393. UTF-8 Validation

393. UTF-8 Validation

Question

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
      (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.

Solutions:

• Method 1: String + Recursion

  class Solution {
      private int[] data;
      public boolean validUtf8(int[] data) {
          this.data = data;
          return valid(0);
      }
      public boolean valid(int index){
          if(index == data.length) return true;
          String cur = Integer.toBinaryString(data[index]);
          if(cur.length() < 8 || cur.startsWith("0")){
              return valid(index + 1);
          }else if(cur.startsWith("10")){
              return false;
          }else{
              int ind = cur.indexOf("0");
              if(ind > 4 || ind == -1 || ind + index > data.length) return false;
              index++;
              for(int i = index; i < index + ind - 1; i++){
                  String s = Integer.toBinaryString(data[i]);
                  if(s.length() < 8 || !s.startsWith("10")) return false;
              }
              return valid(index + ind - 1);
          }
      }
  }
  

• Method 2: Bit operation + recursion

  class Solution {
      private int[] data;
      public boolean validUtf8(int[] data) {
          this.data = data;
          return valid(0);
      }
      private boolean valid(int index){
          if(index == data.length) return true;
          int cur = data[index];
          if((cur & (1 << 7)) == 0){
              return valid(index + 1);
          }else if((cur & (3 << 6)) == 0b10000000){
              return false;
          }else{
              int count = 0;
              int mask = 0b10000000;
              while((cur & mask) > 0){
                  count++;
                  mask >>= 1;
              }
              if(count > 4 || index + count > data.length) return false;
              for(int i = index + 1; i < index + count; i++){
                  if((data[i] & (3 << 6)) != 0b10000000) return false;
              }
              return valid(index + count);
          }
      }
  }