CodeGym/Help with Java Tasks/I wonder if I could've made it shorter

Level 11

Budapest

18.04.2024
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I wonder if I could've made it shorter

Question about the task

Under discussion

I've just written the longest code in my short professional life:) It produces the expected output but I'm wondering if I could make it any simpler. The second issue is that after verification the website says, "Be sure that the toHex(String) method returns an empty string if it receives an empty string as input. Be sure that the toBinary(String) method returns an empty string if it receives an empty string as input", although these methods in my code start with the following lines (correspondingly): (1)

if (binaryNumber.isEmpty())
    return null;

(2)

if (hexNumber.isEmpty())
    return null;

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.isEmpty()) return null; else { for (int i = 0; i < binaryNumber.length(); i++) { if (binaryNumber.charAt(i) != '0' && binaryNumber.charAt(i) != '1') return null; } } 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.isEmpty()) return null; 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) { 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; } public static String toBinary(String hexNumber) { //write your code here if (hexNumber == "" || hexNumber == null) return null; // if the input string is empty -> null else { String binaryNumber = ""; for (int i = 0; i < hexNumber.length(); i++) { char tempChar = hexNumber.charAt(i); if (Character.isDigit(tempChar)) { int tempInt = Character.getNumericValue(tempChar); String tempString = decimalToBinary(tempInt); //string representation of the binary number corresponding to tempInt binaryNumber += tempString; } else if (Character.isLetter(tempChar) && Character.isLowerCase(tempChar)) { if (tempChar >= 'f') return null; else { String tempString = letterToBinary(tempChar);//string rep of the binary number corresponding to the letter binaryNumber += tempString; } } else { return null; } } return binaryNumber; } } public static String decimalToBinary(int decimalNumber) { String binaryNumber = ""; if (decimalNumber == 0) { binaryNumber = "0000"; } while (decimalNumber > 0) { binaryNumber = decimalNumber % 2 + binaryNumber; decimalNumber = decimalNumber / 2; } return binaryNumber; } public static String letterToBinary(char letter) { String result; switch (letter) { 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; default: result = "1111"; } return result; } }

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Guadalupe Gagnon

Level 37 , Tampa, United States

18 April, 12:45

null does not equal an empty String:

String nullString = null;
String emptyString = "";

System.out.println(emptyString.equals(nullString)); // outputs false
System.out.println(nullString.equals(emptyString)); // null pointer exception thrown

Thomas

Level 41 , Bayreuth, Germany

18 April, 12:48

If we both have the same idea then this has to be the problem 😂

Guadalupe Gagnon

Level 37 , Tampa, United States

18 April, 12:50

as far as making the code smaller: Personally I wouldn't worry about that until you finish the task. If you want to go back at a later time and improve your code that would be the time to shorten it. I did this quite a bit when I working through the tasks. When i tried to make the best code before passing the task I spent WAAY too much time on each task. Even professionals write with the goal first to make working code and then optimization; following best practices as they go along.

Evgeniia Shabaeva

Level 11 , Budapest, Hungary

19 April, 08:46

Thank you very much for the clarification. Could you also tell me what I check then when I use

str.isEmpty()?

Is that the same as

if (str == 0)

? For some reason the 'toHex method' only passed validation when I wrote

if (binaryNumber == "" || binaryNumber == null)
           return "";

But it wasn't marked correct when it was

if (binaryNumber.isEmpty() || binaryNumber == null)
            return "";

while IntellijIDEA recommend I use str.isEmpty() instead of str == 0. I'm just wondering why it's like that.

Evgeniia Shabaeva

Level 11 , Budapest, Hungary

19 April, 09:08

)) Thank you very much! Well, now I've got:

public static String toHex(String binaryNumber) {
    if (binaryNumber == "" || binaryNumber == null)
        return "";
(...)
}

and:

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

I also found one more mistake in the toBinary method. I had

if (tempChar >= 'f') return "";

while it had to be (and now is)

if (tempChar >= 'g') return "";

As a result, the toHex method passes validation, while the toBinary one still doesn't. I'm thinking:).

