leetcode 393 - UTF-8 Validation

https://leetcode.com/problems/utf-8-validation/

Given an integer array data representing the data, return whether it is a valid UTF-8 encoding.

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, the n + 1 bit is 0, followed by n - 1 bytes with the 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
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 * 10^4
0 <= data[i] <= 255

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class Solution {
public:
bool validUtf8(vector<int>& data) {
int n = data.size();
int bytes_to_process = 0;

for (int i = 0; i < n; i++) {
if (bytes_to_process == 0) {
int bits = get_leading_bits(data[i]);
if (bits == 1 || bits > 4) return false;

if (bits)
bytes_to_process = bits - 1;
} else {
if (get_leading_bits(data[i]) != 1)
return false;
bytes_to_process--;
}
}

return bytes_to_process == 0;
}

int get_leading_bits(int num) {
int bit = 0;

for (int i = 7; i >= 0; i--) {
if ((num & (1 << i)) == 0)
break;
bit++;
}

return bit;
}
};