Question

The digital hardness of an integer is measured by examining its representation in binary (base 2). Working from the ends of the representation inward, we remove/count corresponding pairs of leading and trailing 1s until we can no longer do so. The total number of 1s (bits) removed is the original number's digital hardness value. For example, suppose that we have an integer whose binary representation is 110101010101 (the binary representation always begins with a leading 1). We remove the leading and trailing 1s to get 1010101010. This value does not begin and end with 1, so we stop. We have only successfully removed one pair of 1s (in the first step), so the digital hardness measure is 2. Complete the hardness() method below, which takes a single String argument containing only 1s and Os (e.g., "111001101010"). The method returns the digital hardness of this argument, using the calculation process described above. You do not need to modify the original argument; just use two variables that track the current "first and "last"index values within your String. public static int hardness (String number)

Answer 1

public class MeasureHardness {

   public static int hardness(String number) {
       int i = 0, j = number.length() - 1;
       int hardness = 0;
       while (i < j) {

           if (number.charAt(i) != number.charAt(j))
               return hardness;
           else if (number.charAt(i) == '1' && number.charAt(j) == '1')
               hardness = hardness + 2;

           ++i;
           --j;

       }
       return hardness;

   }

   public static void main(String[] args) {
       // TODO Auto-generated method stub

       System.out.println("Hardness is : " + hardness("11111000000111111"));

       System.out.println("Hardness is : " + hardness("111001101010"));

       System.out.println("Hardness is : " + hardness("110101010101"));

   }

}

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See Output


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