What is going wrong in my toBinary(String) method?

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I have tested it with various online calculators, and it seems to output the correct solution. Additionally, I looked through the other questions pertaining to this task, but I must be missing something because I cannot figure out where I am going wrong. Any help, even if it is just an input that does not work correctly with my code, would be greatly appreciated. Edit: I ended up solving this task by going at it differently than my code in this post. To solve it I just created if statements for each hexadecimal number/ symbol. Example:

if(hexNumber.charAt(i) == '0') binary += "0000";

I placed all of these in a for loop that went through each character in the given hexadecimal number. I really do not see how this would work and my previous code would not as this code leaves a redundant 0000 at the beginning of the conversion if the hexadecimal begins with 0. Maybe this is why my previous code failed though. I cannot test it now as verification is complete. If anyone can confirm if this is the case that would be great.

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 */ import java.util.Objects; public class Solution { private static final String HEX = "0123456789abcdef"; public static void main(String[] args) { String binaryNumber = "100111010000"; System.out.println("Binary number " + binaryNumber + " is equal to hexadecimal number " + toHex(binaryNumber)); String hexNumber = "9d0"; System.out.println("Hexadecimal number " + hexNumber + " is equal to binary number " + toBinary(hexNumber)); } public static String toHex(String binaryNumber) { //write your code here String hexer = ""; String temp = ""; if((binaryNumber == null) || (Objects.equals(binaryNumber, ""))) { return hexer; } if (binaryNumber.length() % 4 != 0) { while (binaryNumber.length() % 4 > 0) { binaryNumber = "0" + binaryNumber; } } for(int checkValid = 0; checkValid < binaryNumber.length(); checkValid++) { int check = binaryNumber.charAt(checkValid); if(check > '1' || check < '0') { return hexer; } } for(int i = 0; i < binaryNumber.length() / 4; i++) { for(int j = 0; j < binaryNumber.length(); j++) { temp += binaryNumber.charAt(j); if(j >= binaryNumber.length() - 1) { int deci = 0; for (int ij = 0; ij < temp.length(); ij++) { deci += (temp.charAt(temp.length() - ij - 1) - '0') * Math.pow(2, ij); if(ij >= temp.length() -1) { while(deci > 0) { hexer = HEX.charAt(deci % 16) + hexer; deci = deci / 16; } return hexer; } } } } } return hexer; } public static String toBinary(String hexNumber) { //write your code here int deci = 0; if (hexNumber == null || Objects.equals(hexNumber, "") ) { return ""; }for(int check = 0; check < hexNumber.length(); check++) { if((hexNumber.charAt(check) < '0') || (hexNumber.charAt(check) > '9' && hexNumber.charAt(check) < 'a') || hexNumber.charAt(check) > 'f') { return ""; } } for (int i = 0; i < hexNumber.length(); i++) { if (hexNumber.charAt(i) <= '9') { deci = 16 * deci + HEX.charAt(hexNumber.charAt(i) - '0') - '0'; } else { deci = 16 * deci + HEX.charAt(hexNumber.charAt(i) - 'a') - ('a' / 2) + 10; } } String binary = ""; while (deci > 0) { binary = String.valueOf(deci % 2) + binary; deci = deci / 2; } while(binary.length() % 4 !=0) { binary = '0' + binary; } while(binary.length() < 4) { binary = '0' + binary; } return binary; } }

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Lisa L

Level 47 , Nuremberg, Germany

4 June 2022, 13:06

useful

One reason may be that your hex to bin algorithm can handle input only input strings of a limited length. Looks like you first convert to dec and then to binary. The decimal value you save in an int and that again only is able to hold values up to 2.100.000.000 That's not rally much. The hex string in line 10 is to large for it. You could have tried to convert digit per digit. That'll work. My solution is not very elegant but it validated

public static String toBinary(String hexNumber) {
    if (hexNumber == null || hexNumber.isEmpty())
        return "";

    int decimalNumber;
    StringBuilder binaryNumber = new StringBuilder();
    StringBuilder helper = new StringBuilder();

    for (int i = 0; i < hexNumber.length(); i++) {
        char hex = hexNumber.charAt(hexNumber.length() - i - 1);
        // check for valid digits
        if((hex < '0' || hex > '9') && (hex < 'a' || hex > 'f'))
            return "";
        // hex to dec
        decimalNumber = hex <= '9' ? hex - '0' : hex - 'a' + 10;
        // now dec to bin
        while (decimalNumber > 0) {
            helper.insert(0, decimalNumber % 2);
            decimalNumber /= 2;
        }
        while (helper.length() < 4) {
            helper.insert(0,0);
        }
        binaryNumber.insert(0, helper);
        helper.setLength(0);
    }
    return binaryNumber.toString();
}

Anonymous #10987379

Level 22 , United States

5 June 2022, 00:49

Ahhh, ok. I noticed that I was limited to the max int value while I was trying to figure out what was wrong with it, but for some reason it did not occur to me that it may have been the reason for the failure. Thank you for the input.