Lexicographically Minimum String After Removing Stars

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.

For example:

s = "aaba*" should return "aab"

s = "abc" should return "abc"

def remove_stars(s: str) -> str:
    """Removes '*' characters from the string s according to the given rules.

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

    Returns:
        The lexicographically smallest resulting string after removing all '*' characters.
    """
    stack = []
    for char in s:
        if char == '*':
            if stack:
                stack.pop()
        else:
            stack.append(char)
    return "".join(stack)

# Example usage:
# s = "aaba*"
# print(remove_stars(s))  # Output: "aab"
# s = "abc"
# print(remove_stars(s))  # Output: "abc"

Explanation:

The provided code uses a stack to efficiently process the string and remove the * characters along with the smallest non-* character to its left. Here's a breakdown of how it works:

1. Initialization:

   • An empty list called stack is initialized. This stack will store the characters that will eventually form the result string.

2. Iteration:

   • The code iterates through each character char in the input string s.

3. Handling * characters:

   • If the current character char is an asterisk ('*'), the code checks if the stack is not empty.
   • If the stack is not empty, it means there's a character to the left of the * that can be removed. The pop() operation removes the last element added to the stack, effectively deleting the smallest non-"*" character to its left. The problem statement allows us to remove any of the smallest characters if multiple exist, and the stack handles this implicitly by just removing the top character.
   • If the stack is empty when encountering a *, it implies there are no characters to the left, and the * is effectively ignored (as there's nothing to remove).

4. Handling non-* characters:

   • If the current character char is not an asterisk, it's simply appended to the stack.

5. Result:

   • After processing the entire string, the stack contains the characters of the resulting string. The "".join(stack) method concatenates all the characters in the stack into a single string, which is then returned.

Big(O) Runtime Analysis:

The code iterates through the input string s once. The append() and pop() operations on the stack take constant time, O(1). The join() operation takes O(k) time, where k is the number of characters in the stack which is bound by the length of the input string s. Therefore, the overall time complexity is O(n), where n is the length of the input string s.

Big(O) Space Analysis:

The stack can, in the worst case (when there are no * characters in the input string), store all the characters of the input string. Therefore, the space complexity is O(n), where n is the length of the input string s.

Edge Cases:

1. Empty string: If the input string s is empty, the code will correctly return an empty string because the loop will not execute.

2. String with only * characters: If the input string contains only * characters, the stack will remain empty after the loop, and the code will return an empty string.

3. String with no * characters: If the input string contains no * characters, all characters will be added to the stack, and the code will return the original string.

4. Multiple consecutive * characters: The code correctly handles multiple consecutive * characters by popping multiple times from the stack (if the stack is not empty). If the stack is empty, they're ignored.

5. * at the beginning of the string: If the string starts with a *, and stack is empty, then that * will be ignored. No errors.