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.

Source: leetcode | Difficulty: MEDIUM_HARD
Tags: Bit manipulation

Data structures

Starter code: class Solution:
    def validUtf8(self, data: List[int]) -> bool:
        

Solutions:
class Solution:

	def validUtf8(self, data):
		count = 0
		for x in data:
			if count == 0:
				if x >> 5 == 6:
					count = 1
				elif x >> 4 == 14:
					count = 2
				elif x >> 3 == 30:
					count = 3
				elif x >> 7 == 1:
					return False
			else:
				if x >> 6 != 2:
					return False
				count -= 1
		return count == 0


class Solution:

	def countLeadingOne(self, w):
		if w[0] == '0':
			return 1
		elif w[0:3] == '110':
			return 2
		elif w[0:4] == '1110':
			return 3
		elif w[0:5] == '11110':
			return 4
		else:
			return -1

	def checkStartWith10(self, L, l):
		if len(L) != l:
			return False
		for w in L:
			if w.startswith('10') == False:
				return False
		return True

	def validUtf8(self, data):
		A = []
		for d in data:
			A.append(format(d, '08b'))
		i = 0
		while i < len(A):
			l = self.countLeadingOne(A[i])
			if l == -1:
				return False
			if l > 1:
				if self.checkStartWith10(A[i + 1:i + l], l - 1) == False:
					return False
			i += l
		return True


class Solution:

	def validUtf8(self, data):
		n = len(data)
		check10 = 0
		for byte in data:
			if check10:
				if byte & 192 != 128:
					return False
				check10 -= 1
			elif byte & 248 == 240:
				check10 = 3
			elif byte & 240 == 224:
				check10 = 2
			elif byte & 224 == 192:
				check10 = 1
			elif byte & 128 == 0:
				continue
			else:
				return False
		return check10 == 0


class Solution:

	def check(self, nums, start, size):
		for i in range(start + 1, start + size + 1):
			if i >= len(nums) or nums[i] >> 6 != 2:
				return False
		return True

	def validUtf8(self, nums, start=0):
		while start < len(nums):
			first = nums[start]
			if first >> 3 == 30 and self.check(nums, start, 3):
				start += 4
			elif first >> 4 == 14 and self.check(nums, start, 2):
				start += 3
			elif first >> 5 == 6 and self.check(nums, start, 1):
				start += 2
			elif first >> 7 == 0:
				start += 1
			else:
				return False
		return True


class Solution:

	def validUtf8(self, data):

		def numOfFollowingBytes(num):
			if 0 <= num <= 127:
				return 0
			elif 192 <= num <= 223:
				return 1
			elif 224 <= num <= 239:
				return 2
			elif 240 <= num <= 247:
				return 3
			else:
				return -1

		def isFollowingByte(num):
			return 128 <= num <= 191
		if not data:
			return False
		(i, n) = (0, len(data))
		while i < n:
			bytesToFollow = numOfFollowingBytes(data[i])
			if bytesToFollow == -1:
				return False
			i += 1
			if i + bytesToFollow > n:
				return False
			for _ in range(bytesToFollow):
				if not isFollowingByte(data[i]):
					return False
				i += 1
		return True


class Solution:

	def validUtf8(self, data):
		cont = 0
		for n in data:
			s = format(n, '08b')
			if cont > 0:
				if s[:2] == '10':
					cont -= 1
					continue
				return False
			if s[0] == '0':
				continue
			if s[:3] == '110':
				cont = 1
				continue
			if s[:4] == '1110':
				cont = 2
				continue
			if s[:5] == '11110':
				cont = 3
				continue
			return False
		return cont == 0


class Solution:

	def validUtf8(self, data):
		seqs = len(data)
		idx = 0
		while idx < seqs:
			number = data[idx] & 255 | 256
			seq = bin(number)[3:]
			bits = len(seq.split('0')[0])
			if idx + bits > seqs:
				return False
			elif bits > 4:
				return False
			elif bits == 0:
				idx += 1
			elif bits == 1:
				return False
			else:
				for i in range(1, bits):
					num = data[idx + i] & 255
					if num >= int('11000000', 2) or num < int('10000000', 2):
						return False
				idx += bits
		return True


class Solution:

	def validUtf8(self, data):
		current = 0
		for byte in data:
			b = '{0:08b}'.format(byte)
			if current == 0:
				cnt = 0
				for i in b:
					if i == '0':
						break
					else:
						cnt += 1
				if cnt == 1 or cnt > 4:
					return False
				if cnt > 0:
					current = cnt - 1
				else:
					current = 0
			else:
				if b[0:2] != '10':
					return False
				current -= 1
		if current > 0:
			return False
		return True