Permutations in java

Question

given a problem , a = (1,2,3) ,b= (A,B,C) ,c =(a1,b1,c1)

I want to get an output which prints me the value 1Aa1,1Ab1,1Ac1,1Ba1,1Bb1,1Bc1,1Ca1,1Cb1,1Cc1 Similarly with 2, with 3 then should start with A,B,C and then a1,b1,c1 . The given output should print all the permutations and combinations variables of the same .

I've already tried printing the same in for loop, it could only generate the combinations , Tried a few tutorials on printing the permutations

Can Someone please help me on that .

Check [this](http://stackoverflow.com/questions/2799078/permutation-algorithm-without-recursion-java?lq=1) — TheLostMind, Dec 30 '15 at 05:04
is this what u want : String a[] = {"1","2","3"}; String b[] = {"A","B","C"}; String c[] = {"a1","b1","c1"}; for(int i=0;i<3;i++) { for(int j=0;j<3;j++) { for(int k=0;k<3;k++) { System.out.println(a[i]+b[j]+c[k]); } } } — Arpan Paliwal, Dec 30 '15 at 06:32

Sumit Trehan · Answer 1 · 2015-12-30T06:54:10.830

Please try to post some code you have tried on your own.

For your reference I am giving a sample code and some explanation below:

The code uses recursion and is not optimized. It is just to give you an idea of how you could work with recursive problems. The function uses 4 args.

Arg1 : is the String generated so far.

Arg2 : is the array of Strings from which we can select a candidate to fill the next position.

Arg3 : represents the positions filled up till now.

Arg4 : is the length of permutation string, which will help us to determine we have reached the base case and we will just print the permutation and will not recurse further.

Basic Thought Process: The basic idea is that we will extract a character from the given arrays for every position and would form new String which would exclude that selected member and would pass this newly created set to further recurse.

class Test
{
    public static void main (String[] args) throws java.lang.Exception
    {
        String s[]= {"123","ABC","abc"};
        int len = 3; /* Length of the final permutation */
        permute ("", s, 0, len);
    }


    public static void permute (String soFar, String rest[], int pos, int len)
    {

        if (pos == len) {
            System.out.println (soFar);
        } else {

            for (int i = 0; i < rest[pos].length(); i++) {
                String remaining = rest[pos].substring(0, i) + rest[pos].substring(i+1);
                String newRest[] = new String[len];

                for (int j = 0; j < len; j++) {
                    newRest[j] = new String(rest[j]);
                }
                newRest[pos] = new String(remaining);

                permute (soFar + rest[pos].charAt(i), newRest, pos + 1, len);
            }
        }
    }
}

Reference taken from: https://see.stanford.edu/materials/icspacs106b/h19-recbacktrackexamples.pdf

Output:

1Aa 1Ab 1Ac 1Ba 1Bb 1Bc 1Ca 1Cb 1Cc 2Aa 2Ab 2Ac 2Ba 2Bb 2Bc 2Ca 2Cb 2Cc 3Aa 3Ab 3Ac 3Ba 3Bb 3Bc 3Ca 3Cb 3Cc

Permutations in java

1 Answers1