What is the time complexity of Remove Invalid Parentheses?

The optimal solution has a time complexity of O(2^N).

What data structures are used to solve Remove Invalid Parentheses?

This problem uses String, Backtracking, Breadth-First Search. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Remove Invalid Parentheses?

Insight 1: The core idea is to explore all possible combinations of removing parentheses using backtracking or BFS. Since we want to remove the *minimum* number of invalid parentheses, BFS can be efficient, especially in practice, as it explores solutions layer by layer. Insight 2: We need to efficiently identify and remove the *minimum* number of parentheses. First, we can calculate the number of extra left and right parentheses that need to be removed. Then, during the search, we can prune branches that would lead to removing more parentheses than necessary. Early termination is crucial for performance.

Which companies ask LeetCode 301. Remove Invalid Parentheses in interviews?

This problem is frequently asked at Deliveroo, Google, Rubrik, Snowflake.

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Remove Invalid Parentheses - LeetCode 301 Solution

Q: How to solve Remove Invalid Parentheses?

The provided solution uses a Depth-First Search (DFS) approach (also known as backtracking). It calculates the minimum number of left and right parentheses that must be removed to make the string valid. This information is used to constrain the search space during the DFS. The DFS function explores the string character by character, considering two options for each parenthesis: remove it or keep it. The 'open_count' parameter tracks the number of unmatched open parentheses. If we encounter a closing parenthesis and there are no unmatched open parentheses, it must be removed. The base case for the recursion is when we have processed the entire string. The string is valid if all required parentheses have been removed (l_rem == 0 and r_rem == 0) and all open parentheses have been matched (open_count == 0).

Remove Invalid Parentheses - Complete Solution Guide

Remove Invalid Parentheses is LeetCode problem 301, 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 s that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid. Return a list of unique strings that are valid with the minimum number of removals . You may return the answer in any order . Example 1: Input: s = "()())()" Output: ["(())()","()()()"] Example 2: Input: s = "(a)())()" Output: ["(a())()","(a)()()"] Example 3: Input: s = ")(" Output: [""] Constraints: 1 <= s.length <= 25 s consists of lowercase English letters a

Detailed Explanation

The problem requires us to find all possible valid strings that can be obtained by removing the minimum number of invalid parentheses from a given string 's'. The input string consists of lowercase English letters and parentheses '(' and ')'. The output should be a list of unique valid strings. A valid string means that for every opening parenthesis '(', there is a corresponding closing parenthesis ')', and the parentheses are properly nested.

Solution Approach

The provided solution uses a Depth-First Search (DFS) approach (also known as backtracking). It calculates the minimum number of left and right parentheses that must be removed to make the string valid. This information is used to constrain the search space during the DFS. The DFS function explores the string character by character, considering two options for each parenthesis: remove it or keep it. The 'open_count' parameter tracks the number of unmatched open parentheses. If we encounter a closing parenthesis and there are no unmatched open parentheses, it must be removed. The base case for the recursion is when we have processed the entire string. The string is valid if all required parentheses have been removed (l_rem == 0 and r_rem == 0) and all open parentheses have been matched (open_count == 0).

Step-by-Step Algorithm

Step 1: Calculate the minimum number of left parentheses (rem_l) and right parentheses (rem_r) that need to be removed to make the string valid. Iterate through the string, incrementing rem_l for each '(' and decrementing it for each ')' if rem_l > 0; otherwise, increment rem_r.
Step 2: Implement a DFS function that takes the current index in the string, the number of left and right parentheses to remove (l_rem, r_rem), the number of open parentheses (open_count), and the current path (partial string).
Step 3: The base case for the DFS is when the index reaches the end of the string. If l_rem, r_rem, and open_count are all zero, add the current path to the results set.
Step 4: If l_rem + r_rem is greater than the remaining length of the string, then we have already pruned too few characters and must return, this speeds up runtime significantly.
Step 5: For each character in the string, consider the following cases:
a. If the character is '(', consider removing it (if l_rem > 0) or keeping it (increment open_count).
b. If the character is ')', consider removing it (if r_rem > 0) or keeping it (if open_count > 0, decrement open_count).
c. If the character is a letter, append it to the path and proceed to the next character.
Step 6: After the DFS completes, convert the set of results to a list and return it.
Step 7: Backtrack by removing the last character added to path after exploring both removal and keeping paths.

