Solution: Minimum Remove to Make Valid Parentheses

Discover how to use a stack data structure to remove the minimum number of invalid parentheses from a string. This lesson guides you through the process of identifying unmatched parentheses and efficiently validating parenthesization using a Last In First Out approach. You will learn to implement this with optimal time and space complexity.

We'll cover the following...

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

...