What is the time complexity of Maximum Nesting Depth of the Parentheses?

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

What data structures are used to solve Maximum Nesting Depth of the Parentheses?

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

What are the key insights for solving Maximum Nesting Depth of the Parentheses?

Insight 1: We only need to track the count of opening parentheses. The problem's constraint of a valid parentheses string means that a closing parenthesis always corresponds to a previously encountered opening parenthesis. Insight 2: A simple counter is sufficient to solve this problem. No sophisticated data structures like stacks are strictly necessary, although a stack-based solution would also work.

Which companies ask LeetCode 1614. Maximum Nesting Depth of the Parentheses in interviews?

This problem is frequently asked at Intel.

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Maximum Nesting Depth of the Parentheses - LeetCode 1614 Solution

Q: How to solve Maximum Nesting Depth of the Parentheses?

The provided solutions utilize a simple iterative approach. They traverse the input string character by character. For each character, if it's an opening parenthesis '(', the `current_depth` counter is incremented. Simultaneously, `max_depth` (tracking the maximum depth seen so far) is updated to be the maximum of `max_depth` and `current_depth`. If the character is a closing parenthesis ')', `current_depth` is decremented (assuming it's greater than 0 to handle potentially erroneous input though the problem statement guarantees valid input). Finally, `max_depth` is returned, representing the maximum nesting depth.

Maximum Nesting Depth of the Parentheses - Complete Solution Guide

Maximum Nesting Depth of the Parentheses is LeetCode problem 1614, a Easy 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 valid parentheses string s , return the nesting depth of s . The nesting depth is the maximum number of nested parentheses. Example 1: Input: s = "(1+(2*3)+((8)/4))+1" Output: 3 Explanation: Digit 8 is inside of 3 nested parentheses in the string. Example 2: Input: s = "(1)+((2))+(((3)))" Output: 3 Explanation: Digit 3 is inside of 3 nested parentheses in the string. Example 3: Input: s = "()(())((()()))" Output: 3 Constraints: 1 <= s.length <= 100 s consists of digits 0-9 and characters

Detailed Explanation

The problem asks to find the maximum nesting depth of parentheses in a given string. The input string `s` is guaranteed to be a valid parentheses string, meaning that every opening parenthesis '(' has a matching closing parenthesis ')'. The nesting depth at any point is the number of open parentheses encountered without encountering a matching closing parenthesis. The output is the maximum nesting depth found anywhere in the string. For example, in the string '(1+(2*3)+((8)/4))+1', the maximum nesting depth is 3 because the digit '8' is enclosed within three nested pairs of parentheses.

Solution Approach

The provided solutions utilize a simple iterative approach. They traverse the input string character by character. For each character, if it's an opening parenthesis '(', the `current_depth` counter is incremented. Simultaneously, `max_depth` (tracking the maximum depth seen so far) is updated to be the maximum of `max_depth` and `current_depth`. If the character is a closing parenthesis ')', `current_depth` is decremented (assuming it's greater than 0 to handle potentially erroneous input though the problem statement guarantees valid input). Finally, `max_depth` is returned, representing the maximum nesting depth.

Step-by-Step Algorithm

Step 1: Initialize `max_depth` and `current_depth` to 0.
Step 2: Iterate through each character in the input string `s`.
Step 3: If the character is an opening parenthesis '(', increment `current_depth` and update `max_depth` to be the maximum of `max_depth` and `current_depth`.
Step 4: If the character is a closing parenthesis ')', decrement `current_depth` (only if `current_depth > 0` to prevent negative depth).
Step 5: After iterating through all characters, return `max_depth`.

Key Insights

Insight 1: We only need to track the count of opening parentheses. The problem's constraint of a valid parentheses string means that a closing parenthesis always corresponds to a previously encountered opening parenthesis.
Insight 2: A simple counter is sufficient to solve this problem. No sophisticated data structures like stacks are strictly necessary, although a stack-based solution would also work.
Insight 3: The maximum depth is updated only when an opening parenthesis is encountered. Closing parentheses simply decrement the counter. This allows for a single pass solution.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: String, Stack.

Companies

Asked at: Intel.

Maximum Nesting Depth of the Parentheses | LeetCode 1614

Easy

String Stack

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

Solution Code

Available in:

class Solution {
    public int maxDepth(String s) {
        int maxDepth = 0;
        int currentDepth = 0;
        for (char c : s.toCharArray()) {
            if (c == '(') {
                currentDepth++;
                maxDepth = Math.max(maxDepth, currentDepth);
            } else if (c == ')') {
                currentDepth--;
            }
        }
        return maxDepth;
    }
}

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Problem Info

Description

Problem 1614: Maximum Nesting Depth of the Parentheses - Detailed explanation and solution approaches.

Topics

String Stack

Companies

Intel

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:1614

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 `max_depth` and `current_depth` to 0.
2
Step 2: Iterate through each character in the input string `s`.
3
Step 3: If the character is an opening parenthesis '(', increment `current_depth` and update `max_depth` to be the maximum of `max_depth` and `current_depth`.
4
Step 4: If the character is a closing parenthesis ')', decrement `current_depth` (only if `current_depth > 0` to prevent negative depth).
5
Step 5: After iterating through all characters, return `max_depth`.

💡

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 iterates through the input string once, performing constant-time operations for each character. Therefore, the time complexity is linear with respect to the length of the string (n). The space complexity is constant because only a few integer variables are used to store the depth, 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

Use a stack to store opening parentheses. When an opening parenthesis is encountered, push it onto the stack. When a closing parenthesis is encountered, pop from the stack. The maximum stack size at any point represents the maximum nesting depth.

Time: O(n)Space: O(n) in the worst case (nested parentheses throughout the string)

Trade-offs: The stack-based approach is less efficient in space complexity compared to the counter-based approach, especially for strings with deep nesting. The counter-based approach is simpler and more efficient in space, utilizing O(1) auxiliary space.

✨

Key Insights

1Insight 1: We only need to track the count of opening parentheses. The problem's constraint of a valid parentheses string means that a closing parenthesis always corresponds to a previously encountered opening parenthesis.
2Insight 2: A simple counter is sufficient to solve this problem. No sophisticated data structures like stacks are strictly necessary, although a stack-based solution would also work.
3Insight 3: The maximum depth is updated only when an opening parenthesis is encountered. Closing parentheses simply decrement the counter. This allows for a single pass solution.

📚

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 solution correctly handles an empty string by returning 0.
Edge case 2: String with no parentheses. The solution correctly returns 0 as there's no nesting.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the case where a closing parenthesis is encountered before an opening one (though this is not possible in the problem constraint). A robust solution might check for `current_depth > 0` before decrementing.

✗Common mistake 2: Incorrectly updating `max_depth`. It's crucial to update `max_depth` only when an opening parenthesis is encountered and after `current_depth` is incremented.

✅

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 balanced parentheses in code or mathematical expressions. This algorithm forms a core part of syntax checking and error detection in compilers and interpreters.

🏦Real-world scenario 2: Analyzing nested data structures like JSON or XML. Determining the depth of nested elements can be useful for parsing and data manipulation tasks.

💡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

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ 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 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

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

❓

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.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

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 "Maximum Nesting Depth of the Parentheses"?

For this easy-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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