What is the time complexity of Minimum Insertions to Balance a Parentheses String?

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

What data structures are used to solve Minimum Insertions to Balance a Parentheses String?

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

What are the key insights for solving Minimum Insertions to Balance a Parentheses String?

Insight 1: The core idea is to iterate through the string and keep track of the number of ')' needed to balance the encountered '('. Insight 2: Instead of using a stack, we can use a counter to represent the number of ')' required. This avoids the O(n) space complexity associated with stacks in some cases.

Which companies ask LeetCode 1541. Minimum Insertions to Balance a Parentheses String in interviews?

This problem is frequently asked at J.P. Morgan.

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Minimum Insertions to Balance a Parentheses String - LeetCode 1541 Solution

Minimum Insertions to Balance a Parentheses String - Complete Solution Guide

Minimum Insertions to Balance a Parentheses String is LeetCode problem 1541, 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 a parentheses string s containing only the characters '(' and ')' . A parentheses string is balanced if: Any left parenthesis '(' must have a corresponding two consecutive right parenthesis '))' . Left parenthesis '(' must go before the corresponding two consecutive right parenthesis '))' . In other words, we treat '(' as an opening parenthesis and '))' as a closing parenthesis. For example, "())" , "())(())))" and "(())())))" are balanced, ")()" , "()))" and "(()))" are not balanced. You

Detailed Explanation

The problem asks us to find the minimum number of insertions (of '(' or ')') needed to balance a parentheses string `s` consisting only of '(' and ')'. A string is considered balanced if every '(' has two consecutive ')' following it. In other words, '(' acts as an opening parenthesis and '))' acts as a closing parenthesis. The insertions can be made at any position in the string.

Solution Approach

The solution iterates through the input string `s`. It maintains a variable `right_needed` that represents the number of ')' needed to balance the previously encountered '('. When an opening parenthesis '(' is encountered, we check if `right_needed` is odd, meaning there is one ')' short from previous parenthesis. If so, we increment `insertions` and decrease `right_needed`. Then, we increment `right_needed` by 2 because each '(' needs two ')'. When a closing parenthesis ')' is encountered, we decrease `right_needed` by 1. If `right_needed` becomes negative, it means we have more ')' than expected, so we need to insert a '(' to balance, which means we increment `insertions` and set `right_needed` to 1. Finally, after iterating through the whole string, we add the remaining `right_needed` to `insertions`, since these are the number of ')' needed at the end.

Step-by-Step Algorithm

Step 1: Initialize `insertions` and `right_needed` to 0. `insertions` will store the final answer, and `right_needed` will track how many closing parenthesis are needed at any given point.
Step 2: Iterate through the input string `s` character by character.
Step 3: If the current character is '(', check if `right_needed` is odd. If it is, increment `insertions` by 1 and decrement `right_needed` by 1 (simulating insertion of a ')'). Then, increment `right_needed` by 2, because a '(' needs two ')'.
Step 4: If the current character is ')', decrement `right_needed` by 1.
Step 5: If `right_needed` becomes negative, it means we need to insert a '(' to balance. Increment `insertions` by 1 and reset `right_needed` to 1, as we have just 'used' one ')' to pair with the inserted '('.
Step 6: After the loop finishes, add the final `right_needed` to `insertions`. This handles the case where there are remaining '(' that need to be closed with '))' at the end of the string.
Step 7: Return the final value of `insertions`.

Key Insights

Insight 1: The core idea is to iterate through the string and keep track of the number of ')' needed to balance the encountered '('.
Insight 2: Instead of using a stack, we can use a counter to represent the number of ')' required. This avoids the O(n) space complexity associated with stacks in some cases.
Insight 3: We need to handle cases where we encounter a ')' but no '(' is present, meaning we need to add a '(' before. Also, if we end up with remaining '(' without enough '))', we need to add the required '))' at the end.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: String, Stack, Greedy.

Companies

Asked at: J.P. Morgan.

Minimum Insertions to Balance a Parentheses String | LeetCode 1541

Medium

String Stack Greedy

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

Solution Code

Available in:

class Solution {
    public int minInsertions(String s) {
        int insertions = 0;
        int right_needed = 0;

        for (char char_ : s.toCharArray()) {
            if (char_ == '(') {
                if (right_needed % 2 != 0) {
                    insertions += 1;
                    right_needed -= 1;
                }

                right_needed += 2;
            } else {  // char == ')'
                right_needed -= 1;

                if (right_needed < 0) {
                    insertions += 1;
                    right_needed = 1;
                }
            }
        }

        insertions += right_needed;

        return insertions;
    }
}

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Problem Info

Description

Problem 1541: Minimum Insertions to Balance a Parentheses String - Detailed explanation and solution approaches.

