UTF-8 validation¶
Time: O(N); Space: O(N); medium
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.
Notes:
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:
Input: data = [197, 130, 1]
Output: True
Eplanation:
data = [197, 130, 1] represents the octet sequence: 11000101 10000010 00000001
It is a valid utf-8 encoding for a 2-bytes character followed by a 1-byte character.
Example 2:
Input: data = [235, 140, 4]
Output: False
Eplanation:
data = [235, 140, 4] represented the octet sequence: 11101011 10001100 00000100
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.
[1]:
class Solution1(object):
"""
Time: O(N)
Space: O(1)
"""
def validUtf8(self, data) -> bool:
"""
:type data: List[int]
:rtype: bool
"""
count = 0
for c in data:
if count == 0:
if (c >> 5) == 0b110:
count = 1
elif (c >> 4) == 0b1110:
count = 2
elif (c >> 3) == 0b11110:
count = 3
elif (c >> 7):
return False
else:
if (c >> 6) != 0b10:
return False
count -= 1
return count == 0
[2]:
s = Solution1()
data = [197, 130, 1]
assert s.validUtf8(data) == True
data = [235, 140, 4]
assert s.validUtf8(data) == False