What is the time complexity of Longest Valid Parentheses?

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

What data structures are used to solve Longest Valid Parentheses?

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

What are the key insights for solving Longest Valid Parentheses?

Insight 1: A stack can be used to keep track of the positions of unmatched opening parentheses. When a closing parenthesis is encountered, we can try to match it with an opening parenthesis from the stack. Insight 2: The key is to maintain the index of the last unmatched ')' as the base of our current valid parenthesis string. This is achieved by initializing the stack with -1.

Which companies ask LeetCode 32. Longest Valid Parentheses in interviews?

This problem is frequently asked at Adobe, Amazon, Bloomberg, DE Shaw, InMobi and 14 more companies.

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Longest Valid Parentheses - LeetCode 32 Solution

Q: How to solve Longest Valid Parentheses?

The provided solution uses a stack to keep track of the indices of unmatched opening parentheses. The algorithm iterates through the string. If it encounters an opening parenthesis, it pushes its index onto the stack. If it encounters a closing parenthesis, it pops an element from the stack. If the stack becomes empty after the pop, it means we've closed all the previously opened parentheses, and the current closing parenthesis might be the start of a new valid substring. In this case, we push the index of the current closing parenthesis onto the stack. If the stack is not empty after the pop, it means there's still an unmatched opening parenthesis in the stack. We calculate the length of the current valid substring as the difference between the current index and the index at the top of the stack. We keep track of the maximum length encountered so far.

Longest Valid Parentheses - Complete Solution Guide

Longest Valid Parentheses is LeetCode problem 32, a Hard 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 containing just the characters '(' and ')' , return the length of the longest valid (well-formed) parentheses substring . Example 1: Input: s = "(()" Output: 2 Explanation: The longest valid parentheses substring is "()". Example 2: Input: s = ")()())" Output: 4 Explanation: The longest valid parentheses substring is "()()". Example 3: Input: s = "" Output: 0 Constraints: 0 <= s.length <= 3 * 10 4 s[i] is '(' , or ')' .

Detailed Explanation

The problem asks us to find the length of the longest substring within a given string that contains only '(' and ')' characters, where the substring forms a valid (well-formed) parentheses sequence. A valid parentheses sequence means that each opening parenthesis '(' has a corresponding closing parenthesis ')' to its right, and they are properly nested.

Solution Approach

The provided solution uses a stack to keep track of the indices of unmatched opening parentheses. The algorithm iterates through the string. If it encounters an opening parenthesis, it pushes its index onto the stack. If it encounters a closing parenthesis, it pops an element from the stack. If the stack becomes empty after the pop, it means we've closed all the previously opened parentheses, and the current closing parenthesis might be the start of a new valid substring. In this case, we push the index of the current closing parenthesis onto the stack. If the stack is not empty after the pop, it means there's still an unmatched opening parenthesis in the stack. We calculate the length of the current valid substring as the difference between the current index and the index at the top of the stack. We keep track of the maximum length encountered so far.

Step-by-Step Algorithm

Step 1: Initialize a stack with -1 to represent the index before the string starts. This serves as a base for the initial valid substring.
Step 2: Iterate through the input string `s` from left to right.
Step 3: If the current character is an opening parenthesis '(', push its index onto the stack.
Step 4: If the current character is a closing parenthesis ')', pop an element from the stack.
Step 5: After popping, check if the stack is empty. If it is, it means the popped ')' closed the last valid sequence, so push the current index `i` onto the stack. This marks the start of a new potentially valid sequence after this ')'
Step 6: If the stack is not empty after popping, calculate the length of the current valid substring by subtracting the index at the top of the stack from the current index `i` (current_len = i - stack.top()).
Step 7: Update the `max_len` with the maximum value seen so far: max_len = max(max_len, current_len)
Step 8: After iterating through the entire string, return `max_len`.

Key Insights

Insight 1: A stack can be used to keep track of the positions of unmatched opening parentheses. When a closing parenthesis is encountered, we can try to match it with an opening parenthesis from the stack.
Insight 2: The key is to maintain the index of the last unmatched ')' as the base of our current valid parenthesis string. This is achieved by initializing the stack with -1.
Insight 3: When the stack becomes empty after popping a ')' it means that the current ')' closes all the previously opened '(', and we need to push the index of this ')' onto the stack to mark a new base for subsequent valid substrings.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: String, Dynamic Programming, Stack.

Companies

Asked at: Adobe, Amazon, Bloomberg, DE Shaw, InMobi, Intuit, MakeMyTrip, Meta, Microsoft, Oracle, SOTI, Salesforce, Sprinklr, TikTok, Uber, Yahoo, Zeta, Zoho, eBay.

Longest Valid Parentheses | LeetCode 32

Hard

String Dynamic Programming Stack

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

Solution Code

Available in:

import java.util.Stack;

class Solution {
    public int longestValidParentheses(String s) {
        int maxLen = 0;
        Stack<Integer> stack = new Stack<>();
        stack.push(-1);

        for (int i = 0; i < s.length(); i++) {
            char charAtI = s.charAt(i);
            if (charAtI == '(') {
                stack.push(i);
            } else {
                stack.pop();
                if (stack.isEmpty()) {
                    stack.push(i);
                } else {
                    int currentLen = i - stack.peek();
                    maxLen = Math.max(maxLen, currentLen);
                }
            }
        }

        return maxLen;
    }
}

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Problem Info

Description

Problem 32: Longest Valid Parentheses - Detailed explanation and solution approaches.