Topics

String Stack Greedy

Companies

J.P. Morgan

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1541

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

7 steps

1
Step 1: Initialize `insertions` and `right_needed` to 0. `insertions` will store the final answer, and `right_needed` will track how many closing parenthesis are needed at any given point.
2
Step 2: Iterate through the input string `s` character by character.
3
Step 3: If the current character is '(', check if `right_needed` is odd. If it is, increment `insertions` by 1 and decrement `right_needed` by 1 (simulating insertion of a ')'). Then, increment `right_needed` by 2, because a '(' needs two ')'.
4
Step 4: If the current character is ')', decrement `right_needed` by 1.
5
Step 5: If `right_needed` becomes negative, it means we need to insert a '(' to balance. Increment `insertions` by 1 and reset `right_needed` to 1, as we have just 'used' one ')' to pair with the inserted '('.
6
Step 6: After the loop finishes, add the final `right_needed` to `insertions`. This handles the case where there are remaining '(' that need to be closed with '))' at the end of the string.
7
Step 7: Return the final value of `insertions`.

💡

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 each character of the string once, performing constant-time operations in each iteration. Therefore, the time complexity is directly proportional to the length of the string, resulting in O(n). The algorithm uses only a few integer variables, which do not scale with the input size, hence O(1) 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

Stack-based approach

Use a stack to keep track of the open parenthesis '('. When you encounter '))', pop the stack. When you encounter ')', and stack is empty, you need to push '('. If you encounter '(' push it onto the stack. At the end the stack size will correspond to number of '(' that need to be closed with '))'. Then, you calculate additional insertions for cases like '))' when the stack is empty.

Time: O(n)Space: O(n) in the worst case, where n is the length of the string (e.g., '(((((')

Trade-offs: The stack-based approach is more intuitive to understand, especially if you are comfortable with stack data structure. However, it requires O(n) space in the worst case, while the current solution requires O(1) space.

✨

Key Insights

1Insight 1: The core idea is to iterate through the string and keep track of the number of ')' needed to balance the encountered '('.
2Insight 2: Instead of using a stack, we can use a counter to represent the number of ')' required. This avoids the O(n) space complexity associated with stacks in some cases.
3Insight 3: We need to handle cases where we encounter a ')' but no '(' is present, meaning we need to add a '(' before. Also, if we end up with remaining '(' without enough '))', we need to add the required '))' at the end.

📚

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: String starts with ')'. The algorithm correctly handles this by incrementing `insertions` because `right_needed` becomes negative and then resetting `right_needed` to 1, effectively adding a '(' before the ')'.
Edge case 2: String ends with '('. The algorithm handles this by adding the final `right_needed` value to `insertions`, effectively adding the required '))' at the end.
Edge case 3: String contains only ')'. Each ')' will cause an insertion of '(', and the final `right_needed` will be added which will be 1 if the length is odd and 0 if the length is even after inserting required '('.
Edge case 4: String contains only '('. Each '(' causes `right_needed` to increase by 2. At the end the `right_needed` is added to insertion, which accounts for the missing '))'.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the case where `right_needed` is odd when an opening parenthesis '(' is encountered. This means you forget to account for the situation where there was one ')' missing from the previous open parenthesis.

✗Common mistake 2: Not handling the case where there are remaining '(' that need to be closed with '))' at the end of the string. This requires adding the final `right_needed` value to `insertions`.

✗Common mistake 3: Incorrectly handling edge cases where the string begins with ')' or ends with '('.

✅

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: This algorithm can be used in code editors or IDEs to automatically balance parenthesis in code.

🏦Real-world scenario 2: It can be used in compilers or interpreters to check for syntax errors related to unbalanced parenthesis.

🎮Real-world scenario 3: Validation of mathematical expressions or configuration files where correct parenthesis balancing is crucial.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ 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 10, 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

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

@CodeNinjaJun 30

❓

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.

What is a monotonic stack and when should I use it?

A monotonic stack maintains elements in sorted order (increasing or decreasing). Use it for "next greater element," "largest rectangle," or "stock span" problems. It achieves O(n) time by avoiding redundant comparisons through maintaining order invariants.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

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 "Minimum Insertions to Balance a Parentheses String"?

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.

🤖 Need help with this problem?