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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:

  1. For a 1-byte character, the first bit is a 0, followed by its Unicode code.
  2. 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 \(O(1)\).

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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

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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;
    }
}

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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;
    }
};

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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
}

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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;
}

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