# 393 UTF-8 Validation

### Problem:

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
--------------------+---------------------------------------------
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.
``````
``````public class Solution {
public boolean validUtf8(int[] data) {
int count = 0;
for (int i = 0; i < data.length; i ++) {
if (count == 0) {
if (data[i]>>>5 == 0b110) {
count = 1;
}
else if (data[i]>>>4 == 0b1110) {
count = 2;
}
else if (data[i]>>>3 == 0b11110) {
count = 3;
}
else if (data[i]>>>7 == 1) {
return false;
}
}
else {
if (data[i]>>>6 != 2) {
return false;
}
count --;
}
}
return count == 0;
}
}
``````