What is the time complexity of Check if a Parentheses String Can Be Valid?

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

What data structures are used to solve Check if a Parentheses String Can Be 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 Check if a Parentheses String Can Be Valid?

Insight 1: A parentheses string can be valid only if its length is even. Insight 2: We can use a greedy approach to determine if a string can be valid by simulating the process of balancing parentheses from left to right and from right to left.

Which companies ask LeetCode 2116. Check if a Parentheses String Can Be Valid in interviews?

This problem is frequently asked at ServiceNow.

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Check if a Parentheses String Can Be Valid - LeetCode 2116 Solution

Check if a Parentheses String Can Be Valid - Complete Solution Guide

Check if a Parentheses String Can Be Valid is LeetCode problem 2116, 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 a non-empty string consisting only of '(' and ')' . It is valid if any of the following conditions is true : It is () . It can be written as AB ( A concatenated with B ), where A and B are valid parentheses strings. It can be written as (A) , where A is a valid parentheses string. You are given a parentheses string s and a string locked , both of length n . locked is a binary string consisting only of '0' s and '1' s. For each index i of locked , If locked[i] is '1' , you

Detailed Explanation

The problem asks us to determine if a given parentheses string `s` can be made valid, given another string `locked` of the same length. If `locked[i]` is '1', then `s[i]` cannot be changed. If `locked[i]` is '0', then `s[i]` can be changed to either '(' or ')'. A valid parentheses string satisfies the following: it's either "()", or can be written as AB where A and B are valid, or can be written as (A) where A is valid.

Solution Approach

The solution uses a greedy approach with two passes. First, it checks if the length of the string is even. If not, it's impossible to make the string valid. Then, it iterates through the string from left to right, maintaining a `balance` counter. When `locked[i]` is '0' or `s[i]` is '(', we increment `balance`, representing a potential opening parenthesis. When `locked[i]` is '1' and `s[i]` is ')', we decrement `balance`, representing a closing parenthesis. If `balance` becomes negative, it means we have more closing parentheses than opening parentheses at some point, and it's impossible to balance the string up to that point, so we return false. After the first pass, we iterate from right to left, doing the same thing but treating ')' as an opening parenthesis and '(' as a closing parenthesis. This ensures that we can balance the string in both directions.

Step-by-Step Algorithm

Step 1: Check if the length of the string `s` is even. If not, return `false` because an odd-length string cannot be a valid parentheses string.
Step 2: Initialize a `balance` counter to 0.
Step 3: Iterate through the string `s` from left to right (index `i` from 0 to n-1).
Step 4: Inside the loop, check if `locked[i]` is '0' (meaning `s[i]` can be changed) or `s[i]` is '('. If either condition is true, increment `balance`.
Step 5: Otherwise (if `locked[i]` is '1' and `s[i]` is ')'), decrement `balance`.
Step 6: After each increment or decrement, check if `balance` is negative. If it is, return `false` because it means there are too many closing parentheses before encountering enough opening parentheses to balance them.
Step 7: Reset the `balance` counter to 0.
Step 8: Iterate through the string `s` from right to left (index `i` from n-1 to 0).
Step 9: Inside the loop, check if `locked[i]` is '0' or `s[i]` is ')'. If either condition is true, increment `balance`.
Step 10: Otherwise (if `locked[i]` is '1' and `s[i]` is '('), decrement `balance`.
Step 11: After each increment or decrement, check if `balance` is negative. If it is, return `false` for the same reason as step 6.
Step 12: If both loops complete without returning `false`, return `true` because the string can be made valid.

Key Insights

Insight 1: A parentheses string can be valid only if its length is even.
Insight 2: We can use a greedy approach to determine if a string can be valid by simulating the process of balancing parentheses from left to right and from right to left.
Insight 3: The '0's in the `locked` string represent flexible positions that can be treated as either '(' or ')' to help balance the string.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: String, Stack, Greedy.

Companies

Asked at: ServiceNow.

