Solution: Minimum Remove to Make Valid Parentheses

Explore how to remove the minimum number of invalid parentheses from a string to make it valid by using stack techniques. Understand the stack's Last In First Out approach to identify and delete unmatched parentheses, and learn the time and space complexity implications of this method.

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Statement
Solution
- Time complexity
- Space complexity

Statement

Given a string, s, that may have matchedEach opening parenthesis, (, has a closing parenthesis, ), for it. and unmatchedThere is no corresponding closing parenthesis, ), for an opening parenthesis, (. parentheses, remove the minimum number of parentheses so that the resulting string represents a valid parenthesization. For example, the string “a(b)” represents a valid parenthesization while the string “a(b” doesn’t, since the opening parenthesis doesn’t have any corresponding closing parenthesis.

Constraints:

$1 \leq$ s.length

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