Given a number n, list all n-digit numbers such that each number does not have repeating digits

Maribell Pina Griego :

I'm trying to solve the following problem. Given an integer, n, list all n-digits numbers such that each number does not have repeating digits.

For example, if n is 4, then the output is as follows:

0123
0124
0125
...
9875
9876
Total number of 4-digit numbers is 5040

My present approach is by brute-force. I can generate all n-digit numbers, then, using a Set, list all numbers with no repeating digits. However, I'm pretty sure there is a faster, better and more elegant way of doing this.

I'm programming in Java, but I can read source code in C.

Thanks

John Humphreys - w00te :

Mathematically, you have 10 options for the first number, 9 for the second, 8 for the 3rd, and 7 for the 4th. So, 10 * 9 * 8 * 7 = 5040.

Programmatically, you can generate these with some combinations logic. Using a functional approach usually keeps code cleaner; meaning build up a new string recursively as opposed to trying to use a StringBuilder or array to keep modifying your existing string.

Example Code

The following code will generate the permutations, without reusing digits, without any extra set or map/etc.

public class LockerNumberNoRepeats {
    public static void main(String[] args) {
        System.out.println("Total combinations = " + permutations(4));
    }

    public static int permutations(int targetLength) {
        return permutations("", "0123456789", targetLength);
    }

    private static int permutations(String c, String r, int targetLength) {
        if (c.length() == targetLength) {
            System.out.println(c);
            return 1;
        }

        int sum = 0;
        for (int i = 0; i < r.length(); ++i) {
            sum += permutations(c + r.charAt(i), r.substring(0,i) + r.substring(i + 1), targetLength);
        }
        return sum;
    }
}

Output:

...
9875
9876
Total combinations = 5040

Explanation

Pulling this from a comment by @Rick as it was very well said and helps to clarify the solution.

So to explain what is happening here - it's recursing a function which takes three parameters: a list of digits we've already used (the string we're building - c), a list of digits we haven't used yet (the string r) and the target depth or length. Then when a digit is used, it is added to c and removed from r for subsequent recursive calls, so you don't need to check if it is already used, because you only pass in those which haven't already been used.