Valid Parenthesis String
Given a string with (, ), and *****, where ***** can be (, ), or empty, return True if the string is valid.
Example:
(*)
True
- The input string is
(*), containing one*which can be treated as an empty string, an open parenthesis(, or a close parenthesis). - Considering the
*as an empty string, the string becomes `` (empty string), which is a valid parenthesis string since it doesn't contain any unmatched parentheses. - Alternatively, the
*can also be treated as a(or a), resulting in strings(or), but since there's no matching counterpart, these cases would not be valid on their own. - However, because the
*can be considered as an empty string, the overall string(*)is considered valid, leading to the outputTrue.
Constraints:
- 1 <= len(s) <= 100
Background Knowledge
The problem "Valid Parenthesis String" involves stacks, a fundamental data structure in computer science. A stack is a Last-In-First-Out (LIFO) data structure, meaning the last element added to the stack will be the first one to be removed. In the context of this problem, we can use a stack to keep track of the opening parentheses ( and the wildcard character *****, which can be treated as either an opening or closing parenthesis.
To solve this problem, it's essential to understand the concept of valid parentheses. A string of parentheses is considered valid if every opening parenthesis can be matched with a corresponding closing parenthesis. The wildcard character ***** adds complexity to this problem, as it can be treated as either an opening or closing parenthesis, or even an empty string. This means we need to consider all possible interpretations of the ***** character when checking the validity of the string.
The problem also involves string manipulation and character iteration, as we need to iterate through the input string and process each character accordingly. Understanding how to iterate through a string and perform actions based on the current character is crucial to solving this problem.
Algorithm/Approach
The general approach to solving this type of problem involves using a two-pointer technique or a stack-based approach. The two-pointer technique involves using two pointers to track the opening and closing parentheses, while the stack-based approach involves using a stack to keep track of the opening parentheses and the wildcard character *****. We can also use a greedy approach, where we try to match the opening parentheses with the closing parentheses as soon as possible.
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