Key Insights

Insight 1: The core idea is to explore all possible combinations of removing parentheses using backtracking or BFS. Since we want to remove the *minimum* number of invalid parentheses, BFS can be efficient, especially in practice, as it explores solutions layer by layer.
Insight 2: We need to efficiently identify and remove the *minimum* number of parentheses. First, we can calculate the number of extra left and right parentheses that need to be removed. Then, during the search, we can prune branches that would lead to removing more parentheses than necessary. Early termination is crucial for performance.
Insight 3: To avoid duplicates in the results, use a `Set` data structure. This is essential since many different sequences of removals might lead to the same valid string.

Complexity Analysis

Time Complexity: O(2^N)

Space Complexity: O(N)

Topics

This problem involves: String, Backtracking, Breadth-First Search.

Companies

Asked at: Deliveroo, Google, Rubrik, Snowflake.

Remove Invalid Parentheses | LeetCode 301

Hard

String Backtracking Breadth-First Search

Time: O(2^N)Space: O(N)

Solution Code

Available in:

import java.util.*;

class Solution {
    public List<String> removeInvalidParentheses(String s) {
        Set<String> results = new HashSet<>();
        
        int rem_l = 0, rem_r = 0;
        for (char charr : s.toCharArray()) {
            if (charr == '(') {
                rem_l += 1;
            } else if (charr == ')') {
                if (rem_l > 0) {
                    rem_l -= 1;
                } else {
                    rem_r += 1;
                }
            }
        }

        
        dfs(s, 0, rem_l, rem_r, 0, new ArrayList<Character>(), results);
        
        return new ArrayList<>(results);
    }

    private void dfs(String s, int index, int l_rem, int r_rem, int open_count, List<Character> path, Set<String> results) {
        if (index == s.length()) {
            if (l_rem == 0 && r_rem == 0 && open_count == 0) {
                StringBuilder sb = new StringBuilder();
                for(char c : path) {
                    sb.append(c);
                }
                results.add(sb.toString());
            }
            return;
        }

        if (l_rem + r_rem > s.length() - index) {
            return;
        }

        char charr = s.charAt(index);
        
        if (charr == '(') {
            // Option 1: remove it
            if (l_rem > 0) {
                dfs(s, index + 1, l_rem - 1, r_rem, open_count, new ArrayList<>(path), results);
            }
            
            // Option 2: keep it
            path.add('(');
            dfs(s, index + 1, l_rem, r_rem, open_count + 1, new ArrayList<>(path), results);
            path.remove(path.size() - 1);

        } else if (charr == ')') {
            // Option 1: remove it
            if (r_rem > 0) {
                dfs(s, index + 1, l_rem, r_rem - 1, open_count, new ArrayList<>(path), results);
            }
            
            // Option 2: keep it
            if (open_count > 0) {
                path.add(')');
                dfs(s, index + 1, l_rem, r_rem, open_count - 1, new ArrayList<>(path), results);
                path.remove(path.size() - 1);
            }
        } else { // letter
            path.add(charr);
            dfs(s, index + 1, l_rem, r_rem, open_count, new ArrayList<>(path), results);
            path.remove(path.size() - 1);
        }
    }
}

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Problem Info

Description

Problem 301: Remove Invalid Parentheses - Detailed explanation and solution approaches.

Topics

String Backtracking Breadth-First Search

Companies

DeliverooGoogleRubrikSnowflake

Stats

Difficulty:Hard

Languages:4

Companies:4

Problem #:301

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

10 steps

1
Step 1: Calculate the minimum number of left parentheses (rem_l) and right parentheses (rem_r) that need to be removed to make the string valid. Iterate through the string, incrementing rem_l for each '(' and decrementing it for each ')' if rem_l > 0; otherwise, increment rem_r.
2
Step 2: Implement a DFS function that takes the current index in the string, the number of left and right parentheses to remove (l_rem, r_rem), the number of open parentheses (open_count), and the current path (partial string).
3
Step 3: The base case for the DFS is when the index reaches the end of the string. If l_rem, r_rem, and open_count are all zero, add the current path to the results set.
4
Step 4: If l_rem + r_rem is greater than the remaining length of the string, then we have already pruned too few characters and must return, this speeds up runtime significantly.
5
Step 5: For each character in the string, consider the following cases:
6
a. If the character is '(', consider removing it (if l_rem > 0) or keeping it (increment open_count).
7
b. If the character is ')', consider removing it (if r_rem > 0) or keeping it (if open_count > 0, decrement open_count).
8
c. If the character is a letter, append it to the path and proceed to the next character.
9
Step 6: After the DFS completes, convert the set of results to a list and return it.
10
Step 7: Backtrack by removing the last character added to path after exploring both removal and keeping paths.

