Introduction

In this blog post, we will explore how to solve the LeetCode problem "1717" 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 and two integers x and y. You can perform two types of operations any number of times. Remove substring "ab" and gain x points. For example, when removing "ab" from "cabxbae" it becomes "cxbae". Remove substring "ba" and gain y points. For example, when removing "ba" from "cabxbae" it becomes "cabxe". Return the maximum points you can gain after applying the above operations on s. Example 1: Input: s = "cdbcbbaaabab", x = 4, y = 5 Output: 19 Explanation: - Remove the "ba" underlined in "cdbcbbaaabab". Now, s = "cdbcbbaaab" and 5 points are added to the score. - Remove the "ab" underlined in "cdbcbbaaab". Now, s = "cdbcbbaa" and 4 points are added to the score. - Remove the "ba" underlined in "cdbcbbaa". Now, s = "cdbcba" and 5 points are added to the score. - Remove the "ba" underlined in "cdbcba". Now, s = "cdbc" and 5 points are added to the score. Total score = 5 + 4 + 5 + 5 = 19. Example 2: Input: s = "aabbaaxybbaabb", x = 5, y = 4 Output: 20 Constraints: 1 <= s.length <= 105 1 <= x, y <= 104 s consists of lowercase English letters.

Explanation

Here's a breakdown of the solution:

• Prioritization: The core idea is to prioritize removing the substring that yields the most points (either "ab" or "ba") first.
• Stack-based Processing: Use stacks to efficiently identify and remove occurrences of "ab" and "ba" while keeping track of potential removals.

• Iterative Optimization: Iteratively process the string, removing the more valuable substring until no more removals are possible, then move on to removing the less valuable substring.

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

• Storage Complexity: O(n) in the worst case, due to the stacks.

Code

    def max_points(s: str, x: int, y: int) -> int:
    """
    Calculates the maximum points achievable by removing "ab" and "ba" substrings.
    """

    score = 0
    if x > y:
        pattern1, pattern2 = "ab", "ba"
        score1, score2 = x, y
    else:
        pattern1, pattern2 = "ba", "ab"
        score1, score2 = y, x

    stack = []
    for char in s:
        stack.append(char)
        while len(stack) >= 2:
            top = stack[-1]
            second_top = stack[-2]
            if second_top + top == pattern1:
                score += score1
                stack.pop()
                stack.pop()
            else:
                break

    temp_string = "".join(stack)
    stack = []

    for char in temp_string:
        stack.append(char)
        while len(stack) >= 2:
            top = stack[-1]
            second_top = stack[-2]
            if second_top + top == pattern2:
                score += score2
                stack.pop()
                stack.pop()
            else:
                break

    return score