How to count the number of letters that exist without repeated

I want a way to count the letters in an string for example:

My string : "Hello my friends "

The characters in the string : {H,e,l,o, ,m,y,f,r,i,n,d,s}

These letters exist without repeating them (with a blank space)
So the result I want is: 13

The goal of all of this is, I want to convert a string to a table of character without repeating the letters

EX: MY String = "Hello"

The table I want to get {H,e,l,o}

My attempt

 public static int numberRep(String txt) {
    int count = 0;
        boolean v = false;
        for (int i = 0; i != txt.length(); i++) {
            char c = txt.charAt(i);
            for (int j = i + 1; j != txt.length(); j++) {
                if (txt.charAt(j) == c) {
                    count++;
                }
            }
            if(count == txt.length()-1){
                v = true;
            }
        }
        if(v){
            return 1 ;
        }
        else{
        return count;
        }
      
    }

Split the string into characters and store them into a Set. A Set keeps only unique elements. The elements of the Set will be the required characters and the size of the Set will give you the count of these characters.

Do it as follows:

import java.util.Arrays;
import java.util.LinkedHashSet;
import java.util.Set;
import java.util.stream.Collectors;

public class Testing {
	public static void main(String[] args) {
		String myString = "Hello my friends ";
		Set<String> set = Arrays.stream(myString.split("")).collect(Collectors.toCollection(LinkedHashSet::new));
		System.out.println("Unique characters including space: " + set);
		System.out.println("No. of unique characters including space: " + set.size());
	}
}

Output:

Unique characters including space: [H, e, l, o,  , m, y, f, r, i, n, d, s]
No. of unique characters including space: 13

Well, this could be one of the approach to do so:

String str = "Hello my friends "
String noRepeatStr = "";

for(int i = 0; i < str.length; i++) {
    if(noRepeatStr.indexOf(str[i]) == -1) // check if a char already exist, if not exist then return -1
        noRepeatStr = noRepeatStr+str[i];      // add new char
}

System.out.println("Letters: " + noRepeatStr.toCharArray())
System.out.println("No of letters: " + noRepeatStr.length)

What you could do is:

1. Create an empty list a
2. for character in list:
3. If the character is not your list a: add it to your list

This way you won't get any duplicate characters.

You can also use a set like recommended in Arvind's answer, but I think it is better for you to write a function as a task.
Set is the cleaner way to do it once you grasped the concepts of Java.

 public static void main(String[] args) {
        final String test = "Hello my friends";

        final int size = IntStream.range(0, test.length())
                .mapToObj(test::charAt)
                .collect(Collectors.toSet()).size();

        System.out.println(size);
    }

What I've done here is iterated over all of the characters in the input string, mapped them to a char object and then collected them into a Set - instead of keeping the raw Set I've used .size() in order to get the output that you were expecting.