What is the time complexity of Generate Parentheses?

The optimal solution has a time complexity of O(4^n/sqrt(n)).

What is the space complexity of Generate Parentheses?

The space complexity is O(4^n/sqrt(n)).

What data structures are used to solve Generate Parentheses?

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

What are the key insights for solving Generate Parentheses?

Insight 1: The core idea is to use backtracking. Backtracking allows us to explore all possible combinations by recursively adding either an open or a close parenthesis at each step, while ensuring we maintain the well-formed property. Insight 2: We need to keep track of the number of open and close parentheses used so far. This is crucial for determining when to add an open parenthesis (only if open_count < n) or a close parenthesis (only if close_count < open_count).

Which companies ask LeetCode 22. Generate Parentheses in interviews?

This problem is frequently asked at Adobe, Airtel, Amazon, Amdocs, Apple and 33 more companies.

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Generate Parentheses - LeetCode 22 Solution

Q: How to solve Generate Parentheses?

The solution uses a backtracking algorithm. It maintains a stack (or string in C/C++) to build the parentheses string incrementally. At each step, it considers adding either an open or a close parenthesis. The conditions for adding each are crucial: we can only add an open parenthesis if the number of open parentheses is less than 'n', and we can only add a close parenthesis if the number of close parentheses is less than the number of open parentheses. When we have added 'n' open and 'n' close parentheses, we have a valid combination, and we add it to the result list. The stack is then popped to backtrack and try other possibilities.

Q: Which companies ask LeetCode 22. Generate Parentheses in interviews?

This problem is frequently asked at Adobe, Airtel, Amazon, Amdocs, Apple and 33 more companies.

Generate Parentheses - Complete Solution Guide

Generate Parentheses is LeetCode problem 22, 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 n pairs of parentheses, write a function to generate all combinations of well-formed parentheses . Example 1: Input: n = 3 Output: ["((()))","(()())","(())()","()(())","()()()"] Example 2: Input: n = 1 Output: ["()"] Constraints: 1 <= n <= 8

Detailed Explanation

The problem asks us to generate all possible combinations of well-formed (valid) parentheses given a number 'n' representing the number of pairs of parentheses. A well-formed parentheses string follows the rule that for any prefix of the string, the number of open parentheses must be greater than or equal to the number of close parentheses, and the total number of open and close parentheses must be equal to 'n' and 'n', respectively.

Solution Approach

The solution uses a backtracking algorithm. It maintains a stack (or string in C/C++) to build the parentheses string incrementally. At each step, it considers adding either an open or a close parenthesis. The conditions for adding each are crucial: we can only add an open parenthesis if the number of open parentheses is less than 'n', and we can only add a close parenthesis if the number of close parentheses is less than the number of open parentheses. When we have added 'n' open and 'n' close parentheses, we have a valid combination, and we add it to the result list. The stack is then popped to backtrack and try other possibilities.

Step-by-Step Algorithm

Step 1: Initialize an empty list (res) to store the valid parentheses combinations.
Step 2: Initialize an empty stack (or string) to build the current parentheses combination.
Step 3: Define a recursive function backtrack(open_count, close_count).
Step 4: Inside backtrack, check if open_count == n and close_count == n. If true, it means we have a valid combination. Add the current stack content to the result list and return.
Step 5: If open_count < n, add an open parenthesis to the stack, increment open_count, and recursively call backtrack(open_count + 1, close_count). After the recursive call, remove the open parenthesis from the stack (backtracking step).
Step 6: If close_count < open_count, add a close parenthesis to the stack, increment close_count, and recursively call backtrack(open_count, close_count + 1). After the recursive call, remove the close parenthesis from the stack (backtracking step).
Step 7: Call backtrack(0, 0) to start the process.
Step 8: Return the result list (res).

Key Insights

Insight 1: The core idea is to use backtracking. Backtracking allows us to explore all possible combinations by recursively adding either an open or a close parenthesis at each step, while ensuring we maintain the well-formed property.
Insight 2: We need to keep track of the number of open and close parentheses used so far. This is crucial for determining when to add an open parenthesis (only if open_count < n) or a close parenthesis (only if close_count < open_count).
Insight 3: The base case for the recursion is when both the number of open parentheses and the number of close parentheses are equal to 'n'. This means we have a valid combination, which can be added to our result set.

Complexity Analysis

Time Complexity: O(4^n/sqrt(n))

Space Complexity: O(4^n/sqrt(n))

Topics

This problem involves: String, Dynamic Programming, Backtracking.

Companies

Asked at: Adobe, Airtel, Amazon, Amdocs, Apple, Avito, BlackRock, Bloomberg, DE Shaw, Disney, Grammarly, Huawei, IBM, Infosys, Intuit, J.P. Morgan, Lucid, MakeMyTrip, Meta, Microsoft, Morgan Stanley, Netflix, Nvidia, Oracle, PhonePe, Point72, ServiceNow, Texas Instruments, TikTok, Uber, Veeva Systems, Walmart Labs, Yahoo, Yandex, Zenefits, Zoho, eBay, tcs.

