# Solving Leetcode Interviews in Seconds with AI: UTF-8 Validation


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "393" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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  

	# Explanation
	Here's a breakdown of the solution:

*   **Iterate and Validate:** The core idea is to iterate through the `data` array and, for each byte, determine if it's the start of a UTF-8 character or a continuation byte.
*   **Determine Number of Bytes:** If a byte represents the start of a character, we examine its leading bits to determine the number of bytes that character should occupy.
*   **Check Continuation Bytes:** If the character is multi-byte, we then check that the subsequent bytes are valid continuation bytes (starting with "10").

*   **Runtime Complexity:** O(n), where n is the length of the `data` array.
*   **Storage Complexity:** O(1) - constant extra space.

	
	# Code
	```python
	def validUtf8(data: list[int]) -> bool:
    """
    Determines if a given integer array represents a valid UTF-8 encoding.
    """
    n = len(data)
    i = 0
    while i < n:
        byte = data[i]

        if (byte >> 7) == 0:  # 1-byte character
            i += 1
        elif (byte >> 5) == 0b110:  # 2-byte character
            if i + 1 >= n or (data[i + 1] >> 6) != 0b10:
                return False
            i += 2
        elif (byte >> 4) == 0b1110:  # 3-byte character
            if i + 2 >= n or (data[i + 1] >> 6) != 0b10 or (data[i + 2] >> 6) != 0b10:
                return False
            i += 3
        elif (byte >> 3) == 0b11110:  # 4-byte character
            if i + 3 >= n or (data[i + 1] >> 6) != 0b10 or (data[i + 2] >> 6) != 0b10 or (data[i + 3] >> 6) != 0b10:
                return False
            i += 4
        else:
            return False  # Invalid starting byte

    return True
	```