java find the longest non-repeating substring- getting weird result

Question

I am doing one classic leet code problem: finding the longest non-repeating substring in a string. Though there are many similar problems on stack overflow. I scratched my head for hours couldn't know why my code produce this weird result. hoping that someone could tell me why my

I am doing it in Java

    public static void main( String[] args )
    {
        String s= "wwhoamiUrektxineabcdefghijklmno";
        longestNonRepeatStr(s);     
    }

    public static void longestNonRepeatStr(String tstring) {
        String str="";
        String compare="";
        List<String> list= new ArrayList<String>();
        int biggest =0;

        //find the nonrepeating string in each loop and add to list str
       for (int i=0; i<tstring.length(); i++) { 
        for (int j=i; j<tstring.length()-1; j++) {
            str+= tstring.charAt(j);
            compare= Character.toString(tstring.charAt(j+1));
            if (str.contains(compare)){
                list.add(str);
                str="";
                break;
            }
        }
    }

    //find the longest nonrepeating string in the list
       for (int i=0; i<list.size(); i++) {
        if (list.get(biggest).length()< list.get(i).length()) {
        biggest=i;
        }
    }

       System.out.println(list);    
       System.out.println(list.get(biggest));           
    }

For input string

"wwhoamiUrektxineabcdefghijklmno"

output is

"abcdefghijklmnb" 

but it is wrong, the last letter should be "o"

Answer 1

You have several issues (see comments where I corrected the code):

public static void longestNonRepeatStr(String tstring) {
  String str="";
  String compare="";
  List<String> list= new ArrayList<String>();
  int biggest =0;

  for (int i=0; i<tstring.length(); i++) { 
    str = ""; // you must clear the current String before each iteration of the inner loop
    for (int j=i; j<tstring.length(); j++) { // here you were skipping the last character
      str+= tstring.charAt(j);
      // I improved the following condition
      if (j+1 < tstring.length() && str.contains(Character.toString(tstring.charAt(j+1)))){
        list.add(str);
        str="";
        break;
      }
    }
    if (str.length() > 0) { // if you finish the inner loop without breaking, you should
                            // add the current String to the List
      list.add(str);
    }
  }

  for (int i=0; i<list.size(); i++) {
    if (list.get(biggest).length()< list.get(i).length()) {
      biggest=i;
    }
  }

  System.out.println(list);    
  System.out.println(list.get(biggest));           
}

Or, as an alternative, you can add the current String to the List in the last iteration of the inner loop:

public static void longestNonRepeatStr(String tstring) {
  String str="";
  String compare="";
  List<String> list= new ArrayList<String>();
  int biggest =0;

  for (int i=0; i<tstring.length(); i++) { 
    str = "";
    for (int j=i; j<tstring.length(); j++) {
      str+= tstring.charAt(j);
      if (j+1 >= tstring.length() || str.contains(Character.toString(tstring.charAt(j+1)))){
        list.add(str);
        str="";
        break;
      }
    }
  }

  for (int i=0; i<list.size(); i++) {
    if (list.get(biggest).length()< list.get(i).length()) {
      biggest=i;
    }
  }

  System.out.println(list);    
  System.out.println(list.get(biggest));           
}

Further explanation of how you got the output "abcdefghijklmnb":

When i==16, your inner loop builds the String "abcdefghijklmn". Then it skips the "o", since you ended that loop prematurely (due to j<tstring.length()-1). This String is not yet added to the List, since you haven't detected a repeating character. Now when i==17, you append "b" to str and get "abcdefghijklmnb". Now you check if the next character "c" already appears in str, which is true, so you add "abcdefghijklmnb" to your List.