One of the methods in this code is still not accepted by the validator. What am I doing wrong?

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I kind of put aside this task some time ago and moved on. But today I decided to catch up with some unresolved tasks and got back to this one. The code seems to work fine in general, but there is still something wrong with the toBinary(String value) method, even though it produces and outputs the expected result. The validator says it needs to be implemented as outlined in the task conditions. The conditions are these: 1) the inut parameter mustn't be empty: 2) if the input parameter 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. I checked for those two things and made the method return "" in both cases. I also wrote two additional methods (decimalToBinary(int value) and letterToBinary(char value)) to help use the method in question (the toBinary(String value) one). Now it returns a string representation of the binary equivalent to the hexadecimal number given as an input parameter, but something is still wrong, according to the validator. Can you help me out, please?

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 = "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 if (binaryNumber == null || binaryNumber.isEmpty()) return ""; else { for (int i = 0; i < binaryNumber.length(); i++) { if (binaryNumber.charAt(i) != '0' && binaryNumber.charAt(i) != '1') return ""; } } if (binaryNumber.length() % 4 != 0) { //We check if the length of the input string is a multiple of four int extraZeroes = 4 - binaryNumber.length() % 4; for (int i = 0; i < extraZeroes; i++) { binaryNumber = "0" + binaryNumber; // we add extra zeroes at the beginning 0f the input string if necessary } } //I'm not sure if I need to return the altered binaryNumber here String hexNumber = ""; for (int i = 0; i < binaryNumber.length(); i += 4) { //we create a loop to go through the altered input string String fourBits = binaryNumber.substring(i, (i + 4)); //we separate a substring of 4 bits String fourBitsCoded = binaryToDecimal(fourBits); //string to string fourBitsCoded = decimalToAF(fourBitsCoded); //string to string hexNumber += fourBitsCoded; //we concatenate the coded version of each four bits of the binary number to the hexNumber, which was empty in the beginning } return hexNumber; } public static String binaryToDecimal(String binary) { if (binary == null || binary.isEmpty()) return ""; else { int decimalNumber = 0; for (int i = 0; i < binary.length(); i++) { int index = binary.length() - 1 - i; int value = Character.getNumericValue(binary.charAt(index)); decimalNumber += (int) (value * Math.pow(2, i)); } return String.valueOf(decimalNumber); } } public static String decimalToAF(String stringDecimal) { if (stringDecimal == null || stringDecimal.isEmpty()) return ""; int tempNumber = Integer.valueOf(stringDecimal); String result = "";//we need an integer here to compare it to 9 //if the decimal is less than or equal to 9, we return it as a string value if (tempNumber <=9) result = stringDecimal; //if the decimal is greater than 9 and less than 16, we convert it to a letter from a to f else if (tempNumber > 9 && tempNumber < 16) { String toSymbol; switch (tempNumber) { case 10: toSymbol = "a"; break; case 11: toSymbol = "b"; break; case 12: toSymbol = "c"; break; case 13: toSymbol = "d"; break; case 14: toSymbol = "e"; break; default: toSymbol = "f"; } result = toSymbol; } return result; } /*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.*/ public static String toBinary(String hexNumber) { // write your code here StringBuilder binaryNumber = new StringBuilder(); if (hexNumber == null || hexNumber.isEmpty()) return ""; else if (!hexNumber.matches("[0-9a-f]+")) return ""; else if (hexNumber.matches("[0-9a-f]+")) { for (int i = 0; i < hexNumber.length(); i++) { char tempChar = hexNumber.charAt(i); String tempString = ""; if (Character.isDigit(tempChar)) { int tempInt = Character.getNumericValue(tempChar); tempString = decimalToBinary(tempInt); //this method changes an int value into its binary euivalent's string representation } else if (Character.isLetter(tempChar)/* && Character.isLowerCase(tempChar)*/) { tempString = letterToBinary(tempChar); //this method changes a char into its binary euivalent's string representation } binaryNumber.append(tempString); } } return binaryNumber.toString(); } public static String decimalToBinary(int decimalNumber) { String result = ""; if (decimalNumber < 0) return ""; else if (decimalNumber == 0) { result = "0000"; } else if (decimalNumber >= 0) { while (decimalNumber != 0) { result = decimalNumber % 2 + result; decimalNumber = decimalNumber / 2; } } return result; } public static String letterToBinary(char tempChar){ String result = ""; switch (tempChar) { case 'a': result = "1010"; break; case 'b': result = "1011"; break; case 'c': result = "1100"; break; case 'd': result = "1101"; break; case 'e': result = "1110"; break; case 'f': result = "1111"; break; default: result = ""; } return result; } }

