Introduction

In this blog post, we will explore how to solve the LeetCode problem "3170" 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. It may contain any number of '' characters. Your task is to remove all '' characters. While there is a '', do the following operation: Delete the leftmost '' and the smallest non-'' character to its left. If there are several smallest characters, you can delete any of them. Return the lexicographically smallest resulting string after removing all '' characters. Example 1: Input: s = "aaba" Output: "aab" Explanation: We should delete one of the 'a' characters with ''. If we choose s[3], s becomes the lexicographically smallest. Example 2: Input: s = "abc" Output: "abc" Explanation: There is no '' in the string. Constraints: 1 <= s.length <= 105 s consists only of lowercase English letters and ''. The input is generated such that it is possible to delete all '*' characters.

Explanation

Here's the breakdown of the solution:

• Iterate Backwards: Process the string from right to left. This allows efficient identification of characters to keep.
• Stack for Non-Star Characters: Use a stack to maintain the characters that will be part of the final result, ensuring the lexicographically smallest order.

• Conditional Popping from Stack: When encountering a '', pop characters from the stack that are smaller than or equal to the current character available for deletion to the left of the ''.

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

• Storage Complexity: O(n)

Code

    def remove_stars(s: str) -> str:
    """
    Removes '*' characters from the string s and returns the lexicographically smallest resulting string.
    """

    stack = []
    for char in reversed(s):
        if char == '*':
            continue
        else:
            stack.append(char)
            while len(stack) > 1 and stack[-2] == '*':
                star_index = len(stack) - 2
                curr_char = stack[-1]

                # Find the smallest character that can be deleted before star
                smallest_to_left = curr_char # Initialize it to current character, in case the stack doesn't have anything to the left of the '*'

                #Delete the leftmost '*' and the smallest non-'*' character to its left.
                stack.pop()
                stack.pop(star_index)
    return "".join(reversed(stack))