Introduction

In this blog post, we will explore how to solve the LeetCode problem "2825" 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 two 0-indexed strings str1 and str2. In an operation, you select a set of indices in str1, and for each index i in the set, increment str1[i] to the next character cyclically. That is 'a' becomes 'b', 'b' becomes 'c', and so on, and 'z' becomes 'a'. Return true if it is possible to make str2 a subsequence of str1 by performing the operation at most once, and false otherwise. Note: A subsequence of a string is a new string that is formed from the original string by deleting some (possibly none) of the characters without disturbing the relative positions of the remaining characters. Example 1: Input: str1 = "abc", str2 = "ad" Output: true Explanation: Select index 2 in str1. Increment str1[2] to become 'd'. Hence, str1 becomes "abd" and str2 is now a subsequence. Therefore, true is returned. Example 2: Input: str1 = "zc", str2 = "ad" Output: true Explanation: Select indices 0 and 1 in str1. Increment str1[0] to become 'a'. Increment str1[1] to become 'd'. Hence, str1 becomes "ad" and str2 is now a subsequence. Therefore, true is returned. Example 3: Input: str1 = "ab", str2 = "d" Output: false Explanation: In this example, it can be shown that it is impossible to make str2 a subsequence of str1 using the operation at most once. Therefore, false is returned. Constraints: 1 <= str1.length <= 105 1 <= str2.length <= 105 str1 and str2 consist of only lowercase English letters.

Explanation

Here's the solution:

• Greedy Matching: Iterate through str1 and str2. For each character in str2, try to find a matching character in str1 either directly or after a single increment operation.
• Single Increment Check: For each str2 character, first check if it exists in str1 without any increments (standard subsequence check). If not, check if incrementing a str1 character at the current index makes it equal to the str2 character.

• Early Exit: If at any point, it becomes impossible to match the current str2 character with a str1 character even after incrementing, return False.

• Runtime Complexity: O(n + m), where n is the length of str1 and m is the length of str2.

• Storage Complexity: O(1)

Code

    def is_subsequence_with_increment(str1: str, str2: str) -> bool:
    """
    Checks if str2 is a subsequence of str1 after at most one increment operation.
    """
    i = 0  # Index for str1
    j = 0  # Index for str2

    while i < len(str1) and j < len(str2):
        if str1[i] == str2[j]:
            i += 1
            j += 1
        elif chr(((ord(str1[i]) - ord('a') + 1) % 26) + ord('a')) == str2[j]:
            i += 1
            j += 1

        else:
            i += 1

    return j == len(str2)