Solution.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.


"""


class Solution(object):
    def validUtf8(self, data):
        """
        :type data: List[int]
        :rtype: bool
        """
        num_follow = 0
        for n in data:
            s = bin(n)[2:].rjust(8, '0')
            if s.startswith('0'):
                if num_follow > 0:
                    return False
                continue
            elif s.startswith('110'):
                if num_follow > 0:
                    return False
                num_follow = 1
            elif s.startswith('1110'):
                if num_follow > 0:
                    return False
                num_follow = 2
            elif s.startswith('11110'):
                if num_follow > 0:
                    return False
                num_follow = 3
            elif s.startswith('10'):
                if num_follow == 0:
                    return False
                num_follow -= 1
            else:
                return False
        return num_follow == 0