Topics

String Dynamic Programming Stack

Companies

AdobeAmazonBloombergDE ShawInMobiIntuitMakeMyTripMeta

Stats

Difficulty:Hard

Languages:4

Companies:19

Problem #:32

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 a stack with -1 to represent the index before the string starts. This serves as a base for the initial valid substring.
2
Step 2: Iterate through the input string `s` from left to right.
3
Step 3: If the current character is an opening parenthesis '(', push its index onto the stack.
4
Step 4: If the current character is a closing parenthesis ')', pop an element from the stack.
5
Step 5: After popping, check if the stack is empty. If it is, it means the popped ')' closed the last valid sequence, so push the current index `i` onto the stack. This marks the start of a new potentially valid sequence after this ')'
6
Step 6: If the stack is not empty after popping, calculate the length of the current valid substring by subtracting the index at the top of the stack from the current index `i` (current_len = i - stack.top()).
7
Step 7: Update the `max_len` with the maximum value seen so far: max_len = max(max_len, current_len)
8
Step 8: After iterating through the entire string, return `max_len`.

💡

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 O(n) because we iterate through the string once to process each character. The space complexity is O(n) in the worst case because the stack could store up to n/2 indices in the case of a string with many consecutive opening parentheses.

📖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

Dynamic Programming

Another approach is to use dynamic programming. Create a DP array `dp` where `dp[i]` represents the length of the longest valid parentheses substring ending at index `i`. Iterate through the string. If `s[i]` is ')', check if `s[i-1]` is '('. If yes, `dp[i] = dp[i-2] + 2`. If `s[i]` is ')' and `s[i-1]` is ')', check if `s[i - dp[i-1] - 1]` is '('. If so, `dp[i] = dp[i-1] + 2 + dp[i - dp[i-1] - 2]`.

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

Trade-offs: DP might be slightly less intuitive to understand than the stack-based approach. Stack based is generally preferred due to readability. DP can potentially be faster in some implementations due to constant factor differences.

Two Pointers

We can use two counters, `left` and `right`, to count the number of '(' and ')' respectively. First, we move from left to right. If `left == right`, we found a valid substring. If `right > left`, we reset both to 0 because there are more closing parenthesis which makes the sequence invalid. We repeat the process moving from right to left. This is done to handle cases like '(()'.

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

Trade-offs: Two pointers has the best space complexity O(1). It can be slightly harder to understand than the stack based solution. But has the advantage of not using any extra space.

✨

Key Insights

1Insight 1: A stack can be used to keep track of the positions of unmatched opening parentheses. When a closing parenthesis is encountered, we can try to match it with an opening parenthesis from the stack.
2Insight 2: The key is to maintain the index of the last unmatched ')' as the base of our current valid parenthesis string. This is achieved by initializing the stack with -1.
3Insight 3: When the stack becomes empty after popping a ')' it means that the current ')' closes all the previously opened '(', and we need to push the index of this ')' onto the stack to mark a new base for subsequent valid substrings.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

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 returns 0 because max_len is initialized to 0 and the loop doesn't execute.
Edge case 2: String with only opening parentheses - The algorithm returns 0 because it never finds matching parentheses. The stack will contain all the indices but the max length will not change.
Edge case 3: String with only closing parentheses - The algorithm returns 0, as the stack will always be empty (or only contain the index of the last closing parenthesis) and no valid pairs will be formed. It correctly handles multiple closing parenthesis at the start due to the initial -1 on the stack.
Edge case 4: String with mismatched parentheses (e.g., '(()') - The algorithm correctly calculates the length of the longest valid substring, which is 2 ('()').

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where the stack becomes empty after popping a closing parenthesis. This requires remembering to add the index of current ')' onto the stack.

✗Common mistake 2: Failing to initialize the stack with -1. This initial value ensures that we can correctly calculate the length of the substring when a valid pair is found at the beginning of the string.

✗Common mistake 3: Incorrectly calculating the length of the valid substring. It is calculated by subtracting the index at the top of the stack from the current index.

✗Common mistake 4: Not updating the max_len correctly.

✗Common mistake 5: Forgetting to handle edge cases like an empty string, or a string with only open or only closing brackets.

✅

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: Syntax validation in compilers and interpreters. Checking if parentheses, brackets, and braces are properly balanced in code.

🏦Real-world scenario 2: Text editors and IDEs use this logic to highlight matching parentheses and identify errors in code.

🎮Real-world scenario 3: Mathematical expression parsing: Determining the validity and structure of mathematical expressions involving parentheses.

💡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

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ 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: July 3, 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

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

@InterviewProJul 13

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

@CodeNinjaJul 6

❓

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.

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

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 "Longest Valid Parentheses"?

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

🤖 Need help with this problem?