UTF-8 Validation in Java

Last Updated : 28 Jan, 2021

A character in UTF-8 can be from 1 to 4 bytes long, subjected to the following rules:

For a 1-byte character, the first bit is a 0, followed by its Unicode code.
For n-bytes character, the first n-bits are all one’s, the n+1 bit is 0, followed by n-1 bytes with the most significant 2 bits being 10.

This is how the UTF-8 encoding would work:

  Char. number range   |        UTF-8 octet sequence
      (hexadecimal)    |              (binary)
   --------------------+---------------------------------------------
   0000 0000-0000 007F | 0xxxxxxx
   0000 0080-0000 07FF | 110xxxxx 10xxxxxx
   0000 0800-0000 FFFF | 1110xxxx 10xxxxxx 10xxxxxx
   0001 0000-0010 FFFF | 11110xxx 10xxxxxx 10xxxxxx 10xxxxxx

Example:

Given an array of integers representing the data, return whether it is a valid UTF-8 encoding.

The input is an array of integers. Only the least significant 8 bits of each integer is used to store the data. This means each integer represents only 1 byte of data.

data = [235, 140, 4], which represented the octet sequence: 11101011 10001100 00000100.
Return false.
The first 3 bits are all one’s and the 4th bit is 0 means it is a 3-bytes character.
The next byte is a continuation byte which starts with 10 and that’s correct.
But the second continuation byte does not start with 10, so it is invalid.
———————————————————————————————–
data = [197, 130, 1], which represents the octet sequence: 11000101 10000010 00000001.
Return true.
It is a valid utf-8 encoding for a 2-bytes character followed by a 1-byte character.

Approach: As long as every byte in the array is of the right type, it is a valid UTF-8 encoding.

Start from index 0, determine each byte’s type and check its validity.
There are five kinds of valid byte type: 0**, 10**, 110**,1110** and 11110**
Give them type numbers, 0, 1, 2, 3, 4 which are the index of the first 0 from left.
So, the index of the first 0 determines the byte type.
If a byte belongs to one of them: if it is type 0, continue if it is type 2 or 3 or 4, check whether the following 1, 2, and 3 byte(s) are of type 1 or not.
If not, return false; else if a byte is type 1 or not of valid type, return false.

Java

// Java program to check whether the data
// is a valid UTF-8 encoding
  
import java.io.*;
import java.util.*;
  
class Sol {
  
    private int[] masks = { 128, 64, 32, 16, 8 };
  
    public boolean validUtf8(int[] data)
    {
        int len = data.length;
  
        // for each value in the data array we have to take
        // the "and" with the masks array
        for (int i = 0; i < len; i++) {
            int curr = data[i];
  
            // method to check the array if the
            // and with the num and masks array is
            // 0 then return true
            int type = getType(curr);
  
            if (type == 0) {
                continue;
            }
  
            else if (type > 1 && i + type <= len)
            {
                while (type-- > 1) 
                {
                    if (getType(data[++i]) != 1)
                    {
                        return false;
                    }
                }
            }
            else {
                return false;
            }
        }
        return true;
    }
  
    // method to check the type
    public int getType(int num)
    {
        for (int i = 0; i < 5; i++) {
  
            // checking the each input
            if ((masks[i] & num) == 0) {
                return i;
            }
        }
  
        return -1;
    }
}
  
class GFG {
    public static void main(String[] args)
    {
        Sol st = new Sol();
        int[] arr = { 197, 130, 1 };
  
        boolean res = st.validUtf8(arr);
        System.out.println(res);
    }
}

Output

true

Time Complexity: O(n)
Space Complexity: O(1)

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