Check if a Parentheses String Can Be Valid | LeetCode 2116

Medium

String Stack Greedy

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

Solution Code

Available in:

class Solution {
    public boolean canBeValid(String s, String locked) {
        int n = s.length();
        if (n % 2 != 0) {
            return false;
        }

        int balance = 0;
        for (int i = 0; i < n; i++) {
            if (locked.charAt(i) == '0' || s.charAt(i) == '(') {
                balance += 1;
            } else {
                balance -= 1;
            }
            if (balance < 0) {
                return false;
            }
        }

        balance = 0;
        for (int i = n - 1; i >= 0; i--) {
            if (locked.charAt(i) == '0' || s.charAt(i) == ')') {
                balance += 1;
            } else {
                balance -= 1;
            }
            if (balance < 0) {
                return false;
            }
        }
        
        return true;
    }
}

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Problem Info

Description

Problem 2116: Check if a Parentheses String Can Be Valid - Detailed explanation and solution approaches.

Topics

String Stack Greedy

Companies

ServiceNow

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2116

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

12 steps

1
Step 1: Check if the length of the string `s` is even. If not, return `false` because an odd-length string cannot be a valid parentheses string.
2
Step 2: Initialize a `balance` counter to 0.
3
Step 3: Iterate through the string `s` from left to right (index `i` from 0 to n-1).
4
Step 4: Inside the loop, check if `locked[i]` is '0' (meaning `s[i]` can be changed) or `s[i]` is '('. If either condition is true, increment `balance`.
5
Step 5: Otherwise (if `locked[i]` is '1' and `s[i]` is ')'), decrement `balance`.
6
Step 6: After each increment or decrement, check if `balance` is negative. If it is, return `false` because it means there are too many closing parentheses before encountering enough opening parentheses to balance them.
7
Step 7: Reset the `balance` counter to 0.
8
Step 8: Iterate through the string `s` from right to left (index `i` from n-1 to 0).
9
Step 9: Inside the loop, check if `locked[i]` is '0' or `s[i]` is ')'. If either condition is true, increment `balance`.
10
Step 10: Otherwise (if `locked[i]` is '1' and `s[i]` is '('), decrement `balance`.
11
Step 11: After each increment or decrement, check if `balance` is negative. If it is, return `false` for the same reason as step 6.
12
Step 12: If both loops complete without returning `false`, return `true` because the string can be made 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's time complexity is determined by the two linear scans of the input string. The space complexity is constant because the algorithm uses only a fixed number of variables, regardless 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

Instead of using a simple balance counter, you could potentially use a stack to keep track of opening parentheses. However, the flexibility offered by the '0's in the locked string makes it more complex to handle using a stack directly. This wouldn't inherently be better in terms of time or space complexity.

Time: O(n)Space: O(n) in the worst case if the stack holds all opening parentheses.

Trade-offs: The stack approach is generally less efficient and more complex for this particular problem due to the need to handle unlocked positions. The greedy approach provides a simpler and more direct solution.

✨

Key Insights

1Insight 1: A parentheses string can be valid only if its length is even.
2Insight 2: We can use a greedy approach to determine if a string can be valid by simulating the process of balancing parentheses from left to right and from right to left.
3Insight 3: The '0's in the `locked` string represent flexible positions that can be treated as either '(' or ')' to help balance the string.

📚

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 or string with a single character. The length check at the beginning handles these cases, as a single-character string will always have odd length and return false.
Edge case 2: String with only locked parentheses that are unbalanced, e.g., s = "(()))", locked = "11111". The algorithm correctly returns false because it cannot change any parentheses to balance them.
Edge case 3: String with many unlocked positions that can be used to balance the string, e.g., s = ")))))", locked = "000000". The algorithm correctly returns true as it allows us to make enough changes

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not checking for the even length of the string. An odd-length string can never be a valid parentheses string.

✗Common mistake 2: Only iterating from left to right and not considering the right-to-left balance. Failing to check the right-to-left balance might result in incorrectly returning true for strings that are only balanced from the left but not from the right.

✗Common mistake 3: Forgetting to reset the balance counter between the left-to-right and right-to-left iterations.

✅

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: Validating code syntax in text editors or IDEs. Ensuring that parentheses, brackets, and braces are properly balanced is crucial for code execution.

🏦Real-world scenario 2: Parsing mathematical expressions or configuration files. Correctly interpreting expressions requires balanced delimiters.

💡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: October 15, 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

❓

Frequently Asked Questions

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

What problems are best solved with a stack?

Stacks excel at: matching problems (parentheses validation), next greater/smaller element, expression evaluation, backtracking, and undo operations. The key pattern is LIFO (Last In, First Out) - when you need to access the most recently added element.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Check if a Parentheses String Can Be 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.

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