Thread: permutations of a string using iteration

Could you please your code in [key] here enter your code [/key] in order to be more clear to everybody?

in order to work write code instead of key

 
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);
}
}

Actually can you provide a small example.For example the word "Sam" what permutations does it have?I remember that the number of them is 2³,but i am not sure i know what the goal is.Sorry for my noobness :/

if u have a string "abc" then the output should be: abc, acb,bac,bca,cba,cab

Ok,so i thought that if we let the letters of the words to be different among them,then we have to output n! permutations,where n is the number of the word's letters.So i also saw that given a string "abcd" with n=4 then it's permutations are the following

abcd
abdc
acbd
acdb
adbc
adcb
bacd
badc
bcad
bcda
bdac
bdca
cabd
cadb
cbad
cbda
cdab
cdba
dabc
dacb
dbac
dbca
dcab
dcba

So i noticed that at 1st postition of the permutation every letter should appear (n-1)! times,at the 2nd position every letter(without the first included) should appear (n-2)! -in the set where the 1st's position letter remains the same-,at the 3rd position every letter(without the first two) should appear (n-3)! and at the 4rth position every letter(without the first three) should appear (n-4)!=(4-4)!=0!=1 times .So in general at m-th position every letter(without previous appearance of it) should appear (n-m)! times.
.
So this concept led me to a recursion code.But then i remember you wanted an iterative one.So i came around this code

public static void combString(String s) {
        int numberOfPermutations = 0;
        for (int i = 0; i < s.length(); i++) {
            for (int j = 0; j < s.length(); j++) {
                for (int k = 0; k < s.length(); k++) {
                    for (int z = 0; z < s.length(); z++) {
                        if (i != j && j != k && i != k && i != z && j != z && k != z) {
                            System.out.format("%c%c%c%c%n", s.charAt(i), s.charAt(j),
                                    s.charAt(k), s.charAt(z));
                            numberOfPermutations++;
                        }
                    }
                }
            }
        }
        System.out.println(numberOfPermutations);
    }

The problem is that depending the number of letters of the word ,this code has to use the same number of for-loops(so the number of the letters is going to be equal with the number of the variables(with the counter for permutations not included.).This code is for a word with four letters and lets letters to appear as many times they found in the word,iterates of letters are enabled in other words.

The logic of the code is similar as you would do it in a piece of paper .The important thing is the if statement,which has this form because we do not want to permute a letter more than one time in every permutation.Try to remove this statement or some parts of it in order to get yourself convinced how this is need.

Originally Posted by jps

recursion comes to mind

And that's better than the iterative one i provided.If you consider the complexity of what i wrote ,you will be dissapointed.

However he stated that he wants no recursive method