human
I found an interesting problem on https://leetcode.com/problems/utf-8-validation/:
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.

I tried it in Python, but could not do it. Can you solve it?
assistant
#Here is the Python program you requested.

class Solution:
     def validUtf8(self, data):
         """
         :type data: List[int]
         :rtype: bool
         """
         count = 0
         for d in data:
             if count == 0:
                 if d >> 5 == 0b110:
                     count = 1
                 elif d >> 4 == 0b1110:
                     count = 2
                 elif d >> 3 == 0b11110:
                     count = 3
                 elif d >> 7 != 0:
                     return False
             else:
                 if d >> 6 == 0b10:
                     count -= 1
                 else:
                     return False
         return count == 0