<c;j++)
{
for(i=0;i<j;i++)
{
if(a[i]==a[j])
{
count=1;
}
}
if(count==0)
{
b[k++]=a[i];
}
count=0;
}
for(m=0;b[m]!='\0';m++)
{
System.out.println(b[m]);
}
}
}

i wrote this program. Am using 2 char arrays instead. You can define the number of duplicate chars you want to eliminate from the original string and also shows the number of occurances of each character in the string.
public String removeMultipleOcuranceOfChar(String string, int numberOfChars){
char[] word1 = string.toCharArray();
char[] word2 = string.toCharArray();
int count=0;
StringBuilder builderNoDups = new StringBuilder();
StringBuilder builderDups = new StringBuilder();
for(char x: word1){
for(char y : word2){
if (x==y){
count++;
}//end if
}//end inner loop
System.out.println(x + " occurance: " + count );
if (count ==numberOfChars){
builderNoDups.append(x);
}else{
builderDups.append(x);
}//end if else
count = 0;
}//end outer loop
return String.format("Number of identical chars to be in or out of input string: "
+ "%d\nOriginal word: %s\nWith only %d identical chars: %s\n"
+ "without %d identical chars: %s",
numberOfChars,string,numberOfChars, builderNoDups.toString(),numberOfChars,builderDups.toString());
}

Try this simple solution for REMOVING DUPLICATE CHARACTERS/LETTERS FROM GIVEN STRING
import java.util.Scanner;
public class RemoveDuplicateLetters {
public static void main(String[] args) {
Scanner scn=new Scanner(System.in);
System.out.println("enter a String:");
String s=scn.nextLine();
String ans="";
while(s.length()>0)
{
char ch = s.charAt(0);
ans+= ch;
s = s.replace(ch+"",""); //Replacing all occurrence of the current character by a spaces
}
System.out.println("after removing all duplicate letters:"+ans);
}
}

In Java 8 we can do that using
private void removeduplicatecharactersfromstring() {
String myString = "aabcd eeffff ghjkjkl";
StringBuilder builder = new StringBuilder();
Arrays.asList(myString.split(" "))
.forEach(s -> {
builder.append(Stream.of(s.split(""))
.distinct().collect(Collectors.joining()).concat(" "));
});
System.out.println(builder); // abcd ef ghjkl
}

permutations of a string using iteration

I'm trying to find permutation of a given string, but I want to use iteration. The recursive solution I found online and I do understand it, but converting it to an iterative solution is really not working out. Below I have attached my code. I would really appreciate the help:
public static void combString(String s) {
char[] a = new char[s.length()];
//String temp = "";
for(int i = 0; i < s.length(); i++) {
a[i] = s.charAt(i);
}
for(int i = 0; i < s.length(); i++) {
String temp = "" + a[i];
for(int j = 0; j < s.length();j++) {
//int k = j;
if(i != j) {
System.out.println(j);
temp += s.substring(0,j) + s.substring(j+1,s.length());
}
}
System.out.println(temp);
}
}

Following up on my related question comment, here's a Java implementation that does what you want using the Counting QuickPerm Algorithm:
public static void combString(String s) {
// Print initial string, as only the alterations will be printed later
System.out.println(s);
char[] a = s.toCharArray();
int n = a.length;
int[] p = new int[n]; // Weight index control array initially all zeros. Of course, same size of the char array.
int i = 1; //Upper bound index. i.e: if string is "abc" then index i could be at "c"
while (i < n) {
if (p[i] < i) { //if the weight index is bigger or the same it means that we have already switched between these i,j (one iteration before).
int j = ((i % 2) == 0) ? 0 : p[i];//Lower bound index. i.e: if string is "abc" then j index will always be 0.
swap(a, i, j);
// Print current
System.out.println(join(a));
p[i]++; //Adding 1 to the specific weight that relates to the char array.
i = 1; //if i was 2 (for example), after the swap we now need to swap for i=1
}
else {
p[i] = 0;//Weight index will be zero because one iteration before, it was 1 (for example) to indicate that char array a[i] swapped.
i++;//i index will have the option to go forward in the char array for "longer swaps"
}
}
}
private static String join(char[] a) {
StringBuilder builder = new StringBuilder();
builder.append(a);
return builder.toString();
}
private static void swap(char[] a, int i, int j) {
char temp = a[i];
a[i] = a[j];
a[j] = temp;
}

List<String> results = new ArrayList<String>();
String test_str = "abcd";
char[] chars = test_str.toCharArray();
results.add(new String("" + chars[0]));
for(int j=1; j<chars.length; j++) {
char c = chars[j];
int cur_size = results.size();
//create new permutations combing char 'c' with each of the existing permutations
for(int i=cur_size-1; i>=0; i--) {
String str = results.remove(i);
for(int l=0; l<=str.length(); l++) {
results.add(str.substring(0,l) + c + str.substring(l));
}
}
}
System.out.println("Number of Permutations: " + results.size());
System.out.println(results);
Example:
if we have 3 character string e.g. "abc", we can form permuations as below.
1) construct a string with first character e.g. 'a' and store that in results.
char[] chars = test_str.toCharArray();
results.add(new String("" + chars[0]));
2) Now take next character in string (i.e. 'b') and insert that in all possible positions of previously contsructed strings in results. Since we have only one string in results ("a") at this point, doing so gives us 2 new strings 'ba', 'ab'. Insert these newly constructed strings in results and remove "a".
for(int i=cur_size-1; i>=0; i--) {
String str = results.remove(i);
for(int l=0; l<=str.length(); l++) {
results.add(str.substring(0,l) + c + str.substring(l));
}
}
3) Repeat 2) for every character in the given string.
for(int j=1; j<chars.length; j++) {
char c = chars[j];
....
....
}
This gives us "cba", "bca", "bac" from "ba" and "cab", "acb" and "abc" from "ab"

