393. UTF-8 Validation

Problem

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, 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:

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 * 10^4

• 0 <= data[i] <= 255

Solution

class Solution {
    public boolean validUtf8(int[] data) {
        int count = 0;
        for (int d : data) {
            if (count == 0) {
                if (d >> 5 == 0b110) {
                    count = 1;
                } else if (d >> 4 == 0b1110) {
                    count = 2;
                } else if (d >> 3 == 0b11110) {
                    count = 3;
                } else if (d >> 7 == 1) {
                    return false;
                }
            } else {
                if (d >> 6 != 0b10) {
                    return false;
                } else {
                    count--;
                }
            }
        }
        return count == 0;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).