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utf-8-validation.py
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utf-8-validation.py
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# Time: O(n)
# Space: O(1)
# 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 Solution(object):
def validUtf8(self, data):
"""
: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