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

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

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

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

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

Insight 1: We can determine the number of extra '(' or ')' needed by iterating through the string and keeping track of the balance between open and close parentheses. Insight 2: We don't need a stack data structure to solve this problem. Instead, we can maintain counters for the number of open parentheses that need to be closed and the number of close parentheses that need to be opened.

Which companies ask LeetCode 921. Minimum Add to Make Parentheses Valid in interviews?

This problem is frequently asked at Nvidia, Siemens, Walmart Labs.

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Minimum Add to Make Parentheses Valid - LeetCode 921 Solution

Minimum Add to Make Parentheses Valid - Complete Solution Guide

Minimum Add to Make Parentheses Valid is LeetCode problem 921, 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

A parentheses string is valid if and only if: It is the empty string, 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. You are given a parentheses string s . In one move, you can insert a parenthesis at any position of the string. For example, if s = "()))" , you can insert an opening parenthesis to be "( ( )))" or a closing parenthesis to be "()) ) )" . Return the minimum number of moves required to make

Detailed Explanation

The problem asks us to determine the minimum number of parentheses (either '(' or ')') we need to add to a given string `s` of parentheses so that the resulting string is a valid parentheses string. A valid parentheses string is defined as either empty, formed by concatenating two valid parentheses strings, or enclosed in parentheses like `(A)` where `A` is another valid parentheses string. In simpler terms, we need to balance the parentheses in the input string by adding the missing opening or closing parentheses.

Solution Approach

The solution uses a greedy approach. We iterate through the string, character by character. If we encounter an opening parenthesis '(', we increment a counter `close_needed`, indicating that we need a corresponding closing parenthesis. If we encounter a closing parenthesis ')', we check if there's an open parenthesis waiting to be closed (i.e., `close_needed > 0`). If so, we decrement `close_needed`. Otherwise, it means this closing parenthesis has no corresponding opening parenthesis, so we increment `open_needed`, indicating that we need to add an opening parenthesis to balance it. Finally, we return the sum of `open_needed` and `close_needed`, which represents the total number of parentheses we need to add.

Step-by-Step Algorithm

Step 1: Initialize two integer variables, `open_needed` and `close_needed`, to 0. These will keep track of the number of missing opening and closing parentheses, respectively.
Step 2: Iterate through the input string `s` character by character.
Step 3: If the current character is '(', increment `close_needed`. This means we need one more closing parenthesis later.
Step 4: If the current character is ')', check if `close_needed` is greater than 0. If it is, decrement `close_needed` because this closing parenthesis has found a matching opening parenthesis. If `close_needed` is 0, it means we have an unmatched closing parenthesis, so increment `open_needed`. This signifies the need for an opening parenthesis.
Step 5: After iterating through the entire string, return the sum of `open_needed` and `close_needed`. This sum represents the minimum number of parentheses that need to be added to make the string valid.

Key Insights

Insight 1: We can determine the number of extra '(' or ')' needed by iterating through the string and keeping track of the balance between open and close parentheses.
Insight 2: We don't need a stack data structure to solve this problem. Instead, we can maintain counters for the number of open parentheses that need to be closed and the number of close parentheses that need to be opened.
Insight 3: The problem can be solved in a single pass through the input string using a greedy approach, making it efficient.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: String, Stack, Greedy.

Companies

Asked at: Nvidia, Siemens, Walmart Labs.

Minimum Add to Make Parentheses Valid | LeetCode 921

Medium

String Stack Greedy

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int minAddToMakeValid(String s) {
        int open_needed = 0;
        int close_needed = 0;

        for (char char_ : s.toCharArray()) {
            if (char_ == '(') {
                close_needed += 1;
            } else if (char_ == ')') {
                if (close_needed > 0) {
                    close_needed -= 1;
                } else {
                    open_needed += 1;
                }
            }
        }
        
        return open_needed + close_needed;
    }
}

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Problem Info

Description

Problem 921: Minimum Add to Make Parentheses Valid - Detailed explanation and solution approaches.

