Introduction

In this blog post, we will explore how to solve the LeetCode problem "3316" 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 source of size n, a string pattern that is a subsequence of source, and a sorted integer array targetIndices that contains distinct numbers in the range [0, n - 1]. We define an operation as removing a character at an index idx from source such that: idx is an element of targetIndices. pattern remains a subsequence of source after removing the character. Performing an operation does not change the indices of the other characters in source. For example, if you remove 'c' from "acb", the character at index 2 would still be 'b'. Return the maximum number of operations that can be performed. Example 1: Input: source = "abbaa", pattern = "aba", targetIndices = [0,1,2] Output: 1 Explanation: We can't remove source[0] but we can do either of these two operations: Remove source[1], so that source becomes "a_baa". Remove source[2], so that source becomes "ab_aa". Example 2: Input: source = "bcda", pattern = "d", targetIndices = [0,3] Output: 2 Explanation: We can remove source[0] and source[3] in two operations. Example 3: Input: source = "dda", pattern = "dda", targetIndices = [0,1,2] Output: 0 Explanation: We can't remove any character from source. Example 4: Input: source = "yeyeykyded", pattern = "yeyyd", targetIndices = [0,2,3,4] Output: 2 Explanation: We can remove source[2] and source[3] in two operations. Constraints: 1 <= n == source.length <= 3 * 103 1 <= pattern.length <= n 1 <= targetIndices.length <= n targetIndices is sorted in ascending order. The input is generated such that targetIndices contains distinct elements in the range [0, n - 1]. source and pattern consist only of lowercase English letters. The input is generated such that pattern appears as a subsequence in source.

Explanation

Here's the solution, addressing the problem's core aspects for efficiency and clarity:

• Greedy Approach with Two Pointers: Iterate through targetIndices. For each index, tentatively remove the character at that index from source. Then, use two pointers to check if pattern is still a subsequence of the modified source.
• Subsequence Check Optimization: The subsequence check is done efficiently by incrementing the pattern pointer only when a match is found in the source.

• Early Exit: If removing a character makes pattern not a subsequence, skip to the next index in targetIndices.

• Time Complexity: O(m * n), where n is the length of targetIndices and m is the length of source.

• Space Complexity: O(m), where m is the length of source (due to string slicing).

Code

    def max_operations(source: str, pattern: str, targetIndices: list[int]) -> int:
    """
    Calculates the maximum number of operations (removals) that can be performed such that
    the pattern remains a subsequence of the source after each removal.
    """
    count = 0
    n = len(source)

    for index in targetIndices:
        temp_source = source[:index] + source[index + 1:]

        pattern_ptr = 0
        source_ptr = 0

        while pattern_ptr < len(pattern) and source_ptr < len(temp_source):
            if pattern[pattern_ptr] == temp_source[source_ptr]:
                pattern_ptr += 1
            source_ptr += 1

        if pattern_ptr == len(pattern):
            source = temp_source
            count += 1
        else:
            pass # Pattern is not a subsequence after removal, so skip

    return count