What is the time complexity of Reverse Prefix of Word?

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

What data structures are used to solve Reverse Prefix of Word?

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

What are the key insights for solving Reverse Prefix of Word?

The problem can be efficiently solved using string manipulation techniques and finding the index of the first occurrence of a character. The choice of using either in-place reversal (like in the Java and C solutions) or creating a new reversed substring (like in the Python solution) impacts space complexity.

Which companies ask LeetCode 2000. Reverse Prefix of Word in interviews?

This problem is frequently asked at Optum.

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Reverse Prefix of Word - LeetCode 2000 Solution

Reverse Prefix of Word - Complete Solution Guide

Reverse Prefix of Word is LeetCode problem 2000, 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

Given a 0-indexed string word and a character ch , reverse the segment of word that starts at index 0 and ends at the index of the first occurrence of ch ( inclusive ). If the character ch does not exist in word , do nothing. For example, if word = "abcdefd" and ch = "d" , then you should reverse the segment that starts at 0 and ends at 3 ( inclusive ). The resulting string will be " dcba efd" . Return the resulting string . Example 1: Input: word = " abcd efd", ch = "d" Output: " dcba efd" Expl

Detailed Explanation

The problem asks you to reverse a prefix of a given string. The prefix extends from the beginning of the string up to and including the first occurrence of a specified character. If the specified character is not found in the string, the string remains unchanged. The input consists of a string `word` and a character `ch`. The output is the modified string with the reversed prefix.

Solution Approach

The solutions generally follow these steps: First, find the index of the first occurrence of character `ch` in the string `word`. If `ch` is not found, return the original string. Otherwise, reverse the substring from the beginning of `word` up to and including the found index. Finally, concatenate the reversed prefix with the remaining suffix of the original string to form the result.

Step-by-Step Algorithm

Find the index of the first occurrence of character `ch` in the string `word` using either `word.index(ch)` (Python) or `word.indexOf(ch)` (Java) or a manual loop (C++ and C).
If the character `ch` is not found (index is -1), return the original string `word`.
If `ch` is found, reverse the substring from index 0 to the found index (inclusive). This can be done either in-place using two pointers (Java, C, C++) or by creating a reversed substring and concatenating (Python).
Concatenate the reversed prefix with the remaining part of the original string (from the index after the found character to the end) to get the final result.

Key Insights

The problem can be efficiently solved using string manipulation techniques and finding the index of the first occurrence of a character.
The choice of using either in-place reversal (like in the Java and C solutions) or creating a new reversed substring (like in the Python solution) impacts space complexity.
Handling the case where the specified character `ch` is not present in the input string `word` is crucial to avoid errors.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Two Pointers, String, Stack.

Companies

Asked at: Optum.

Reverse Prefix of Word | LeetCode 2000

Easy

Two Pointers String Stack

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

Solution Code

Available in:

class Solution {
    public String reversePrefix(String word, char ch) {
        int index = word.indexOf(ch);
        if (index == -1) {
            return word;
        }
        char[] chars = word.toCharArray();
        int left = 0;
        int right = index;
        while (left < right) {
            char temp = chars[left];
            chars[left] = chars[right];
            chars[right] = temp;
            left++;
            right--;
        }
        return new String(chars);
    }
}

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Problem Info

Description

Problem 2000: Reverse Prefix of Word - Detailed explanation and solution approaches.

Topics

Two Pointers String Stack

Companies

Optum

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:2000

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

4 steps

1
Find the index of the first occurrence of character `ch` in the string `word` using either `word.index(ch)` (Python) or `word.indexOf(ch)` (Java) or a manual loop (C++ and C).
2
If the character `ch` is not found (index is -1), return the original string `word`.
3
If `ch` is found, reverse the substring from index 0 to the found index (inclusive). This can be done either in-place using two pointers (Java, C, C++) or by creating a reversed substring and concatenating (Python).
4
Concatenate the reversed prefix with the remaining part of the original string (from the index after the found character to the end) to get the final result.

💡

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 dominant operation is iterating through the string. In the Python solution, creating substrings requires additional space proportional to the length of the string. The in-place reversal methods modify the string directly without requiring extra space beyond a few temporary variables.

📖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 solution could be implemented to reverse the prefix, though it would likely be less efficient than the iterative approaches.

Time: O(n)Space: O(n) due to recursive call stack

Trade-offs: Recursive solutions are generally less efficient for this problem due to function call overhead. The iterative approaches are preferred for their better performance.

✨

Key Insights

1The problem can be efficiently solved using string manipulation techniques and finding the index of the first occurrence of a character.
2The choice of using either in-place reversal (like in the Java and C solutions) or creating a new reversed substring (like in the Python solution) impacts space complexity.
3Handling the case where the specified character `ch` is not present in the input string `word` is crucial to avoid errors.

📚

Key Terms Explained

For Students

Two Pointers

Using two indices to traverse data from different positions simultaneously.

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

If the character `ch` is not found in the string, the function should return the original string without modifications. All solutions handle this correctly.
If `ch` is the first character of the string, the entire string should be reversed. All solutions handle this edge case correctly.

⚠️

Common Mistakes to Avoid

✗Forgetting to handle the case where the character `ch` is not present in the input string.

✗Off-by-one errors when determining the indices for reversing the prefix. Care must be taken to include or exclude the index of `ch` correctly based on problem requirements.

✗Incorrectly handling the in-place reversal using two pointers, leading to incorrect swapping of characters.

✅

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?

💼String manipulation tasks in text processing or data cleaning. For instance, reversing the prefix of words could be a preprocessing step in a natural language processing pipeline.

🏦Cryptography or data encoding/decoding algorithms, where reversing parts of strings might be a part of the encoding/decoding process.

💡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

The two-pointer technique is powerful for sorted arrays. The key insight is recognizing when elements can be processed from both ends simultaneously.

🎯 Interview Tip

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

⚠️ Watch Out

The most common mistake is not handling the case when pointers meet. Always consider the termination condition.

👨‍💻 From Experience

This is one of the first problems I solved when learning Two Pointers. 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: June 21, 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

When is the two-pointer technique most effective?

Two pointers work best on sorted arrays or when searching for pairs/triplets that satisfy a condition. Common patterns include: moving pointers inward from both ends (like in two-sum variants), fast/slow pointers for cycle detection, or left/right pointers for partitioning.

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 "Reverse Prefix of Word"?

For this easy-level Two Pointers problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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