java - Implement a function to check if a string/byte array follows utf-8 format

Question

I am trying to solve this interview question.

After given clearly definition of UTF-8 format. ex: 1-byte : 0b0xxxxxxx 2- bytes:.... Asked to write a function to validate whether the input is valid UTF-8. Input will be string/byte array, output should be yes/no.

I have two possible approaches.

First, if the input is a string, since UTF-8 is at most 4 byte, after we remove the first two characters "0b", we can use Integer.parseInt(s) to check if the rest of the string is at the range 0 to 10FFFF. Moreover, it is better to check if the length of the string is a multiple of 8 and if the input string contains all 0s and 1s first. So I will have to go through the string twice and the complexity will be O(n).

Second, if the input is a byte array (we can also use this method if the input is a string), we check if each 1-byte element is in the correct range. If the input is a string, first check the length of the string is a multiple of 8 then check each 8-character substring is in the range.

I know there are couple solutions on how to check a string using Java libraries, but my question is how I should implement the function based on the question.

Thanks a lot.

Answer 1

Let's first have a look at a visual representation of the UTF-8 design.

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Now let's resume what we have to do.

• Loop over all character of the string (each character being a byte).
• We will need to apply a mask to each byte depending on the codepoint as the x characters represent the actual codepoint. We will use the binary AND operator (&) which copy a bit to the result if it exists in both operands.
• The goal of applying a mask is to remove the trailing bits so we compare the actual byte as the first code point. We will do the bitwise operation using 0b1xxxxxxx where 1 will appear "Bytes in sequence" time, and other bits will be 0.
• We can then compare with the first byte to verify if it is valid, and also determinate what is the actual byte.
• If the character entered in none of the case, it means the byte is invalid and we return "No".
• If we can get out of the loop, that means each of the character are valid, hence the string is valid.
• Make sure the comparison that returned true correspond to the expected length.

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The method would look like this :

public static final boolean isUTF8(final byte[] pText) {

    int expectedLength = 0;

    for (int i = 0; i < pText.length; i++) {
        if ((pText[i] & 0b10000000) == 0b00000000) {
            expectedLength = 1;
        } else if ((pText[i] & 0b11100000) == 0b11000000) {
            expectedLength = 2;
        } else if ((pText[i] & 0b11110000) == 0b11100000) {
            expectedLength = 3;
        } else if ((pText[i] & 0b11111000) == 0b11110000) {
            expectedLength = 4;
        } else if ((pText[i] & 0b11111100) == 0b11111000) {
            expectedLength = 5;
        } else if ((pText[i] & 0b11111110) == 0b11111100) {
            expectedLength = 6;
        } else {
            return false;
        }

        while (--expectedLength > 0) {
            if (++i >= pText.length) {
                return false;
            }
            if ((pText[i] & 0b11000000) != 0b10000000) {
                return false;
            }
        }
    }

    return true;
}

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Edit : The actual method is not the original one (almost, but not) and is stolen from here. The original one was not properly working as per @EJP comment.