Work queue allows us to create an elegant iterative solution for this problem.
static List<String> permutations(String string) {
List<String> permutations = new LinkedList<>();
Deque<WorkUnit> workQueue = new LinkedList<>();
// We need to permutate the whole string and haven't done anything yet.
workQueue.add(new WorkUnit(string, ""));
while (!workQueue.isEmpty()) { // Do we still have any work?
WorkUnit work = workQueue.poll();
// Permutate each character.
for (int i = 0; i < work.todo.length(); i++) {
String permutation = work.done + work.todo.charAt(i);
// Did we already build a complete permutation?
if (permutation.length() == string.length()) {
permutations.add(permutation);
} else {
// Otherwise what characters are left?
String stillTodo = work.todo.substring(0, i) + work.todo.substring(i + 1);
workQueue.add(new WorkUnit(stillTodo, permutation));
}
}
}
return permutations;
}
A helper class to hold partial results is very simple.
/**
* Immutable unit of work
*/
class WorkUnit {
final String todo;
final String done;
WorkUnit(String todo, String done) {
this.todo = todo;
this.done = done;
}
}
You can test the above piece of code by wrapping them in this class.
import java.util.*;
public class AllPermutations {
public static void main(String... args) {
String str = args[0];
System.out.println(permutations(str));
}
static List<String> permutations(String string) {
...
}
}
class WorkUnit {
...
}
Try it by compiling and running.
$ javac AllPermutations.java; java AllPermutations abcd
The below implementation can also be easily tweaked to return a list of permutations in reverse order by using a LIFO stack of work instead of a FIFO queue.

import java.util.List;
import java.util.Set;
import java.util.ArrayList;
import java.util.HashSet;
public class Anagrams{
public static void main(String[] args)
{
String inpString = "abcd";
Set<String> combs = getAllCombs(inpString);
for(String comb : combs)
{
System.out.println(comb);
}
}
private static Set<String> getAllCombs(String inpString)
{
Set<String> combs = new HashSet<String>();
if( inpString == null | inpString.isEmpty())
return combs;
combs.add(inpString.substring(0,1));
Set<String> tempCombs = new HashSet<String>();
for(char a : inpString.substring(1).toCharArray())
{
tempCombs.clear();
tempCombs.addAll(combs);
combs.clear();
for(String comb : tempCombs)
{
combs.addAll(getCombs(comb,a));
}
}
return combs;
}
private static Set<String> getCombs(String comb, char a) {
Set<String> combs = new HashSet<String>();
for(int i = 0 ; i <= comb.length(); i++)
{
String temp = comb.substring(0, i) + a + comb.substring(i);
combs.add(temp);
//System.out.println(temp);
}
return combs;
}
}

Just posting my approach to the problem:
import java.util.ArrayDeque;
import java.util.Queue;
public class PermutationIterative {
public static void main(String[] args) {
permutationIterative("abcd");
}
private static void permutationIterative(String str) {
Queue<String> currentQueue = null;
int charNumber = 1;
for (char c : str.toCharArray()) {
if (currentQueue == null) {
currentQueue = new ArrayDeque<>(1);
currentQueue.add(String.valueOf(c));
} else {
int currentQueueSize = currentQueue.size();
int numElements = currentQueueSize * charNumber;
Queue<String> nextQueue = new ArrayDeque<>(numElements);
for (int i = 0; i < currentQueueSize; i++) {
String tempString = currentQueue.remove();
for (int j = 0; j < charNumber; j++) {
int n = tempString.length();
nextQueue.add(tempString.substring(0, j) + c + tempString.substring(j, n));
}
}
currentQueue = nextQueue;
}
charNumber++;
}
System.out.println(currentQueue);
}
}

package vishal villa;
import java.util.Scanner;
public class Permutation {
static void result( String st, String ans)
{
if(st.length() == 0)
System.out.println(ans +" ");
for(int i = 0; i<st.length(); i++)
{
char ch = st.charAt(i);
String r = st.substring(0, i) + st.substring(i + 1);
result(r, ans + ch);
}
}
public static void main(String[] args)
{
Scanner Sc = new Scanner(System.in);
System.out.println("enter the string");
String st = Sc.nextLine();
Permutation p = new Permutation();
p.result(st,"" );
}
}

// Java program to print all permutations of a
// given string.
public class Permutation
{
public static void main(String[] args)
{
String str = "ABC";
int n = str.length();
Permutation permutation = new Permutation();
permutation.permute(str, 0, n-1);
}
/**
* permutation function
* #param str string to calculate permutation for
* #param s starting index
* #param e end index
*/
private void permute(String str, int s, int e)
{
if (s == e)
System.out.println(str);
else
{
for (int i = s; i <= s; i++)
{
str = swap(str,l,i);
permute(str, s+1, e);
str = swap(str,l,i);
}
}
}
/**
* Swap Characters at position
* #param a string value
* #param i position 1
* #param j position 2
* #return swapped string
*/
public String swap(String a, int i, int j)
{
char temp;
char[] charArray = a.toCharArray();
temp = charArray[i] ;
charArray[i] = charArray[j];
charArray[j] = temp;
return String.valueOf(charArray);
}
}