java - Generate all possible string combinations by replacing the hidden “#” number sign

Question

My task is to generates all possible combinations of that rows without the hidden # number sign. The input is XOXX#OO#XO and here is the example of what the output should be:

XOXXOOOOXO 
XOXXOOOXXO 
XOXXXOOOXO 
XOXXXOOXXO

I am only allowed to solve this solution iteratively and I am not sure how to fix this and have been working on this code for a week now.

Here is my code:

import java.lang.Math;

public class help {
    public static void main(String[] args) {
        String str = new String("XOXX#OO#XO");
        UnHide(str);
    }

    public static void UnHide(String str) {
        //converting string to char 
        char[] chArr = str.toCharArray();
        //finding all combinations for XO 
        char[] xo = new char[]{'X', 'O'};

        int count = 0;
        char perm = 0;
        String s = "";

        //finding amount of times '#' appears in string
        for (int i = 0; i < str.length(); i++) {
            if (chArr[i] == '#')
                count++;
        }

        int[] combo = new int[count];
        int pMax = xo.length;

        while (combo[0] < pMax) {
            // print the current permutation
            for (int k = 0; k < count; k++) {
                //print each character
                //System.out.print(xo[combo[i]]);
                perm = xo[combo[k]];
                s = String.valueOf(perm);

                char[] xoArr = s.toCharArray();
                String strChar = new String(xoArr);
                //substituting '#' to XO combo
                for (int i = 0; i < chArr.length; i++) {
                    for (int j = 0; j < s.length(); j++) {
                        if (chArr[i] == '#') {
                            chArr[i] = xoArr[j];
                            strChar = String.copyValueOf(chArr);
                            i++;
                        }
                    }
                    i++;
                    if (i == chArr.length - 1) {
                        System.out.println(strChar);
                        i = 0;
                    }
                }
            }

            System.out.println(); //print end of line

            // increment combo
            combo[count - 1]++; // increment the last index
            //// if increment overflows
            for (int i = count - 1; combo[i] == pMax && i > 0; i--) {
                combo[i - 1]++;  // increment previous index
                combo[i] = 0;   // set current index to zero  
            }
        }
    }
}

Answer 1

Since your input has 2 #'s, there are 2ⁿ = 4 permutations.

If you count from 0 to 3, and look at the numbers in binary, you get 00, 01, 10, and 11, so if you use that, inserting O for 0 and X for 1, you can do this using simple loops.

public static void unHide(String str) {
    int count = 0;
    for (int i = 0; i < str.length(); i++)
        if (str.charAt(i) == '#')
            count++;
    if (count > 30)
        throw new IllegalArgumentException("Too many #'s found. " + count + " > 30");
    char[] buf = str.toCharArray();
    for (int permutation = 0, end = 1 << count; permutation < end; permutation++) {
        for (int i = buf.length - 1, bit = 0; i >= 0; i--)
            if (str.charAt(i) == '#')
                buf[i] = "OX".charAt(permutation >>> bit++ & 1);
        System.out.println(buf);
    }
}

Test

unHide("XOXX#OO#XO");

Output

XOXXOOOOXO
XOXXOOOXXO
XOXXXOOOXO
XOXXXOOXXO