Return how many unique digitis a number contains

Question

I have a string of digits and I want to count how many unique digits the string contains.

Example:

111222
1002345
000000

Expected output:

111222 2
1002345 6
000000 1

I have achieved this using the following code:

private static int Counter(string ID)
{
    char[] numbers = new char[]{'0','1','2','3','4','5','6','7','8','9'};
    List<int> listofmatched = new List<int>();
    var split = ID.ToArray();
    foreach (var num in split)
    {
        if (numbers.Contains(num))
        {
            if (listofmatched.Contains(num))
            {
                continue;
            }
            else
            {
                listofmatched.Add(num);
            }
        }
    }
    return listofmatched.Count;
}

Is there any way to improve the code above? I feel like there's unnecessary loops.

Answer 1

Don't know if it fits your defintion of "improvement", but you could do it like this:

str.Where(x => char.IsDigit(x)).GroupBy(x => x).Count();

See it here:

https://dotnetfiddle.net/t5OW6T

Edit:

As noticed in the comments, you could also use Distinct isntead of GroupBy(x => x).

GroupBy(x => x) can be useful if you want to know which digit occurs how often in the string, this can be done like this:

str.Where(x => char.IsDigit(x)).GroupBy(x => x).Select(x => (x.Key, x.Count()))

Where instead of calling Count on the "whole" grouping result, you evaluate the Count for each individual group of the grouping result.

You may find an example here:

https://dotnetfiddle.net/D60ygZ