What is the time complexity of Minimum Remove to Make Valid Parentheses?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Minimum Remove to Make Valid Parentheses?

This problem uses String, Stack. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimum Remove to Make Valid Parentheses?

Insight 1: Use a stack to keep track of unmatched open parentheses. The index of each open parenthesis is pushed onto the stack. Insight 2: When encountering a closing parenthesis, check if the stack is empty. If it's not empty, pop an open parenthesis from the stack, indicating a match. If it is empty, this closing parenthesis is unmatched and should be removed.

Which companies ask LeetCode 1249. Minimum Remove to Make Valid Parentheses in interviews?

This problem is frequently asked at GE Digital, Netflix, Snap, Tencent.

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Minimum Remove to Make Valid Parentheses - LeetCode 1249 Solution

Minimum Remove to Make Valid Parentheses - Complete Solution Guide

Minimum Remove to Make Valid Parentheses is LeetCode problem 1249, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given a string s of '(' , ')' and lowercase English characters. Your task is to remove the minimum number of parentheses ( '(' or ')' , in any positions ) so that the resulting parentheses string is valid and return any valid string. Formally, a parentheses string is valid if and only if: It is the empty string, contains only lowercase characters, or It can be written as AB ( A concatenated with B ), where A and B are valid strings, or It can be written as (A) , where A is a valid string. Exampl

Detailed Explanation

The problem asks us to take a string `s` that can contain open parentheses '(', close parentheses ')', and lowercase English letters. The goal is to remove the minimum number of parentheses to make the string a 'valid parentheses string'. A valid parentheses string is defined as either empty, contains only lowercase characters, can be written as AB (A concatenated with B) where A and B are valid strings, or can be written as (A), where A is a valid string. Essentially, every open parenthesis must have a corresponding closing parenthesis, and the closing parenthesis must come after the open one.

Solution Approach

The solution uses a stack to efficiently identify and remove the minimum number of invalid parentheses. It iterates through the string once. When an opening parenthesis is encountered, its index is pushed onto the stack. When a closing parenthesis is encountered, the stack is checked. If the stack is not empty, it means there is a matching open parenthesis, so we pop from the stack. If the stack is empty, the closing parenthesis is invalid and is marked for removal. After the first pass, any remaining open parentheses in the stack are also invalid and are marked for removal. Finally, a new string is constructed by excluding characters marked for removal.

Step-by-Step Algorithm

Step 1: Initialize an empty stack to store the indices of unmatched open parentheses.
Step 2: Convert the input string to a list (or equivalent mutable structure) to allow character modification in place (Python), or create a string builder object (Java). C and C++ examples directly use vectors/arrays for the string representation.
Step 3: Iterate through the string.
Step 4: If the current character is an opening parenthesis '(', push its index onto the stack.
Step 5: If the current character is a closing parenthesis ')', check if the stack is empty. If the stack is not empty, pop an index from the stack (indicating a match). If the stack is empty, mark the current character in the list/builder for removal.
Step 6: After the loop, iterate through the remaining indices in the stack. For each index, mark the corresponding character in the list/builder for removal.
Step 7: Construct a new string by iterating through the modified list/builder and appending only the characters that are not marked for removal.
Step 8: Return the resulting valid string.

Key Insights

Insight 1: Use a stack to keep track of unmatched open parentheses. The index of each open parenthesis is pushed onto the stack.
Insight 2: When encountering a closing parenthesis, check if the stack is empty. If it's not empty, pop an open parenthesis from the stack, indicating a match. If it is empty, this closing parenthesis is unmatched and should be removed.
Insight 3: After iterating through the string, any remaining indices in the stack represent unmatched open parentheses that should also be removed.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: String, Stack.

Companies

Asked at: GE Digital, Netflix, Snap, Tencent.

Minimum Remove to Make Valid Parentheses | LeetCode 1249

Medium

String Stack

Time: O(n)Space: O(n)

Solution Code

Available in:

import java.util.Stack;

class Solution {
    public String minRemoveToMakeValid(String s) {
        StringBuilder sb = new StringBuilder(s);
        Stack<Integer> stack = new Stack<>();

        for (int i = 0; i < sb.length(); i++) {
            if (sb.charAt(i) == '(') {
                stack.push(i);
            } else if (sb.charAt(i) == ')') {
                if (!stack.isEmpty()) {
                    stack.pop();
                } else {
                    sb.setCharAt(i, Character.MIN_VALUE);
                }
            }
        }

        while (!stack.isEmpty()) {
            sb.setCharAt(stack.pop(), Character.MIN_VALUE);
        }

        StringBuilder result = new StringBuilder();
        for (int i = 0; i < sb.length(); i++) {
            if (sb.charAt(i) != Character.MIN_VALUE) {
                result.append(sb.charAt(i));
            }
        }

        return result.toString();
    }
}

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Problem Info

Description

Problem 1249: Minimum Remove to Make Valid Parentheses - Detailed explanation and solution approaches.

