Question

I am exceptionally new to programming, but I am working on improving my skills as a programmer. Recently, I gave myself the challenge to determine what multiples of a given number are made up of distinct digits. I have gotten most of it to work, but I still need to make the code apply for every number that is a multiple of the input one. The code I have working so far is as follows:

Integer numberA = 432143;
Integer numberB = numberA;
Integer[] digitArray = new Integer[numberA.toString().length()];
int index;
for (index = 0; index < digitArray.length; index++) {
    digitArray[index] = (numberA % 10);
numberA /= 10;
}
int repeats = 0;
for (int i = 0; i < digitArray.length; i++) {
    for (int j = 0; j < digitArray.length; j++) {
    if ((i != j) && (digitArray[i]==digitArray[j])) repeats = repeats + 1;
    }
}
if (repeats == 0) {
  System.out.println(numberB);
}

This will determine if the number is made up of distinct digits, and, if it is, print it out. I have spent quite a bit of time trying to make the rest of the code work, and this is what I’ve come up with:

Integer number = 1953824;
Integer numberA = number;
Integer numberB = numberA;
for (Integer numberC = number; numberC.toString().length() < 11; 
            numberC = numberC + number) {
    Integer[] digitArray = new Integer[numberA.toString().length()];
    int index;
    for (index = 0; index < digitArray.length; index++) {
        digitArray[index] = (numberA % 10);
        numberA /= 10;
    }
    int repeats = 0;
    for (int i = 0; i < digitArray.length; i++) {
        for (int j = 0; j < digitArray.length; j++) {
            if ((i != j) && (digitArray[i]==digitArray[j])) repeats = repeats + 1;
        }
    }
    if (repeats == 0) {
        System.out.println(numberB); 
    }
}

I can’t figure out why, but his just prints whatever the number is a bunch of times if it is made up of distinct digits, and leaves it blank if it is not. If anyone could tell me why this is occurring, or even tell me what I need to do to fix it, that would be superb. Remember, I am very new to programming, so please give a short explanation for any terms you use that are at all out of the ordinary. I am eager to learn, but I currently know very little. Thank you for your time, and I greatly appreciate any and all help you can give me.

Answer 1

You assign the value of numberA to numberB (which is the value of number) right before the for loop. After that, numberB is never modified or assigned to a new value, so for every pass through the for loop, you’re simply printing the value of numberB, which is always 1953824 in this case.

There are several corrections that can be made to achieve the result you desire, while cleaning up the code a little. The first thing is to change the print statement to print the correct number:

System.out.println(numberC);

Since numberC is the variable that is being updated by the for loop, that’s what you’ll want to conditionally print out if there are no repeat digits. Since we’ve replaced numberB with numberC, that means numberB is not longer needed, you can delete the declaration for it.

Now, the next issue is when you’re defining the digital array – you should use the length of numberC, not numberA. Also, inside the for loop, you should assign numberA the value of numberC, or else eventually nothing but 0s will be stored in your digitArray. Overall, here’s what it should look like.

    Integer number = 1953824;
    Integer numberA = number;

    for (Integer numberC = number; numberC.toString().length() < 11; 
                    numberC = numberC + number) {

        Integer[] digitArray = new Integer[numberC.toString().length()];
        numberA = numberC;
        int index;
        for (index = 0; index < digitArray.length; index++) {
            digitArray[index] = (numberA % 10);
            numberA /= 10;
        }
        int repeats = 0;
        for (int i = 0; i < digitArray.length; i++) {
            for (int j = 0; j < digitArray.length; j++) {
                if ((i != j) && (digitArray[i] == digitArray[j]))
                    repeats = repeats + 1;
            }
        }
        if (repeats == 0) {
            System.out.println(numberC);
        }
    }

This should produce the desired result. It seems to work on my machine 🙂

If you want, you can take Jeffrey’s suggestion and change the Integer to the primitive int to avoid the overhead of boxing. However, you still need to use the Integer class to use the toString() method, but you can accomplish that using Integer.valueOf():

Integer.valueOf(numberC).toString()