Topics

String Stack Greedy

Companies

NvidiaSiemensWalmart Labs

Stats

Difficulty:Medium

Languages:4

Companies:3

Problem #:921

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

5 steps

1
Step 1: Initialize two integer variables, `open_needed` and `close_needed`, to 0. These will keep track of the number of missing opening and closing parentheses, respectively.
2
Step 2: Iterate through the input string `s` character by character.
3
Step 3: If the current character is '(', increment `close_needed`. This means we need one more closing parenthesis later.
4
Step 4: If the current character is ')', check if `close_needed` is greater than 0. If it is, decrement `close_needed` because this closing parenthesis has found a matching opening parenthesis. If `close_needed` is 0, it means we have an unmatched closing parenthesis, so increment `open_needed`. This signifies the need for an opening parenthesis.
5
Step 5: After iterating through the entire string, return the sum of `open_needed` and `close_needed`. This sum represents the minimum number of parentheses that need to be added to make the string valid.

💡

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(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The algorithm involves iterating through the string `s` once, performing a constant number of operations for each character. Thus, the time complexity is directly proportional to the length of the string. The space complexity is constant as only a few integer variables are used, whose memory usage is independent of the input size.

📖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

Stack-based approach

We could use a stack to keep track of unmatched opening parentheses. When we encounter a closing parenthesis, we check if the stack is empty or if the top of the stack is an opening parenthesis. If the stack is not empty and the top is an opening parenthesis, we pop it from the stack (matching the parentheses). If the stack is empty or the top is not an opening parenthesis, the closing parenthesis is unmatched, and we increment a counter. After processing the string, the number of elements remaining in the stack represents the number of unmatched opening parentheses, which is added to the closing parentheses counter to get the final result.

Time: O(n)Space: O(n) in the worst case (e.g., "((("), where all characters are opening parentheses and are pushed onto the stack.

Trade-offs: The stack-based approach has the same time complexity as the given solution but requires O(n) space in the worst case, whereas the given solution uses only O(1) space. Thus, the given solution is more space-efficient.

✨

Key Insights

1Insight 1: We can determine the number of extra '(' or ')' needed by iterating through the string and keeping track of the balance between open and close parentheses.
2Insight 2: We don't need a stack data structure to solve this problem. Instead, we can maintain counters for the number of open parentheses that need to be closed and the number of close parentheses that need to be opened.
3Insight 3: The problem can be solved in a single pass through the input string using a greedy approach, making it efficient.

📚

Key Terms Explained

For Students

Stack

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

Greedy Algorithm

Making the best choice at each step without looking ahead.

String

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

Edge Cases

Edge case 1: Empty string. The code correctly returns 0 because both `open_needed` and `close_needed` remain 0.
Edge case 2: String with only opening parentheses, e.g., "(((". The code correctly counts the number of missing closing parentheses and returns the correct result. In this case, `open_needed` will be 0 and `close_needed` will be 3, resulting in a total of 3.
Edge case 3: String with only closing parentheses, e.g., ")))". The code correctly counts the number of missing opening parentheses and returns the correct result. In this case, `open_needed` will be 3 and `close_needed` will be 0, resulting in a total of 3.
Edge case 4: String with balanced parentheses, e.g., "(())". The code correctly returns 0 because all parentheses are balanced, `open_needed` and `close_needed` will be 0 at the end.
Edge case 5: String with mixed and unbalanced parentheses, e.g., "()))((". The code correctly identifies and accounts for the unbalanced parentheses, summing the number of needed open and closed parentheses.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where there are more closing parentheses than opening parentheses. For example, in the string "))", the code must correctly identify that two opening parentheses are needed.

✗Common mistake 2: Incorrectly assuming that all opening parentheses must come before all closing parentheses. The solution should work for interleaved patterns like "(()))(()".

✗Common mistake 3: Using recursion, which might lead to stack overflow errors for large input strings. Iterative solutions are generally preferred for this type of problem.

✅

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 and IDEs use parentheses matching algorithms to highlight matching parentheses and detect syntax errors related to mismatched parentheses.

🏦Real-world scenario 2: Compilers and interpreters use similar algorithms to parse code and ensure that parentheses, brackets, and braces are correctly balanced. This is essential for the code to be executed correctly.

🎮Real-world scenario 3: Mathematical expression evaluation. Making sure expressions are syntactically correct before evaluating them.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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: June 22, 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

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

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 7

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 30

❓

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 Add to Make Parentheses Valid"?

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.

🤖 Need help with this problem?