393. UTF-8 Validation (Medium)

A character in UTF8 can be from 1 to 4 bytes long, subjected to the following rules:

1. For 1-byte character, the first bit is a 0, followed by its unicode code.
2. 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.

Solution:

bool validUtf8(vector<int>& data) {
    int cnt = 0;
    for (int& d : data) {
        if (cnt == 0) {
            if ((d >> 3) == 0b11110) cnt = 3;
            else if ((d >> 4) == 0b1110) cnt = 2;
            else if ((d >> 5) == 0b110) cnt = 1;
            else if (d >> 7) return false;
        } else {
            if ((d >> 6) != 0b10) return false;
            --cnt;
        }
    }
    return cnt == 0;
}

bool validUtf8(vector<int>& data) {
    int mask1 = 0b, mask2 = 192, cnt = 0; // 192 = 11000000

    for (int i = 0; i < data.size(); i++) {
        int cur = data[i];

        if (cnt == 0) {                                         // count leading 1s
            while ((cur & mask1) != 0) {
                cur <<= 1;
                cnt++;
            }
            if (cnt == 1) { return false; }                     // cnt cannot be 1
            cnt = max(0, cnt - 1);                              // decrease cnt by 1 if cnt > 1
        } else {                            
            if ((data[i] & mask2) != mask1) { return false; }   // make sure most significant 2 bits are 10
            cnt--;
        }
    }

    return cnt == 0;
}