Check if String contains part of other string Java

In Java 8 you can proceed as below :

    String s1 = "T,V,E";
    String s2 = "T,W,E";

    List<String> l1 = Arrays.asList(s1.split(","));
    List<String> l2 = Arrays.asList(s2.split(","));

    List<String> result = l1.stream().filter(l2::contains).collect(Collectors.toList());

    System.out.println(String.join(",", result));

The result is "T,E" as expected.

You can achieve this in many ways. One of the ways is using Set.

First, split m1 the characters by (comma) and add it to HashSet. Then split the m2 characters and add it to ArrayList. Now by the for loop try to add the ArrayList characters to HashSet. You will get false from the add method (Set.add()) if it is not added (because the character is already there). If you get false print the character or add it to another ArrayList.

        String m3 = "T, V, E";
        String m2 = "T, W, E";

        Set<String> set = new HashSet<>(Arrays.asList(m3.split(",")));
        List<String> list = Arrays.asList(m2.split(","));
        for (String s : list) {
            if (!set.add(s)) {
                System.out.println(s);
            }
        }

Result will be T and E

The appropriate data structure is Set.

    Set<String> m3 = new TreeSet<>();
    Collections.addAll(m3, "T", "V", "E");
    Collections.addAll(m3, "T, V, E".split(",\\s*")); // Alternatively.

    Set<String> m2 = new HashSet<>();
    Collections.addAll(m2, "T", "W", "E");

    Set<String> m5 = new TreeSet<>(m2);
    m5.retainAll(m3);

Java 9:

    Set<String> m3 = Set.of("T", "V", "E");
    Set<String> m2 = Set.of("T", "W", "E");

you can use the split() function as the following

String a="A, B, C, D";
        String b="B, C, D";
        String[] a_chars =a.split(", "); //returns array with A B C D
        String[] b_chars =b.split(", "); //returns array with B C D

this whay you have 2 arrays of strings now you can compare them using 2 (for) loops

String result="";
        for(int c=0;c<a_chars.length;c++)
        {
            for(int d=0;d<b_chars.length;d++)
            {
                if(a_chars[c].equals(b_chars[d]))
                {
                    result+=a_chars[c]+", ";
                }
            }
        }

now you have the result string like this result="B, C, D, "

check of the length of result is greater than zero if so erase the lase 2 characters which are ,

if(result.length()>0)result=result.substring(0,result.length()-2);

if the length of the result string is zero that means there is no matching letters so no need to modify it

If all your String follows this pattern : ..., ..., .... , you could split them and filter only each String that is contained in the two arrays.
You can then collect them into a List and join them with , :

List<String> commonStrings = Arrays.stream(m2.split(",\\s"))
      .flatMap(s-> Arrays.stream(m3.split(",\\s"))
      .filter(s2.equals(s)))          
      .collect(Collectors.toList());

String joinedString = String.join(",", commonStrings);

Note that this code doesn't return the exact number of equals occurrences in the two Strings.
So if one String contains two A and the other one A, you will get two A in the result.
If it doesn' matter and you want to get only distinct String, you can invoke distinct() before collect().

Otherwise, to return the exact number of equals occurrences, during the processing, as soon as a String part is consumed (A for example) as the two parts are equal in the two Strings, you could create new Strings from the actual Strings but by removing this String part .

String s1 = "T,V,E"; String s2 = "T,W,E";

List<String> l1 = Arrays.asList(s1.split(","));
List<String> l2 = Arrays.asList(s2.split(","));

List<String> result = l1.stream().filter(l2::contains).collect(Collectors.toList());

System.out.println(String.join(",", result));

result = T&E This is good answer I will tell you too this