What is the time complexity of Faulty Keyboard?

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

What data structures are used to solve Faulty Keyboard?

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

What are the key insights for solving Faulty Keyboard?

Insight 1: The problem can be solved iteratively, processing each character one by one. Insight 2: Using a string builder (or similar mutable string structure) is efficient for repeated string modifications. Repeated string concatenation can be inefficient.

Which companies ask LeetCode 2810. Faulty Keyboard in interviews?

This problem is frequently asked at Samsung.

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Faulty Keyboard - LeetCode 2810 Solution

Faulty Keyboard - Complete Solution Guide

Faulty Keyboard is LeetCode problem 2810, 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

Your laptop keyboard is faulty, and whenever you type a character 'i' on it, it reverses the string that you have written. Typing other characters works as expected. You are given a 0-indexed string s , and you type each character of s using your faulty keyboard. Return the final string that will be present on your laptop screen. Example 1: Input: s = "string" Output: "rtsng" Explanation: After typing first character, the text on the screen is "s". After the second character, the text is "st". A

Detailed Explanation

The problem describes a faulty keyboard that reverses the typed string whenever the character 'i' is pressed. The input is a string `s` representing the sequence of characters typed. The output is the final string displayed on the screen after all characters are processed, considering the faulty behavior of the keyboard. The problem constraints specify that the input string's length is between 1 and 100, inclusive, consisting only of lowercase English letters, and the first character is not 'i'.

Solution Approach

The provided solutions use an iterative approach. They iterate through the input string `s`, character by character. If the character is 'i', the current string is reversed; otherwise, the character is appended to the current string. This directly simulates the faulty keyboard's behavior. The use of `StringBuilder` in Java and `string` (which is mutable in C++) improves efficiency compared to repeated string concatenation in Python's original string, which is immutable.

Step-by-Step Algorithm

Step 1: Initialize an empty string (or StringBuilder) to store the currently displayed text.
Step 2: Iterate through each character in the input string `s`.
Step 3: If the character is 'i', reverse the current string.
Step 4: If the character is not 'i', append the character to the current string.
Step 5: After iterating through all characters, return the final string.

Key Insights

Insight 1: The problem can be solved iteratively, processing each character one by one.
Insight 2: Using a string builder (or similar mutable string structure) is efficient for repeated string modifications. Repeated string concatenation can be inefficient.
Insight 3: Reversing a string can be done efficiently in-place using techniques like reversing pointers or using built-in string reversal functions.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: String, Simulation.

Companies

Asked at: Samsung.

Faulty Keyboard | LeetCode 2810

Easy

String Simulation

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

Solution Code

Available in:

class Solution {
    public String finalString(String s) {
        StringBuilder sb = new StringBuilder();
        for (char c : s.toCharArray()) {
            if (c == 'i') {
                sb.reverse();
            } else {
                sb.append(c);
            }
        }
        return sb.toString();
    }
}

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Problem Info

Description

Problem 2810: Faulty Keyboard - Detailed explanation and solution approaches.

Topics

String Simulation

Companies

Samsung

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:2810

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 an empty string (or StringBuilder) to store the currently displayed text.
2
Step 2: Iterate through each character in the input string `s`.
3
Step 3: If the character is 'i', reverse the current string.
4
Step 4: If the character is not 'i', append the character to the current string.
5
Step 5: After iterating through all characters, return the final string.

💡

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 loop iterates n times. Each iteration takes constant time except for the string reversal, which in general can take O(m) time, where m is the current string length. In the worst case, m can be as large as n. The space complexity comes from storing the intermediate string which is at most of length n.

📖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

Using a Deque

Instead of a string builder, use a deque (double-ended queue) data structure. Appending is O(1) and reversing is also O(1) for the whole deque because you can just swap the pointers to the front and back.

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

Trade-offs: This approach might be slightly faster in practice due to the O(1) reversal, especially for long strings with frequent reversals, although the asymptotic complexity remains the same. However, it increases code complexity slightly.

✨

Key Insights

1Insight 1: The problem can be solved iteratively, processing each character one by one.
2Insight 2: Using a string builder (or similar mutable string structure) is efficient for repeated string modifications. Repeated string concatenation can be inefficient.
3Insight 3: Reversing a string can be done efficiently in-place using techniques like reversing pointers or using built-in string reversal functions.

📚

Key Terms Explained

For Students

String

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

Edge Cases

Edge case 1: The input string is empty. The solution correctly handles this by returning an empty string.
Edge case 2: The input string contains only 'i's. The solution would repeatedly reverse an empty/single character string, resulting in an empty string (which is consistent with the problem description that `s[0] != 'i'`.)
Edge case 3: The input string contains multiple consecutive 'i's. The solution correctly handles this by reversing the string for each 'i'.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Using string concatenation repeatedly in Python (or other languages with immutable strings). This leads to O(n^2) time complexity because each concatenation creates a new string.

✗Common mistake 2: Not handling the edge case of an empty input string or a string consisting entirely of 'i's.

✅

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: Simulating a system with a faulty component that introduces unexpected inversions or transformations to data.

🏦Real-world scenario 2: Processing user input where a bug in the input handling reverses the input string under specific conditions (though this scenario is less realistic).

💡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

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

👨‍💻 From Experience

This is one of the first problems I solved when learning String. It's a great foundation for more complex problems.

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.

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 "Faulty Keyboard"?

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 simulation techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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