What is the time complexity of Reverse Substrings Between Each Pair of Parentheses?

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

What data structures are used to solve Reverse Substrings Between Each Pair of Parentheses?

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

What are the key insights for solving Reverse Substrings Between Each Pair of Parentheses?

Insight 1: The problem can be effectively solved using a stack data structure to keep track of the nesting of parentheses. Insight 2: When encountering a closing parenthesis, the characters accumulated since the last opening parenthesis need to be reversed.

Which companies ask LeetCode 1190. Reverse Substrings Between Each Pair of Parentheses in interviews?

This problem is frequently asked at Agoda.

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Reverse Substrings Between Each Pair of Parentheses - LeetCode 1190 Solution

Reverse Substrings Between Each Pair of Parentheses - Complete Solution Guide

Reverse Substrings Between Each Pair of Parentheses is LeetCode problem 1190, 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

You are given a string s that consists of lower case English letters and brackets. Reverse the strings in each pair of matching parentheses, starting from the innermost one. Your result should not contain any brackets. Example 1: Input: s = "(abcd)" Output: "dcba" Example 2: Input: s = "(u(love)i)" Output: "iloveu" Explanation: The substring "love" is reversed first, then the whole string is reversed. Example 3: Input: s = "(ed(et(oc))el)" Output: "leetcode" Explanation: First, we reverse the su

Detailed Explanation

The problem requires reversing substrings enclosed within matching parentheses in a given string. The reversals should be performed from the innermost parentheses outwards. The final output string should not contain any parentheses.

Solution Approach

The solution utilizes a stack to simulate the nested structure of parentheses. When an opening parenthesis '(' is encountered, a new list (or vector/ArrayList) is pushed onto the stack. When a closing parenthesis ')' is encountered, the top list is popped, reversed, and then appended to the list at the top of the stack. Characters other than parentheses are added to the list at the top of the stack. This process effectively reverses the substrings enclosed by each pair of matching parentheses, starting from the innermost ones.

Step-by-Step Algorithm

Step 1: Initialize a stack with an empty list (or vector/ArrayList) as the initial element.
Step 2: Iterate through the input string character by character.
Step 3: If the character is an opening parenthesis '(', push a new empty list onto the stack.
Step 4: If the character is a closing parenthesis ')', pop the top list from the stack, reverse it, and then extend the list at the top of the stack with the reversed list.
Step 5: If the character is a lowercase English letter, append it to the list at the top of the stack.
Step 6: After processing all characters, the remaining list at the bottom of the stack contains the final reversed string. Join the characters in the list to form a string and return it.

Key Insights

Insight 1: The problem can be effectively solved using a stack data structure to keep track of the nesting of parentheses.
Insight 2: When encountering a closing parenthesis, the characters accumulated since the last opening parenthesis need to be reversed.
Insight 3: Each stack level represents a level of nested parentheses.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: String, Stack.

Companies

Asked at: Agoda.

Reverse Substrings Between Each Pair of Parentheses | LeetCode 1190

Medium

String Stack

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

Solution Code

Available in:

import java.util.Stack;
import java.util.ArrayList;
import java.util.Collections;

class Solution {
    public String reverseParentheses(String s) {
        Stack<ArrayList<Character>> stack = new Stack<>();
        stack.push(new ArrayList<>());
        for (char chara : s.toCharArray()) {
            if (chara == '(') {
                stack.push(new ArrayList<>());
            } else if (chara == ')') {
                ArrayList<Character> reversed_segment = stack.pop();
                Collections.reverse(reversed_segment);
                stack.peek().addAll(reversed_segment);
            } else {
                stack.peek().add(chara);
            }
        }

        StringBuilder sb = new StringBuilder();
        for (char chara : stack.get(0)) {
            sb.append(chara);
        }
        return sb.toString();
    }
}

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Problem Info

Description

Problem 1190: Reverse Substrings Between Each Pair of Parentheses - Detailed explanation and solution approaches.

Topics

String Stack

Companies

Agoda

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1190

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

6 steps

1
Step 1: Initialize a stack with an empty list (or vector/ArrayList) as the initial element.
2
Step 2: Iterate through the input string character by character.
3
Step 3: If the character is an opening parenthesis '(', push a new empty list onto the stack.
4
Step 4: If the character is a closing parenthesis ')', pop the top list from the stack, reverse it, and then extend the list at the top of the stack with the reversed list.
5
Step 5: If the character is a lowercase English letter, append it to the list at the top of the stack.
6
Step 6: After processing all characters, the remaining list at the bottom of the stack contains the final reversed string. Join the characters in the list to form a string and return it.

💡

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 linear because each character in the string is processed a constant number of times. The space complexity is linear because the stack can grow to a size proportional to the length of the input string in the worst-case scenario due to nested 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

Recursive Approach

A recursive approach can also be used. Find the innermost parenthesis, reverse it, and recursively call the function on the resulting string until no more parentheses are left.

Time: O(n^2) in the worst case, where n is the length of the input string. Finding the innermost parenthesis takes O(n) and reversing a string takes O(n). The recursion could potentially be O(n) deep.Space: O(n) due to the recursion depth in the worst case, and the string copying that may occur at each level.

Trade-offs: The recursive approach might be easier to understand for some, but it typically has a higher time complexity due to repeated string operations. The stack-based solution is generally more efficient.

✨

Key Insights

1Insight 1: The problem can be effectively solved using a stack data structure to keep track of the nesting of parentheses.
2Insight 2: When encountering a closing parenthesis, the characters accumulated since the last opening parenthesis need to be reversed.
3Insight 3: Each stack level represents a level of nested parentheses.

📚

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 algorithm correctly handles an empty string by returning an empty string.
Edge case 2: String with no parentheses - The algorithm correctly handles a string with no parentheses by simply returning the original string.
Edge case 3: Deeply nested parentheses - The algorithm correctly reverses the substrings enclosed within deeply nested parentheses due to the stack-based approach.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly handling nested parentheses - Forgetting to use a stack or not properly managing the stack leads to incorrect reversal of substrings within nested parentheses.

✗Common mistake 2: Modifying the input string directly without proper handling of indices - This can lead to errors and make the code harder to understand.

✗Common mistake 3: Forgetting to reverse the substring - The core requirement of reversing the substring enclosed in the parentheses must be carefully implemented.

✗Common mistake 4: Memory leaks or incorrect memory management, especially in C, can lead to program crashes or unexpected behavior.

✅

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: Text editors and IDEs often use similar parsing techniques for code folding, syntax highlighting, and bracket matching.

🏦Real-world scenario 2: Expression evaluation in compilers and interpreters requires parsing and evaluating expressions based on operator precedence, which can involve handling nested structures similar to parentheses.

🎮Real-world scenario 3: Data validation and sanitization, where properly formatted input is essential, and nested structures need to be parsed to identify potential security vulnerabilities.

💡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.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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 19, 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

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 7

❓

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.

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 "Reverse Substrings Between Each Pair of 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 stack techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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