# Utf 8Validation

Utf 8Validation bit manipulation

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.
``````

Utf 8Validation Solution

``````public class Utf8Validation {
public boolean validUtf8(int[] data) {
int count = 0;

for(int i : data) {
if(count == 0) {
if((i >> 5) == 0b110) {
count = 1;
} else if((i >> 4) == 0b1110) {
count = 2;
} else if((i >> 3) == 0b11110) {
count = 3;
} else if((i >> 7) == 0b1) {
return false;
}
} else {
if((i >> 6) != 0b10) {
return false;
}

count--;
}
}

return count == 0;
}
}``````