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Grokking the Coding Interview Patterns in Go

Minimum Remove to Make Valid Parentheses

Explore how to fix strings by removing the fewest parentheses needed to achieve valid parenthesization using stacks. This lesson helps you understand the problem constraints, identify unmatched parentheses, and apply stack-based solutions to solve interview-style questions in Go.

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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:

Go
usercode > main.go
package main



func minRemoveParentheses(s string) string {

   

    // Replace this placeholder return statement with your code

    return ""

}
Minimum Remove to Make Valid Parentheses