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, 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

    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