toBinary method works in IDE but will not validate

Question about the task

Resolved

Any helpful tips would be much appreciated. Edit: I found a logic error so I remedied it with the code below but it is still failing to validate.

if(hexNumber != null && !hexNumber.equals("")) {
            for (int i = 0; i < hexNumber.length(); i++) {
                placeHolder = hex.indexOf(hexNumber.substring(i, i + 1));
                if(placeHolder != 0 && placeHolder != 1 && placeHolder != 2 && placeHolder != 3 && placeHolder != 4 && placeHolder != 5 && placeHolder != 6
                && placeHolder != 7) {
                    //Stores the remainder of the function, then adds it to the output variable in the reverse order that it is calculated
                    while (placeHolder != 0) {
                        ans = placeHolder % 2;
                        binary = ans + binary ;
                        placeHolder = placeHolder / 2;
                    }
                    binaryFinal = binaryFinal + binary;
                    binary = "";
                }else {
                    if(placeHolder == 1){
                        binaryFinal = binaryFinal + "0001";
                    }else if(placeHolder == 2){
                        binaryFinal = binaryFinal + "0010";
                    }else if(placeHolder == 3){
                        binaryFinal = binaryFinal + "0011";
                    }else if(placeHolder == 4){
                        binaryFinal = binaryFinal + "0100";
                    }else if(placeHolder == 5){
                        binaryFinal = binaryFinal + "0101";
                    }else if(placeHolder == 6){
                        binaryFinal = binaryFinal + "0110";
                    }else if(placeHolder == 7){
                        binaryFinal = binaryFinal + "0111";
                    }else {
                        binaryFinal = binaryFinal + "0000";
                    }
                }
            }
            while(binaryFinal.length() % 4 != 0) binaryFinal += "0";
            return binaryFinal;
        }

The public static toHex(String) method must convert the string representation of a binary number, received as an input parameter, from the binary numeral system to hexadecimal and return its string representation.
And conversely, the public static toBinary(String) method converts from the string representation of a hexadecimal number to the string representation of a binary number.

The methods only work with non-empty strings.
If the input parameter is an empty string or null, then both methods return an empty string.
If the input parameter of the toHex(String) method contains any character other than 0 or 1, then the method returns an empty string.
If the input parameter of the toBinary(String) method contains any character other than digits from 0 to 9 or lowercase Latin letters from a to f, then the method returns an empty string.

Your task is to implement these methods.

One algorithm for converting the string representation of a binary number to the string representation of a hexadecimal number is as follows:

We check the length of the string received in the input parameter. It must be a multiple of 4.
If this is not the case, then add the required number of 0s to the beginning of the string.
We take every four characters (bits) and check which hexadecimal character they correspond to.

For example:

the binary representation is "100111010000", where "1001" is "9", "1101" is "d", and "0000" - "0"
the hexadecimal representation is "9d0".

One algorithm for converting the string representation of a hexadecimal number to the string representationof a binary number is as follows:
We take each character and check which binary number (4 bits) it corresponds to.

For example:

the hexadecimal representation is "9d0", where "9" is "1001", "d" is "1101", and "0" is "0000"
the binary representation is "100111010000".

The main() method is not involved in testing.

Requirements:

The toHex(String) method needs to be implemented as outlined in the task conditions.
The toBinary(String) method needs to be implemented as outlined in the task conditions.
The Integer.toBinaryString(int) and Long.toBinaryString(int) methods cannot be used.
The Integer.toHexString(int) and Long.toHexString(int) methods cannot be used.
The Integer.parseInt(String, int) and Long.parseLong(String, int) methods cannot be used.
The Integer.toString(int, int) and Long.toString(long, int) methods cannot be used.
You cannot use a BigInteger object.
You cannot use a BigDecimal object.

