Introduction

In this blog post, we will explore how to solve the LeetCode problem "3412" 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

You are given a string s. We define the mirror of a letter in the English alphabet as its corresponding letter when the alphabet is reversed. For example, the mirror of 'a' is 'z', and the mirror of 'y' is 'b'. Initially, all characters in the string s are unmarked. You start with a score of 0, and you perform the following process on the string s: Iterate through the string from left to right. At each index i, find the closest unmarked index j such that j < i and s[j] is the mirror of s[i]. Then, mark both indices i and j, and add the value i - j to the total score. If no such index j exists for the index i, move on to the next index without making any changes. Return the total score at the end of the process. Example 1: Input: s = "aczzx" Output: 5 Explanation: i = 0. There is no index j that satisfies the conditions, so we skip. i = 1. There is no index j that satisfies the conditions, so we skip. i = 2. The closest index j that satisfies the conditions is j = 0, so we mark both indices 0 and 2, and then add 2 - 0 = 2 to the score. i = 3. There is no index j that satisfies the conditions, so we skip. i = 4. The closest index j that satisfies the conditions is j = 1, so we mark both indices 1 and 4, and then add 4 - 1 = 3 to the score. Example 2: Input: s = "abcdef" Output: 0 Explanation: For each index i, there is no index j that satisfies the conditions. Constraints: 1 <= s.length <= 105 s consists only of lowercase English letters.

Explanation

Here's the breakdown of the solution:

• Iterate and Search: Iterate through the string, and for each character, search for the nearest unmarked mirror character to its left.
• Mark and Score: If a suitable mirror is found, mark both indices and update the total score.

• Efficient Lookup: Use an array/list to keep track of the unmarked status of characters.

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

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

Code

    def mirror_string(s: str) -> int:
    """
    Calculates the total score based on the given process on the string s.

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

    Returns:
        The total score at the end of the process.
    """

    n = len(s)
    marked = [False] * n
    score = 0

    for i in range(n):
        if marked[i]:
            continue

        for j in range(i - 1, -1, -1):
            if not marked[j] and chr(219 - ord(s[i])) == s[j]:
                marked[i] = True
                marked[j] = True
                score += i - j
                break

    return score