Introduction

In this blog post, we will explore how to solve the LeetCode problem "467" 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

We define the string base to be the infinite wraparound string of "abcdefghijklmnopqrstuvwxyz", so base will look like this: "...zabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcd....". Given a string s, return the number of unique non-empty substrings of s are present in base. Example 1: Input: s = "a" Output: 1 Explanation: Only the substring "a" of s is in base. Example 2: Input: s = "cac" Output: 2 Explanation: There are two substrings ("a", "c") of s in base. Example 3: Input: s = "zab" Output: 6 Explanation: There are six substrings ("z", "a", "b", "za", "ab", and "zab") of s in base. Constraints: 1 <= s.length <= 105 s consists of lowercase English letters.

Explanation

Here's the approach, complexity analysis, and Python code:

• Identify Relevant Substrings: The crucial idea is that for each position in the input string s, we only need to consider the longest substring ending at that position that is a substring of the infinite base string.
• Dynamic Programming (Implicit): We'll iterate through the string, maintaining a max_len array (or dictionary to be memory-efficient) that tracks the maximum length of a substring ending at each character 'a' to 'z'. This avoids redundant substring checks.

• Count Unique Substrings: Finally, we sum up the values in the max_len structure to get the total number of unique substrings.

• Time Complexity: O(n), where n is the length of the string s.

• Space Complexity: O(1). We only need space for 26 characters.

Code

    def find_substrings_in_wraparound_string(s: str) -> int:
    """
    Calculates the number of unique non-empty substrings of s that are present in the infinite wraparound string.

    Args:
        s: The input string consisting of lowercase English letters.

    Returns:
        The number of unique non-empty substrings of s that are present in the base.
    """

    max_len = {}  # Dictionary to store the max length of substring ending with each character
    current_len = 0

    for i in range(len(s)):
        if i > 0 and (ord(s[i]) - ord(s[i - 1]) == 1 or (s[i - 1] == 'z' and s[i] == 'a')):
            current_len += 1
        else:
            current_len = 1

        char = s[i]
        max_len[char] = max(max_len.get(char, 0), current_len)

    return sum(max_len.values())