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Thomas

Level 41 , Bayreuth, Germany

19 July, 21:27

I wrote a quick test for your code and I saw where the problem is:

String[] tests = {"0", "1", "2", "3", "4", "5", "6", "7", "8", "9", "a", "b", "c", "d", "e", "f"};
Stream.of(tests).forEach(test -> System.out.format("source: %s, result: %s, validation: %s%n",
                test,
                toBinary(test),
                new BigInteger(test, 16).toString(2))
);

0 and a-f display correctly as a full nibble (4 digits) while 1-9 do not. There are also a lot of redundancies and you sometimes use api methods not correctly, eg:

else if (!hexNumber.matches("[0-9a-f]+")) return "";

else if (hexNumber.matches("[0-9a-f]+")) {

if you already drop out of the method if the hex is not 0-9a-f, then why check this another time? It has to be 0-9a-f after the first check. Similar here:

if (decimalNumber < 0) return "";
else if (decimalNumber == 0) {
    result = "0000";
} else if (decimalNumber >= 0) {

if decimalNumber is not <= 0 then it has to be >(=) 0. The last else if can be written as a plain else Character.getNumericValue(c) this not just gets the numeric value of '0'-'9' but also of 'a'-'f', you do not need an if-else and devide that up and handle that in different ways. In your first question to that task I already told you how I'd approach that (and you already adopted a few ideas). Let me show you a simple approach for just one char:

Thomas

Level 41 , Bayreuth, Germany

19 July, 21:28

public static void main(String[] args) {
    String[] tests = {"0", "1", "2", "3", "4", "5", "6", "7", "8", "9", "a", "b", "c", "d", "e", "f"};
    Stream.of(tests).forEach(test -> System.out.format("source: %s, result: %s, validation: %s%n",
            test,
            toBinary(test),
            new BigInteger(test, 16).toString(2))
    );
}

private static String toBinary(String s) {
    if (s == null || !s.matches("[0-9a-f]+")) {
        return "";
    }

    char hex = s.charAt(0);

    // simple hex to dec
    int decimalNumber = hex <= '9' ? hex - '0' : hex - 'a' + 10;

    // now dec to bin
    StringBuilder nibble = new StringBuilder();
    while (decimalNumber > 0) {
        nibble.insert(0, decimalNumber % 2);
        decimalNumber /= 2;
    }

    // pad with zeros
    while (nibble.length() < 4) {
        nibble.insert(0, '0');
    }

    return nibble.toString();
}

you just would need to add another loop to go to an entire string (and not just the first char) and a second StringBuilder to hold the complete result.

Evgeniia Shabaeva

Level 32 , Budapest, Hungary

23 July, 12:26

Thank you so much for all the valuable comments! I deleted all the redundant code and, what makes me really happy, finally understood my main mistake. Thank you for pointing it put in your code. It was that I didn't write the second loop to add the missing zeroes in the sub-method that was supposed to turn int values to their binary equivalents and return strings (of four characters, not just any strings). So, I kept the String decimalToBinary(int decimalNumber) method and added another while-loop to it:

while (result.length() < 4) {
                result = "0" + result;
            }

I also tried rewriting it using StringBuilder, but then I thought that I probably hadn't been supposed to know StringBuilder at that point, so I just used the same String variable called "result". It worked, so I'm glad. Thank you once again!