Solution: Minimum Remove to Make Valid Parentheses

Explore how to remove invalid parentheses from a string to ensure a valid expression using stack methods. Understand the logic of matching parentheses through a Last In First Out approach, learn to track indices, and implement an optimal solution with O(n) time complexity.

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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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