393 UTF-8 Validation.py

"""
A character in UTF8 can be from 1 to 4 bytes long, subjected to the following rules:

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

Note:
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.

Example 1:

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.
Example 2:

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.
Author: Rajeev Ranjan

"""

class Solution(object):
    def validUtf8(self, data):
        """
        starts with 0, then skip
        start with 1, check number of numbers
        :type data: List[int]
        :rtype: bool
        """
        required = 0
        for d in data:
            if d & 0x80 == 0:
                if required != 0:
                    return False
            else:
                one_cnt = 0
                while d & 0x80 == 0x80:
                    one_cnt += 1
                    d <<= 1

                if required != 0:
                    if one_cnt != 1:
                        return False
                    required -= 1
                else:
                    if one_cnt == 1:
                        return False
                    required += (one_cnt - 1)

        return required == 0


if __name__ == "__main__":
    assert Solution().validUtf8([197, 130, 1]) == True
    assert Solution().validUtf8([235, 140, 4]) == False