393-UTF8Validation.swift

// 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.

class Solution {
    func validUtf8(_ data: [Int]) -> Bool {
        var octetSeq = data.map { i -> String in
            var s = String(i, radix: 2)
            if s.count < 8 { s = String(repeating: "0", count: 8 - s.count) + s }
            return s
        }
        
        var i = 0
        while i < octetSeq.count {  // loop thru each utf-8 char
            var byteCount = 0
            while byteCount < 8 && octetSeq[i][byteCount] == "1" { byteCount += 1 } // get # of leading 1's in first byte
            // byteCount should be 0, 2, 3, 4. Start byte has 10xxxx is invalid
            if byteCount == 1 || byteCount > 4 || i + byteCount - 1 >= octetSeq.count { return false }
            if byteCount > 0 {
                for j in i + 1 ..< i + byteCount {  // loop thru rest bytes in a char
                    if !octetSeq[j].hasPrefix("10") { return false }
                }
                i += byteCount
            } else {     // char with single byte
                i += 1
            }
        }
        return true
    }
}

extension String {
    subscript (i: Int) -> Character {
        return self[index(startIndex, offsetBy: i)]
    }
}