text: """Leetcode 393. UTF-8 Validation
  Medium
  
  URL: 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.
  """
  
  class SolutionBitShiftNumRhsBytes(object):
      def validUtf8(self, data):
          """
          :type data: List[int]
          :rtype: bool
  
          Time complexity: O(n), where n is length of data.
          Space complexity: O(1).
          """
          n_rhs_bytes = 0
  
          for d in data:
              # In 1st iteration check number of RHS bytes.
              if n_rhs_bytes == 0:
                  if d >> 3 == 0b11110:
                      n_rhs_bytes = 3
                  elif d >> 4 == 0b1110:
                      n_rhs_bytes = 2
                  elif d >> 5 == 0b110:
                      n_rhs_bytes = 1
                  elif d >> 7 != 0:
                      return False
              else:
                  # For RHS bytes, check if marker matches pattern.
                  if n_rhs_bytes < 0 or d >> 6 != 0b10:
                      return False
                  n_rhs_bytes -= 1
  
          return n_rhs_bytes == 0
  
  
  def main():
      # Output: True
      data = [197, 130, 1]
      print SolutionBitShiftNumRhsBytes().validUtf8(data)
  
      # Output: False
      data = [235, 140, 4]
      print SolutionBitShiftNumRhsBytes().validUtf8(data)
  
  
  if __name__ == '__main__':
      main()
  