Finding and printing the longest unbroken alphabetical string hidden in another string!

[Varuna Seneviratna]

 String s = "snabdgaabcdefahsbdgjrj";
    	   //String s = "abcdefghijklmnopqrstuvwxyz";
    	    String subStr = s.substring(0,1);
    	     //s = " ";
    	     //s = "";
    	    
    	    //System.out.println(subStr.charAt(subStr.length()-1));
    	    
    	    //if(subStr.charAt(subStr.length()-1) > s.charAt(1)) {
    	    	//System.out.println(true);
    	    //}
    	    
    	    
    	    
    	    String finalStr = "";
    	    
    	    //for(int i=1; i<s.length(); i++) {
>
    	    	//if(subStr.charAt(subStr.length()-1) < s.charAt())) {
    	    		
    	    	//}
    	   // if(s.charAt(7) < s.charAt(8))
    	    	//System.out.println(s.charAt(7));
    	  if(s.isBlank() || s.isEmpty())
    		  System.out.println("String Empty");
    	  else {
    	    for(int i=1; i < s.length(); i++) {
    	    	//System.out.println(s.charAt(i));
    	    	if(subStr.charAt(subStr.length()-1) < s.charAt(i)){
    	    		//System.out.println(s.charAt(i));
    	    		subStr = subStr + s.charAt(i);
    	    		//System.out.println(subStr);
    	    		if(subStr.length() > finalStr.length()) {
    	    			finalStr = "";
    	    			 finalStr = subStr;
    	    			 
    	    		}
    	    		        
    	    		System.out.println(finalStr);
    	    	} else {
    	    		subStr = "";
    	    		subStr = subStr + s.charAt(i);
    	    	}
    	    }
    	  }
             
       

Can the code above be improved? What it does is, traverse through an unordered string, find the longest hidden alphabetically ordered string and print it.

[Norm Radder]

Line 34 is redundant.
Line 41 is redundant.
Use the String class's valueOf method on line 42 to convert a char to a String.

[Campbell Ritchie]

Please explain what you are looking for. I can see, “abcdef,” which I think is an “unbroken alphabetical” substring, and I can also see, “bd'” which I think probably isn't “unbroken alphabetical”.
Please only post the code you are actually running; there is no point in posting commented‑out code. That makes the code very difficult to read.
Please format and indent the code correctly.
After an isBlank() call, there is no need for an isEmpty() call (line 24).
Line 29 is confusing. It looks as though you had stumbled upon a very inefficient way to iterate the String.

[Varuna Seneviratna]

@NormRadder
                       Line 34 is finalStr = "";
                       Line 41 is subStr = "";

        Why do you say they are redundant?

       

Use the String class's valueOf method on line 42 to convert a char to a String

                          Line 42 is

 subStr = subStr + s.charAt(i);

                                            The character in s.charAt(i) gets concataneted to the subStr. So why do you say to use the valueOf() of the String class to convert char to String?

@CampbellRitchie
                                 

After an isBlank() call, there is no need for an isEmpty() call (line 24).

                                               Are they are the same?
                             

 Line 29 is confusing.

                                            Line 29 is

 if(subStr.charAt(subStr.length()-1) < s.charAt(i))

                                                      What line 29 does is compare the last character in the subStr for alphabetical order with a character in the string that's currently going through the loop.

[Carey Brown]

Let's say we were dealing with int variables 'i' and 'j', if you wrote this

i = 0;
i = j;

What would be the point of having line 1 if you are just going to throw the value away and replace it with the contents of 'j' on the very next line?

You have the same issue but with String variables.

[Carey Brown]

See Campbell's comments about cleaning up your code.

[Varuna Seneviratna]

@CareyBrown
                        Line 2 is a test to see how the program works if the whole string is alphabetical

[Carey Brown]

I don't understand why you are going through all those painful steps with sub-strings. You could more easily use two indexes into the input String: begin and end. Increment these as needed and use charAt(idx) to compare the characters. The longest distance between begin and end would be your answer.

[Carey Brown]

Varuna Seneviratna wrote:@CareyBrown
                        Line 2 is a test to see how the program works if the whole string is alphabetical

Huh?
Line 2 is
i = j;

[Varuna Seneviratna]

Carey Brown

Huh?
Line 2 is
i = j;

Line 2 was

String s = "abcdefghijklmnopqrstuvwxyz";

        This is just a test I ran to check if the program works correctly if the whole string is alphabetical

[Carey Brown]

Carey Brown wrote:Let's say we were dealing with int variables 'i' and 'j', if you wrote this

i = 0;
i = j;

What would be the point of having line 1 if you are just going to throw the value away and replace it with the contents of 'j' on the very next line?

You have the same issue but with String variables.

[Varuna Seneviratna]

Carey Brown wrote:I don't understand why you are going through all those painful steps with sub-strings. You could more easily use two indexes into the input String: begin and end. Increment these as needed and use charAt(idx) to compare the characters. The longest distance between begin and end would be your answer.

Can you please explain what you mean in

You could more easily use two indexes into the input String: begin and end. Increment these as needed and use charAt(idx) to compare the characters.

How to use two indexes into the input String?

[Carey Brown]

begin = 0;
end = 0;
LOOP THROUGH STRING
    if( s.charAt(end) + 1 == s.charAt(end+1)  ) // IF next char one more than this char
        INCREMENT END
    else
        MOVE BEGIN TO WHERE END WAS

This is just to show you how to use indexes and charAt().

[Campbell Ritchie]

Varuna Seneviratna wrote:. . . Are they are the same?

Look at their documentation.

. . . What line 29 does is compare the last character in the subStr for alphabetical order with a character in the string that's currently going through the loop.

That doesn't add to the explanation, I am afraid.

[Varuna Seneviratna]

Campbell Ritchie wrote:Please don't edit old posts; that makes the thread difficult to follow and makes the line numbers appear wrong. I am refusing the first edit.

Please post the cleaned up code as a new post.

I was responding to your reply. This is your reply I was referring to.

Campbell Ritchie wrote:Please explain what you are looking for. I can see, “abcdef,” which I think is an “unbroken alphabetical” substring, and I can also see, “bd'” which I think probably isn't “unbroken alphabetical”.
Please only post the code you are actually running; there is no point in posting commented‑out code. That makes the code very difficult to read.
Please format and indent the code correctly.
After an isBlank() call, there is no need for an isEmpty() call (line 24).
Line 29 is confusing. It looks as though you had stumbled upon a very inefficient way to iterate the String.

[Varuna Seneviratna]

Campbell Ritchie wrote:

Varuna Seneviratna wrote:. . . Are they are the same?

Look at their documentation.

. . . What line 29 does is compare the last character in the subStr for alphabetical order with a character in the string that's currently going through the loop.

That doesn't add to the explanation, I am afraid.

This is line 29

 if(subStr.charAt(subStr.length()-1) < s.charAt(i))

What line 29 does is check for alphabetical order the last character in the subStr against a character in the string.  The character of the string which is being compared is the character currently going through the loop to be checked for alphabetical order.

The Documentation for String.isEmpty says Returns true if and only if the length() is 0 and for String.isBlank() says Returns true if the string is empty or contains only whitespace codepoints,otherwise false.  Thanks, Campbell

At the time I didn't know whether for whitespace the length() is 0 or not.  The length() is not 0 for whitespace, isn't it? and isEmpty() returns false for whitespace. I was not sure how both methods work at the time, that's why I included them both

Maybe the description I gave is confusing what about the following

[Campbell Ritchie]

Varuna Seneviratna wrote:. . . I was responding to your reply.

Sorry if I wasn't clear.