💡

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(2^N)

💾

Space Complexity

O(N)

📝

Why?

The time complexity is exponential because the DFS algorithm explores all possible combinations of removing parentheses. The space complexity primarily depends on the maximum depth of the recursion, which is proportional to the length of the input string 's'. The results set can also grow linearly with 's' in certain worst-case scenarios, thus contributing linearly to the overall space complexity.

📖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

Breadth-First Search (BFS)

An alternative approach is to use BFS. Start with the original string and generate all possible strings by removing one parenthesis at a time. Use a set to track visited strings to avoid cycles. In each iteration, check if any of the generated strings are valid. If yes, return all valid strings at that level. If not, continue to the next level.

Time: O(N * C(N, L)), where N is the length of the string and L is the number of parentheses to remove. C(N, L) is the binomial coefficient, representing the number of combinations of removing L parentheses from N characters.Space: O(N * C(N, L)), to store the strings in the queue and the visited set.

Trade-offs: BFS is generally more efficient if the minimum number of removals is small. It explores solutions layer by layer, so it can find the shortest path (minimum removals) faster. However, BFS requires more memory to store the queue of strings to explore.

✨

Key Insights

1Insight 1: The core idea is to explore all possible combinations of removing parentheses using backtracking or BFS. Since we want to remove the *minimum* number of invalid parentheses, BFS can be efficient, especially in practice, as it explores solutions layer by layer.
2Insight 2: We need to efficiently identify and remove the *minimum* number of parentheses. First, we can calculate the number of extra left and right parentheses that need to be removed. Then, during the search, we can prune branches that would lead to removing more parentheses than necessary. Early termination is crucial for performance.
3Insight 3: To avoid duplicates in the results, use a `Set` data structure. This is essential since many different sequences of removals might lead to the same valid string.

📚

Key Terms Explained

For Students

String

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

Edge Cases

Edge case 1: Empty string. The code correctly handles this case as the loop to determine rem_l and rem_r will not execute, and the dfs function will return an empty string.
Edge case 2: String with no parentheses. The code correctly handles this because rem_l and rem_r will be 0, and the dfs function will correctly return the original string as the result because every index goes through the 'else' condition.
Edge case 3: String with only opening parentheses. The code will correctly identify all opening parenthesis as invalid, and generate an empty string as the only valid answer.
Edge case 4: String with only closing parentheses. The code will correctly identify all closing parenthesis as invalid, and generate an empty string as the only valid answer.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling duplicate results. Failing to use a `Set` to store results can lead to redundant valid strings in the output, which violates the problem constraint.

✗Common mistake 2: Not calculating the minimum number of parentheses to remove beforehand. This can result in exploring unnecessary branches and significantly increasing the execution time. Calculating rem_l and rem_r allows for early pruning.

✗Common mistake 3: Not validating the parenthesis string correctly. Incorrectly evaluating the balance of parentheses can lead to invalid results.

✗Common mistake 4: Not considering the case where the input string is already valid. The code should handle this case and return the original string as the result.

✗Common mistake 5: Mutating the path incorrectly in backtracking. The path needs to be properly restored (backtracked) after each recursive call to explore other possibilities. This usually involves popping the last added element.

✅

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 user input. Ensuring that user-entered expressions or code snippets have properly balanced parentheses is a common requirement in many applications.

🏦Real-world scenario 2: Code linters and IDEs. These tools often use similar algorithms to check for syntax errors, including unbalanced parentheses, in programming code.

🎮Real-world scenario 3: Parsing mathematical expressions. Before evaluating a mathematical expression, it's crucial to ensure that the parentheses are balanced to maintain the correct order of operations.

💡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

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ 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: September 30, 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 13

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

@InterviewProJul 6

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

@CodeNinjaJun 29

❓

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.

How do I avoid duplicate solutions in backtracking?

Sort the input first, then skip consecutive duplicates at the same recursion level: if (i > start && nums[i] == nums[i-1]) continue. For permutations, use a visited array or swap-based approach. Always consider if order matters (permutations vs combinations).

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

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 "Remove Invalid 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 backtracking techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?