Topics

String Stack

Companies

GE DigitalNetflixSnapTencent

Stats

Difficulty:Medium

Languages:4

Companies:4

Problem #:1249

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Initialize an empty stack to store the indices of unmatched open parentheses.
2
Step 2: Convert the input string to a list (or equivalent mutable structure) to allow character modification in place (Python), or create a string builder object (Java). C and C++ examples directly use vectors/arrays for the string representation.
3
Step 3: Iterate through the string.
4
Step 4: If the current character is an opening parenthesis '(', push its index onto the stack.
5
Step 5: If the current character is a closing parenthesis ')', check if the stack is empty. If the stack is not empty, pop an index from the stack (indicating a match). If the stack is empty, mark the current character in the list/builder for removal.
6
Step 6: After the loop, iterate through the remaining indices in the stack. For each index, mark the corresponding character in the list/builder for removal.
7
Step 7: Construct a new string by iterating through the modified list/builder and appending only the characters that are not marked for removal.
8
Step 8: Return the resulting valid string.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is linear because the algorithm iterates through the input string a fixed number of times. The space complexity is linear due to the stack potentially storing up to n indices and the creation of a modified string.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Two-Pass Approach without Stack

Instead of using a stack, iterate through the string twice. In the first pass, remove extra closing parentheses by maintaining a balance counter. In the second pass (from right to left), remove extra opening parentheses.

Time: O(n)Space: O(n) to store the new string

Trade-offs: This approach avoids using a stack, but it involves two passes through the string and may be slightly less intuitive than the stack-based approach. It is also likely to be less efficient in practice, because of the string modification.

✨

Key Insights

1Insight 1: Use a stack to keep track of unmatched open parentheses. The index of each open parenthesis is pushed onto the stack.
2Insight 2: When encountering a closing parenthesis, check if the stack is empty. If it's not empty, pop an open parenthesis from the stack, indicating a match. If it is empty, this closing parenthesis is unmatched and should be removed.
3Insight 3: After iterating through the string, any remaining indices in the stack represent unmatched open parentheses that should also be removed.

📚

Key Terms Explained

For Students

Stack

LIFO structure - Last In, First Out. Like a stack of plates!

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Empty string. The algorithm correctly handles an empty input string by returning an empty string, as an empty string is considered a valid parentheses string.
Edge case 2: String with only closing parentheses (e.g., ")))"). The algorithm will mark all closing parentheses for removal, resulting in an empty string, which is the correct output.
Edge case 3: String with only opening parentheses (e.g., "((("). The algorithm will mark all opening parentheses for removal, resulting in an empty string, which is the correct output.
Edge case 4: String with nested parentheses and invalid characters (e.g., "lee(t(c)o)de)"). The algorithm will correctly identify and remove the minimum number of parentheses to make the string valid.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the edge case where only opening parentheses are present or only closing parentheses are present. This will lead to an unbalanced parenthesis string.

✗Common mistake 2: Removing too many parentheses or removing the wrong parentheses, leading to a string that is still invalid or not the shortest valid string possible. Always ensure to remove the *minimum* number of parenthesis.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Code editors often use parenthesis matching and validation to help developers write correct code. This algorithm can be used as a basis for that validation process.

🏦Real-world scenario 2: Parsing mathematical expressions or programming language syntax. Ensuring balanced parentheses is a crucial step in correctly interpreting these expressions.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 17, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 29

❓

Frequently Asked Questions

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

What is a monotonic stack and when should I use it?

A monotonic stack maintains elements in sorted order (increasing or decreasing). Use it for "next greater element," "largest rectangle," or "stock span" problems. It achieves O(n) time by avoiding redundant comparisons through maintaining order invariants.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Minimum Remove to Make Valid Parentheses"?

For this medium-level String problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether stack techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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