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

Provided statement:

public String starOut(String str) {

//code goes here

}

// class definition Solution

public class Solution {

   // main method to begin program execution

   public static void main(String args[]) {

       // the starOut method is called with argument

       starOut("ab*cd");

       starOut("ab*cd");

       starOut("sm*eilly");

   }

   // the starOut method is declared and it returns a string

   // it receive a string as the argument

   public static String starOut(String str) {

       // The index of the star is assigned to starIndex

       int starIndex = str.indexOf("*");

       // Index of character before star is known

       int charBeforeStar = starIndex - 1;

       // Index of character after star is known

       int charAfterStar = starIndex + 1;

       // The newString is declared empty

       String newString = "";

       // loop through the string

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

           // if i is either of the following index, the loop continue

          // without concatenating with the newString

           if (i == charBeforeStar || i == charAfterStar || i == starIndex){

               continue;

           }

           // the new string is concatenated with a character

           newString += str.charAt(i);

       }

   // the new string is printed

       System.out.println(newString);

       // the newString is returned

       return newString;

   }

}

Explanation:

The code logic is stated in the comment inside the code.

answer; i believe its compare (b3: b4) ;

i don't think so. although it would be nice. i am pretty sure that was an option an update or two ago. i guess they want you to stay 14 forever. haha.