package en.codegym.task.pro.task09.task0908; /* Binary to hexadecimal converter */ public class Solution { public static void main(String[] args) { String binaryNumber = "100011001111"; System.out.println("Binary number " + binaryNumber + " is equal to hexadecimal number " + toHex(binaryNumber)); String hexNumber = "ff01"; System.out.println("Hexadecimal number " + hexNumber + " is equal to binary number " + toBinary(hexNumber)); } public static String toHex(String binaryNumber) { //write your code here //Index variable, two placeholders, and an output variable String hex = "0123456789abcdef"; String placeHolder; double ans = 0; String hexFinal = ""; if(binaryNumber != null && !binaryNumber.equals("")) { while(binaryNumber.length() % 4 != 0) binaryNumber = "0" + binaryNumber; //Grabs a section of the binary string in groups of four for (int i = 0; i < (binaryNumber.length()); i += 4) { placeHolder = binaryNumber.substring(i, 4 + i); //Iterates through each # in the section of four multiplies it depending on its position and value for (int j = 0; j <= placeHolder.length() - 1; j++) { if (placeHolder.substring(j, j + 1).equals("1")) { ans = Math.pow(2, (3 - j)) + ans; } } hexFinal = hexFinal + hex.substring((int) ans, ((int) ans + 1)); ans = 0; } return hexFinal; } return ""; } public static String toBinary(String hexNumber) { //write your code here //First three variables are placeholders and the last is the output String hex = "0123456789abcdef"; int ans = 0; String binary = ""; int placeHolder; String binaryFinal = ""; if(hexNumber != null && !hexNumber.equals("")) { for (int i = 0; i < hexNumber.length(); i++) { placeHolder = hex.indexOf(hexNumber.substring(i, i + 1)); if(placeHolder != 0 && placeHolder != 1) { //Stores the remainder of the function, then adds it to the output variable in the reverse order that it is calculated while (placeHolder != 0) { ans = placeHolder % 2; binary = ans + binary ; placeHolder = placeHolder / 2; } binaryFinal = binaryFinal + binary; binary = ""; }else { if(placeHolder == 1){ binaryFinal = binaryFinal + "0001"; }else { binaryFinal = binaryFinal + "0000"; } } } return binaryFinal; } return ""; } }

Comments (7)

Popular
New
Old

You must be signed in to leave a comment

Guadalupe Gagnon

Level 37 , Tampa, United States

19 July 2022, 18:30

The method returns a blank String if "0" is passed as the argument.

Wyatt Wilhelmi

Level 20 , United States of America, United States

19 July 2022, 21:33

This is what I got when I passed "0" as an argument on my end.

Guadalupe Gagnon

Level 37 , Tampa, United States

20 July 2022, 13:33

disregard. I may have copied the code wrong yesterday as I get "0000" now.

Guadalupe Gagnon

Level 37 , Tampa, United States

20 July 2022, 14:55

useful

Ok. I have to put my thinking cap on. Some things to try: #1 try removing leading zeros, or if the whole Sting is zeros, such as "0000", try removing all but 1 so: - instead of something like this "000000010101101", return "10101101" - instead of "0000" return just simply "0" but if that was the problem then the toHex method would also be failing.... so #2 Try making your toBinary() method work with capital letters as well. I would suggest just converting the string to lowercase before the conversion algorithm.

Wyatt Wilhelmi

Level 20 , United States of America, United States

20 July 2022, 16:20

I believe I owe you an apology for wasting your time. I should have closed this discussion yesterday because I rewrote the code and it failed again so I just looked up the solution. I am really sorry, I wish I could at least try your suggestions but I can't after the task has been completed.

Matt.P.72

Level 26 , Killen, United States

19 July 2022, 01:28

try running through the algorithm like this the back through for the decimal String binaryNumber = ""; if (decimalNumber > 0) { while (decimalNumber > 0) { binaryNumber = decimalNumber % 2 + binaryNumber; decimalNumber = decimalNumber / 2; } return binaryNumber; } else { return binaryNumber;

Wyatt Wilhelmi

Level 20 , United States of America, United States

19 July 2022, 15:34

I have to say that I am not quite following you. I do not have a decimal number in this program that I can see, and my toHex method validates. The only problem that I have is the toBinary method. Could you please explain your suggestion a little more?