# Solving Leetcode Interviews in Seconds with AI: Length of the Longest Alphabetical Continuous Substring


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2414" 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
	> An alphabetical continuous string is a string consisting of consecutive letters in the alphabet. In other words, it is any substring of the string "abcdefghijklmnopqrstuvwxyz".  For example, "abc" is an alphabetical continuous string, while "acb" and "za" are not.  Given a string s consisting of lowercase letters only, return the length of the longest alphabetical continuous substring.   Example 1:  Input: s = "abacaba" Output: 2 Explanation: There are 4 distinct continuous substrings: "a", "b", "c" and "ab". "ab" is the longest continuous substring.  Example 2:  Input: s = "abcde" Output: 5 Explanation: "abcde" is the longest continuous substring.    Constraints:  1 <= s.length <= 105 s consists of only English lowercase letters.  

	# Explanation
	Here's the breakdown of the approach, complexities, and the Python code for the solution:

*   **High-Level Approach:**
    *   Iterate through the input string `s`.
    *   Maintain a `current_length` of the current continuous substring and a `max_length` to store the longest continuous substring found so far.
    *   If the current character extends the continuous substring (i.e., `ord(s[i]) == ord(s[i-1]) + 1`), increment `current_length`. Otherwise, reset `current_length` to 1. Update `max_length` in each iteration.

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

	
	# Code
	```python
	def longest_continuous_substring(s: str) -> int:
    """
    Given a string s consisting of lowercase letters only, return the length of the longest alphabetical continuous substring.
    """
    if not s:
        return 0

    max_length = 1
    current_length = 1

    for i in range(1, len(s)):
        if ord(s[i]) == ord(s[i - 1]) + 1:
            current_length += 1
        else:
            current_length = 1
        max_length = max(max_length, current_length)

    return max_length
	```