393. UTF-8 Validation
Description
Given an integer array data
representing the data, return whether it is a valid UTF-8 encoding (i.e. it translates to a sequence of valid UTF-8 encoded characters).
A character in UTF8 can be from 1 to 4 bytes long, subjected to the following rules:
- For a 1-byte character, the first bit is a
0
, followed by its Unicode code. - For an n-bytes character, the first
n
bits are all one's, then + 1
bit is0
, followed byn - 1
bytes with the most significant2
bits being10
.
This is how the UTF-8 encoding would work:
Number of Bytes | UTF-8 Octet Sequence | (binary) --------------------+----------------------------------------- 1 | 0xxxxxxx 2 | 110xxxxx 10xxxxxx 3 | 1110xxxx 10xxxxxx 10xxxxxx 4 | 11110xxx 10xxxxxx 10xxxxxx 10xxxxxx
x
denotes a bit in the binary form of a byte that may be either 0
or 1
.
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:
Input: data = [197,130,1] Output: true Explanation: data 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 Explanation: data 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.
Constraints:
1 <= data.length <= 2 * 104
0 <= data[i] <= 255
Solution
utf-8-validation.py
def is_k_bytes_long(data, index, k):
if k == 1:
k = 0
return data[index] >> (7 - k) == ((1 << k) - 1) << 1 and all(
index2 < len(data)
and data[index2] >> 6 == 0b10
for index2 in range(index + 1, index + k)
)
class Solution:
def validUtf8(self, data: List[int]) -> bool:
index = 0
while index < len(data):
for k in range(1, 5):
if is_k_bytes_long(data, index, k):
index += k
break
else:
return False
return True