Question

Write a class called StringCode with the following functions. (Do not change these function names or return types, else tests will fail). public static String blowup (String str) Returns a version of the original string as follows: each digit 0-9 that appears in the original string is replaced by that many occurrences of the character to the right of the digit. So the string "a3tx2z" yields "attttxzzz" and "12x" yields "2xxx". A digit not followed by a character (i.e. at the- end of the string) is replaced by nothing. public static int maxRun(String str) Given a string, returns the length of the largest run in the string. A "run" is a series of zero or more adjacent characters that are the same. So the max run of "xxyyyz" is 3. and the max run of "xyz" is 1.

Answer 1

Screenshot of output:

un Enter the string: x3tyyyzz Test 1 Result for test l xttttyyyzz Test 2 Result for test 23 BUILD SUCCESSFUL (total time 10 seconds)

Screenshot of code:

import java.util.Scanner; import java. lang. StringBuilder; /Class StringCode public class StringCode public static String bloiup (String str) //stb is used for holding the desired string str. StringBuilder stnew StringBuilder (str): //substr is used for the substring which replaces the number in the string as desired. String sub str = ""; //getting the ascii value of the characters of str in an arrayb. bytet] b -str.getBytes ( //The following ї00p iterates up to the second last char acter as. //there is no need to check the following condition for the last character for (int í = 0; ǐ < str.length ( )-1; 1++) //Checking whether there is any number in the string. if (bi 48 & b[i] < 57) //Get the integer value of the number int n Integer.parselnt(str.substring(i, i+1)) substr ": //Repeat the next character of the digit for that many times. for (int j = 0; j < n; j++) //substr consists of the repiting character. substr str.substring(i + 1, i + 2)

//Replace the digit by substır stb.replace(i, i + 1, substr): //Return the string value of stb. return stb.toString) public static int maxRun(String str) //Default value of count. int count, max1; //The following loop iterates upto the second last character as... //there is no need to check the following condition for the last character for (int í = 0; ǐ < str.length()一1: i++) count 1; //Compare a character with the next to check the occurrences. if (str.charkt (i) - str.charkt (i + 1)) // Using do-while loop to calculate the length of the run dof count++ / Increase count i++// Move to next character }vhile ( î < str.lengt h ()-1 “ str.charAt(1)-str.charat (1 + 1)); // Checking if the run continues if (max < count) / Updating maximum run max - count:

return max; public static void main(Stringl] args]) scanner sc = new scanner (System·in), System.out.println("Enter the string: "); String str = sc.nextLine(); System.out.println("Test 1" String result blowup (str): System. out.println("Result for test 1- " + result); System.out.println("Test 2") int num = maxRun (str); System.out.println("Result for test 2 "num);

Code:

import java.util.Scanner;
import java.lang.StringBuilder;
//Class StringCode
public class StringCode
{
    public static String blowup(String str)
    {
        //stb is used for holding the desired string str.
        StringBuilder stb = new StringBuilder(str);
        //substr is used for the substring which replaces the number in the string as desired.
        String substr = "";
        //getting the ascii value of the characters of str in an array b.
        byte[] b = str.getBytes();
        //The following loop iterates upto the second last character as...
        //there is no need to check the following condition for the last character.
        for(int i = 0; i < str.length()-1; i++)
        {
            //Checking whether there is any number in the string.
            if(b[i] >= 48 && b[i] <= 57)
            {
                //Get the integer value of the number.
                int n = Integer.parseInt(str.substring(i,i+1));
                substr = "";
                //Repeat the next character of the digit for that many times.
                for(int j = 0; j < n; j++)
                {
                    //substr consists of the repiting character.
                    substr += str.substring(i + 1, i + 2);
                }
                //Replace the digit by substr
                stb.replace(i, i + 1, substr);
            }
        }
        //Return the string value of stb.
        return stb.toString();
    }
    public static int maxRun(String str)
    {
        //Default value of count.
        int count, max = 1;
        //The following loop iterates upto the second last character as...
        //there is no need to check the following condition for the last character.
        for(int i = 0; i < str.length()-1; i++)
        {
            count = 1;
            //Compare a character with the next to check the occurrences.
            if(str.charAt(i) == str.charAt(i + 1))
            {
                // Using do-while loop to calculate the length of the run
                do{
                    count++; // Increase count
                    i++; // Move to next character
                }while(i < str.length()-1 && str.charAt(i) == str.charAt(i + 1)); // Checking if the run continues
                if(max < count) // Updating maximum run
                    max = count;
            }
        }
        return max;
    }
    public static void main(String[] args)
    {
        Scanner sc = new Scanner(System.in);
        System.out.println("Enter the string:");
        String str = sc.nextLine();
        System.out.println("Test 1");
        String result = blowup(str);
        System.out.println("Result for test 1 = " + result);
        System.out.println("Test 2");
        int num = maxRun(str);
        System.out.println("Result for test 2 = " + num);
     
    }
}