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Minimum Remove to Make Valid Parentheses

Understand how to identify and remove the minimum number of invalid parentheses in a string to create a valid representation. Explore the use of stack data structures to methodically solve this problem, improving your skills in parentheses validation challenges common in coding interviews.

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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≤1 \leq s.length ≤103\leq 10^3
  • s[i] is either an opening parenthesis (( , a closing parenthesiss )), or a lowercase English letter.

Examples

Understand the problem

Let’s take a moment to make sure you’ve correctly understood the problem. The quiz below helps you check if you’re solving the correct problem:

Minimum Remove to Make Valid Parentheses

1.

What is the output if the following string is given as input?

“(((abc)(to)((q)()(”

A.

“(((abc)(to)((q)()”

B.

“(abc)(to)((q)()”

C.

“(abc)(to)(q)()”

D.

“((abc)(to)((q)()”

1 / 4

Figure it out!

We have a game for you to play. Rearrange the logical building blocks to develop a clearer understanding of how to solve this problem.

Sequence - Vertical
Drag and drop the cards to rearrange them in the correct sequence.
1
2
3
4

Try it yourself

Implement your solution in the following coding playground:

Java
usercode > Main.java
import java.util.*;



public class Main{

    public static String minRemoveParentheses(String s) {



        // Replace this placeholder return statement with your code

        return "";

    }

}
Minimum Remove to Make Valid Parentheses