Generate Parentheses | LeetCode 22

Medium

String Dynamic Programming Backtracking

Time: O(4^n/sqrt(n))Space: O(4^n/sqrt(n))

Solution Code

Available in:

import java.util.ArrayList;
import java.util.List;
import java.util.Stack;

class Solution {
    public List<String> generateParenthesis(int n) {
        List<String> res = new ArrayList<>();
        Stack<Character> stack = new Stack<>();

        function backtrack(res,stack,n,0,0);
        return res;
    }

    public void backtrack(List<String> res, Stack<Character> stack, int n, int open_count, int close_count){
        if (open_count == n && close_count == n) {
            StringBuilder sb = new StringBuilder();
            for (Character c : stack) {
                sb.append(c);
            }
            res.add(sb.toString());
            return;
        }

        if (open_count < n) {
            stack.push('(');
            backtrack(res,stack,n, open_count + 1, close_count);
            stack.pop();
        }

        if (close_count < open_count) {
            stack.push(')');
            backtrack(res,stack,n, open_count, close_count + 1);
            stack.pop();
        }
    }
}

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Problem Info

Description

Problem 22: Generate Parentheses - Detailed explanation and solution approaches.

Topics

String Dynamic Programming Backtracking

Companies

AdobeAirtelAmazonAmdocsAppleAvitoBlackRockBloomberg

Stats

Difficulty:Medium

Languages:4

Companies:38

Problem #:22

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 list (res) to store the valid parentheses combinations.
2
Step 2: Initialize an empty stack (or string) to build the current parentheses combination.
3
Step 3: Define a recursive function backtrack(open_count, close_count).
4
Step 4: Inside backtrack, check if open_count == n and close_count == n. If true, it means we have a valid combination. Add the current stack content to the result list and return.
5
Step 5: If open_count < n, add an open parenthesis to the stack, increment open_count, and recursively call backtrack(open_count + 1, close_count). After the recursive call, remove the open parenthesis from the stack (backtracking step).
6
Step 6: If close_count < open_count, add a close parenthesis to the stack, increment close_count, and recursively call backtrack(open_count, close_count + 1). After the recursive call, remove the close parenthesis from the stack (backtracking step).
7
Step 7: Call backtrack(0, 0) to start the process.
8
Step 8: Return the result list (res).

💡

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(4^n/sqrt(n))

💾

Space Complexity

O(4^n/sqrt(n))

📝

Why?

The Catalan number represents the number of valid combinations. The recursion explores a tree of possibilities, where at each level (up to 2n), we have two choices (add '(' or ')'). However, the constraints on adding parentheses prune the search space significantly, resulting in the Catalan number.

📖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

Dynamic programming can also be used to solve this problem. We can build up the solution iteratively, starting with n=1 and gradually increasing n. For each n, we combine the solutions for smaller n values to generate the new solutions.

Time: O(n * C_n), where C_n is the nth Catalan number, which is approximately O(n * 4^n / sqrt(n)).Space: O(C_n), which is approximately O(4^n / sqrt(n)).

Trade-offs: While dynamic programming guarantees not to recompute overlapping subproblems, the backtracking approach, in practice, is often more intuitive and easier to implement for this specific problem. The time complexity is asymptotically the same but the constant factors might differ.

✨

Key Insights

1Insight 1: The core idea is to use backtracking. Backtracking allows us to explore all possible combinations by recursively adding either an open or a close parenthesis at each step, while ensuring we maintain the well-formed property.
2Insight 2: We need to keep track of the number of open and close parentheses used so far. This is crucial for determining when to add an open parenthesis (only if open_count < n) or a close parenthesis (only if close_count < open_count).
3Insight 3: The base case for the recursion is when both the number of open parentheses and the number of close parentheses are equal to 'n'. This means we have a valid combination, which can be added to our result set.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

String

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

Edge Cases

Edge case 1: n = 1. The solution correctly returns ['()'].
Edge case 2: n = 2. The solution correctly returns ['(())', '()()'].
Edge case 3: n = 3. The solution correctly handles generating all valid combinations as specified in the example.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to backtrack properly. After a recursive call, it's crucial to remove the added parenthesis from the stack to explore other possibilities. Failing to do so will lead to incorrect results.

✗Common mistake 2: Adding a close parenthesis before an open parenthesis. The condition close_count < open_count must be checked before adding a close parenthesis to ensure the parentheses are well-formed.

✗Common mistake 3: Not allocating sufficient memory in the C solution for storing the generated strings. The C solution requires careful memory management to avoid buffer overflows.

✅

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: Parsing expressions in compilers. The ability to generate and validate parentheses is fundamental to parsing and interpreting programming languages and mathematical expressions.

🏦Real-world scenario 2: Validating data structures. Parentheses can be used to represent nested data structures, such as JSON or XML. Validating the parentheses helps ensure the structure's integrity.

💡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: June 23, 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."

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 "Generate 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 dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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