# Solving Leetcode Interviews in Seconds with AI: Removing Stars From a String


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2390" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), 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, which contains stars *. In one operation, you can:  Choose a star in s. Remove the closest non-star character to its left, as well as remove the star itself.  Return the string after all stars have been removed. Note:  The input will be generated such that the operation is always possible. It can be shown that the resulting string will always be unique.    Example 1:  Input: s = "leet**cod*e" Output: "lecoe" Explanation: Performing the removals from left to right: - The closest character to the 1st star is 't' in "leet**cod*e". s becomes "lee*cod*e". - The closest character to the 2nd star is 'e' in "lee*cod*e". s becomes "lecod*e". - The closest character to the 3rd star is 'd' in "lecod*e". s becomes "lecoe". There are no more stars, so we return "lecoe". Example 2:  Input: s = "erase*****" Output: "" Explanation: The entire string is removed, so we return an empty string.    Constraints:  1 <= s.length <= 105 s consists of lowercase English letters and stars *. The operation above can be performed on s.  

	# Explanation
	Here's the approach to solve the problem:

*   **Stack-based approach:** Use a stack (implemented using a Python list) to keep track of non-star characters.
*   **Iterate and process:** Iterate through the string. If a character is not a star, push it onto the stack. If it's a star, pop the last element from the stack (if the stack is not empty).
*   **Result construction:** After processing the entire string, the stack will contain the remaining non-star characters. Join them to form the final result.

*   **Runtime Complexity:** O(n), where n is the length of the string. **Storage Complexity:** O(n) in the worst case (when there are no stars in the input string).

	
	# Code
	```python
	def remove_stars(s: str) -> str:
    """
    Removes stars and their preceding non-star characters from a string.

    Args:
        s: The input string containing stars and lowercase English letters.

    Returns:
        The string after all stars have been removed.
    """
    stack = []
    for char in s:
        if char == '*':
            if stack:
                stack.pop()
        else:
            stack.append(char)
    return "".join(stack)
	```