Leetcode 393. UTF-8 Validation

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dendiz
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Leetcode 393. UTF-8 Validation

Post by dendiz » Fri Jan 18, 2019 8:56 am

Bit magic question, highly unlikely but one of my friends did get this question at a tech screen with google.
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:

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   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.

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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.

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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.
I actually didn't see the note about the input being the last significant bits, so I took me some trial error to realize that. So the numbers will be
in the range of 0 - 255.

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    boolean testBit(int n, int x) {
        return ((n >>> x) & 1) == 1;
    }
this utility function returns true if the Xth bit of integer N is set.
Next we just need to probe for the rules.

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        int fourByteMask =  0b000000000000000000000011110000;
        int threeByteMask = 0b000000000000000000000011100000;
        int twoByteMask =   0b000000000000000000000011000000;
        int nextMask =      0b000000000000000000000010000000;
these masks will test the 2-4 byte characters.

Code: Select all

int req = 0;
        int i =0;
        while(i < data.length) {
            int cur = data[i];
            if ((fourByteMask & cur) == fourByteMask && !testBit(cur, 3)) {
                req = 3;
            }
            else if ((threeByteMask & cur) == threeByteMask && !testBit(cur, 4)) {
                req = 2;
            }

            else if ((twoByteMask & cur) == twoByteMask && !testBit(cur, 5)) {
                req = 1;
            }
            else {
                req = 0;
            }
            
            if (req > 0) {
                while(req > 0) {
                    i++;
                    if (i > data.length - 1) return false;
                    boolean c = testBit(data[i], 7) && !testBit(data[i], 6);
                    if (!c) return false;
                    req--;
                }
                i++;
            } else {
                if (testBit(cur, 7)) return false;
                i++;
            }
            
        }
        return true;

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