Thomas

Level 41 , Bayreuth, Germany

19 April, 12:01

good catch, checking boundaries often introdunces mistakes. For your other topic first let us look again at object delaration and instantiation. For that process you first declare a variable of a certain type, eg. an instance variable private Point point; The compiler reserves memory for the reference, for the variable itself but at that time not yet for the object. An object does not exist at that time, only the reference and it's value is null. Now you create the object using a constructor. Its return value is the reference to that created object. eg. point = new Point(0, 0); will save the memory address of the created Point object in the point variable. So every point varible can have two states, the value after declaring the variable (null) and the state after assigning a reference to an object, a valid address of that object. This has a major impact on the execution of methods. If you try to invoke a method on the null reference then a NullPointerException will be thrown. That's something happening all the time and it is our responsibility to take care of that. And one way is to check method arguments for null BEFORE you invoke methods on that references. When there eg. is a parameter String hexValue then the first thing you do is checking it for null. If you ignore that rule and check eg. for isEmpty() and hexValue is null, then that NullPointerException will be thrown. Just try that yourself and pass null to that method. This is what you have to do:

if (binaryNumber == null || binaryNumber.isEmpty()) {
    return "";
}

== compares references and isEmpty() is special way to write equals(""). The latter requires a valid object. The first makes sure there is a valid object.

Guadalupe Gagnon

Level 37 , Tampa, United States

22 April, 15:53

A little clarification on top of everything that Thomas said: The double equals operand (==) checks the memory address of objects and not the value. With String you must use .equals() method if you want to check the equality of the values. You can see the difference with this bit of code:

String a = "0";
String b = new String("0");


System.out.println(a == b);
System.out.println(a.equals(b));

In this code two Strings are created and their values are set to "0". The first output line, with the double equals, will output false while the second output line, the one with the equals method, will output true. The difference is that both a and b are different objects but I forced Java to give b a separate memory address. The first output line checks the memory addresses and correctly outputs false.

Guadalupe Gagnon

Level 37 , Tampa, United States

22 April, 16:11

The .isEmpyty() method check whether a String has no characters in it whatsoever. It is the same exact thing as .equals(""). You can use either of those for the same exact results. Using a double equals, such as Str == "" , is invalid and would be a bug in the program as the double equals operand checks memory addresses as explained in the above code. Now the way that you had your first example results in a null point exception being thrown. You can see what happens if you run this code:

String str = null;

if(str.isEmpty() || str == null); // .isEmpty() thorws exception because str is null
// this happens before str == null is checked and the exception ends the program

System.out.println("REACHED");

The correct way to do what you want to achieve was given by Thomas, you first check if the String is null and then check if it .isEmpty(). So with my above example you just need to change the order of statements you check in the if statement:

if (str == null || str.isEmpty()) ;

This way if the String is null then the expression whole expression is true and it never evaluates str.isEmpty() The reason your second example:

if (binaryNumber == "" || binaryNumber == null)
           return "";

passed validation is because the double equals method is null safe, so the first check doesn't result in an exception. Even if this code resulted in false when an empty String was passed in, the rest of your code still resulted in a blank String being returned so it passed. Checking Strings against a number is not a valid operation, so:

if(str == 0)

is invalid and wont even compile.

Thomas

Level 41 , Bayreuth, Germany

18 April, 12:42

There are possibilities to shorten your code. However it is more important to first get things going and to learn how to approach things ;) For your problem. I'd try to check for null first and then for isEmpty() and return an empty string for both cases. I do not know if that is the problem because I didn't go the pro route and don't know this task. For later two easy improvements: String.matches(regexExpression) may help you simplifying the code. Eg. "0001110010101".matches("[01]+") will make sure there is at least one 0 or 1 and no other char "abcdef123".matches("[0-9a-fA-F]+") will make sure there is at least one character in the range of 0 to 9 or a to f or A to F. All other chars will make it return false. To get the decimal value of a hex char you have to easy possiblilities a) char[] hexVals = "0123456789abcdef".toCharArray(); now eg 1 has the index 1 and the value 1 eg. a has the index 10 and the value 10 eg. d has the index 13 and the value 13 You get the index either using a loop or some Arrays method like int index = Arrays.binarySearch(hexVals, 'd'); b) if the char is 0 to 9 then the value is char - '0', if the char is a to f, then the value is char - 'a' + 10 So in version a you'd need to lowercase the string in version b you could add a variant for uppercase letters, too. If the char is eg. uppercase D, then then value is 'D' - 'A' + 10 In code using two ternary operators that'll look similar to (but you need the prior validation using matches):

int hexValue = hex <= '9' ? hex - '0' : hex <= 'F' ? hex - 'A' + 10 : hex - 'a' + 10;