393. UTF-8 Validation 

Description

Given an integer array data representing the data, return whether it is a valid UTF-8 encoding (i.e. it translates to a sequence of valid UTF-8 encoded characters).

A character in UTF8 can be from 1 to 4 bytes long, subjected to the following rules:

For a 1-byte character, the first bit is a 0, followed by its Unicode code.
For an n-bytes character, the first n bits are all one's, the n + 1 bit is 0, followed by n - 1 bytes with the most significant 2 bits being 10.

This is how the UTF-8 encoding would work:

     Number of Bytes   |        UTF-8 Octet Sequence
                       |              (binary)
   --------------------+-----------------------------------------
            1          |   0xxxxxxx
            2          |   110xxxxx 10xxxxxx
            3          |   1110xxxx 10xxxxxx 10xxxxxx
            4          |   11110xxx 10xxxxxx 10xxxxxx 10xxxxxx

x denotes a bit in the binary form of a byte that may be either 0 or 1.

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:

Input: data = [197,130,1]
Output: true
Explanation: data represents the octet sequence: 11000101 10000010 00000001.
It is a valid utf-8 encoding for a 2-bytes character followed by a 1-byte character.

Example 2:

Input: data = [235,140,4]
Output: false
Explanation: data represented the octet sequence: 11101011 10001100 00000100.
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.

Constraints:

1 <= data.length <= 2 * 10⁴
0 <= data[i] <= 255

Solutions

Solution 1: Single Pass

We use a variable $cnt$ to record the current number of bytes that need to be filled starting with $10$, initially $cnt = 0$.

For each integer $v$ in the array:

If $cnt > 0$, then check if $v$ starts with $10$. If not, return false, otherwise decrement $cnt$.
If the highest bit of $v$ is $0$, then $cnt = 0$.
If the highest two bits of $v$ are $110$, then $cnt = 1$.
If the highest three bits of $v$ are $1110$, then $cnt = 2$.
If the highest four bits of $v$ are $11110$, then $cnt = 3$.
Otherwise, return false.

Finally, if $cnt = 0$, return true, otherwise return false.

The time complexity is $O(n)$, where $n$ is the length of the array data. The space complexity is $O(1)$.

Python3

class Solution:
    def validUtf8(self, data: List[int]) -> bool:
        cnt = 0
        for v in data:
            if cnt > 0:
                if v >> 6 != 0b10:
                    return False
                cnt -= 1
            elif v >> 7 == 0:
                cnt = 0
            elif v >> 5 == 0b110:
                cnt = 1
            elif v >> 4 == 0b1110:
                cnt = 2
            elif v >> 3 == 0b11110:
                cnt = 3
            else:
                return False
        return cnt == 0

Java

class Solution {
    public boolean validUtf8(int[] data) {
        int cnt = 0;
        for (int v : data) {
            if (cnt > 0) {
                if (v >> 6 != 0b10) {
                    return false;
                }
                --cnt;
            } else if (v >> 7 == 0) {
                cnt = 0;
            } else if (v >> 5 == 0b110) {
                cnt = 1;
            } else if (v >> 4 == 0b1110) {
                cnt = 2;
            } else if (v >> 3 == 0b11110) {
                cnt = 3;
            } else {
                return false;
            }
        }
        return cnt == 0;
    }
}

C++

class Solution {
public:
    bool validUtf8(vector<int>& data) {
        int cnt = 0;
        for (int& v : data) {
            if (cnt > 0) {
                if (v >> 6 != 0b10) {
                    return false;
                }
                --cnt;
            } else if (v >> 7 == 0) {
                cnt = 0;
            } else if (v >> 5 == 0b110) {
                cnt = 1;
            } else if (v >> 4 == 0b1110) {
                cnt = 2;
            } else if (v >> 3 == 0b11110) {
                cnt = 3;
            } else {
                return false;
            }
        }
        return cnt == 0;
    }
};

Go

func validUtf8(data []int) bool {
	cnt := 0
	for _, v := range data {
		if cnt > 0 {
			if v>>6 != 0b10 {
				return false
			}
			cnt--
		} else if v>>7 == 0 {
			cnt = 0
		} else if v>>5 == 0b110 {
			cnt = 1
		} else if v>>4 == 0b1110 {
			cnt = 2
		} else if v>>3 == 0b11110 {
			cnt = 3
		} else {
			return false
		}
	}
	return cnt == 0
}

TypeScript

function validUtf8(data: number[]): boolean {
    let cnt = 0;
    for (const v of data) {
        if (cnt > 0) {
            if (v >> 6 !== 0b10) {
                return false;
            }
            --cnt;
        } else if (v >> 7 === 0) {
            cnt = 0;
        } else if (v >> 5 === 0b110) {
            cnt = 1;
        } else if (v >> 4 === 0b1110) {
            cnt = 2;
        } else if (v >> 3 === 0b11110) {
            cnt = 3;
        } else {
            return false;
        }
    }
    return cnt === 0;
}