Minimum Remove to Make Valid Parentheses
Recognize the pattern
Brute force idea
The naive version of Minimum Remove to Make Valid Parentheses sounds like this: Try removing every subset and check validity. Exponential. That direct path helps you understand the question, but it tends to treat every possibility as brand new instead of learning from earlier steps.
Better approach
A calmer way to see Minimum Remove to Make Valid Parentheses is this: Stack of indices: push '(' indices. On ')': if stack has '(', pop (matched). Else mark this ')' for removal. After processing, remaining '(' on stack are also removed. The goal is not to be clever for its own sake, but to remember the one relationship that keeps the solution grounded as you move forward.
Key invariant
The truth you want to protect throughout Minimum Remove to Make Valid Parentheses is this: The stack tracks unmatched '(' indices. Unmatched ')' are detected immediately. At the end, both types of unmatched parentheses are removed. If that remains true after every update, the rest of the reasoning has a stable place to stand.
Watch out for
The trap in Minimum Remove to Make Valid Parentheses usually looks like this: Only removing unmatched ')' — you must also remove unmatched '(' that remain on the stack after processing the entire string. When the code becomes mechanical before the idea is clear, small edge cases start breaking the whole story.