Return a version of the given string, where for every star (*) in the string the star and the chars immediately to its left and right are gone. So "ab*cd" yields "ad" and "ab**cd" also yields "ad". starOut("ab*cd") → "ad" starOut("ab**cd") → "ad" starOut("sm*eilly") → "silly"

Explanation:

The following code is written in Java and uses a for loop with a series of IF ELSE statements to check the next and previous characters in a string. Checking to make sure that there are no asterix. If so it adds that character to the String variable output. Which is returned to the user at the end of the method. Two test cases have been created and the output can be seen in the attached image below.

class {

   public static void main(String[] args) {

       System.out.println(starOut("sm*eilly"));

       System.out.println(starOut("ab**cd"));

   }

   public static String starOut(String str) {

       String output = "";

       for (int i = 0; i < str.length(); i++) {

           if ((i != 0) && (i != str.length()-1)) {

               if ((str.charAt(i-1) != '*') && (str.charAt(i+1) != '*') && (str.charAt(i) != '*')) {

                   output += str.charAt(i);

               }

           } else {

               if ((i == 0) && (str.charAt(i) != '*') && (str.charAt(i+1) != '*')) {

                   output += str.charAt(i);

               } else if ((i == str.length()-1) && (str.charAt(i) != '*') && (str.charAt(i-1) != '*')) {

                   output += str.charAt(i);

               }

           }

       }

       return output;

   }

}

the answer is a

you welcome

answer6-9 screws;