What is the time complexity of Reverse Vowels of a String?

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

What data structures are used to solve Reverse Vowels of a String?

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

What are the key insights for solving Reverse Vowels of a String?

Insight 1: Identifying vowels efficiently: A fast way to check if a character is a vowel is crucial for performance. Using a string containing all vowels and checking for membership (e.g., using `in` in Python, `indexOf` in Java, `find` in C++, or `strchr` in C) is efficient. Insight 2: Two-pointer approach: Using two pointers, one at the beginning and one at the end of the string, allows for in-place swapping of vowels without needing extra space proportional to the string length. This is significantly more efficient than creating a new string.

Which companies ask LeetCode 345. Reverse Vowels of a String in interviews?

This problem is frequently asked at Accenture, Accolite, Twitch, USAA, Zoho.

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Reverse Vowels of a String - LeetCode 345 Solution

Reverse Vowels of a String - Complete Solution Guide

Reverse Vowels of a String is LeetCode problem 345, 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 string s , reverse only all the vowels in the string and return it. The vowels are 'a' , 'e' , 'i' , 'o' , and 'u' , and they can appear in both lower and upper cases, more than once. Example 1: Input: s = "IceCreAm" Output: "AceCreIm" Explanation: The vowels in s are ['I', 'e', 'e', 'A'] . On reversing the vowels, s becomes "AceCreIm" . Example 2: Input: s = "leetcode" Output: "leotcede" Constraints: 1 <= s.length <= 3 * 10 5 s consist of printable ASCII characters.

Detailed Explanation

The problem asks you to reverse only the vowels (a, e, i, o, u, and their uppercase counterparts) within a given string while keeping the consonants and their positions unchanged. The input is a string, and the output is a modified string with the vowels reversed. Constraints specify that the input string length is between 1 and 3 * 10<sup>5</sup> characters and consists of printable ASCII characters.

Solution Approach

The provided solutions employ a two-pointer approach. Two pointers, `left` and `right`, start at the beginning and end of the string, respectively. The pointers move towards each other. At each step, they check if the character pointed to is a vowel. If both pointers point to vowels, the vowels are swapped. The process continues until the pointers cross each other. This ensures only vowels are swapped, preserving the order of consonants.

Step-by-Step Algorithm

Step 1: Initialize two pointers, `left` to 0 and `right` to the last index of the string.
Step 2: While `left` is less than `right`:
Step 3: Move `left` pointer to the right until it points to a vowel.
Step 4: Move `right` pointer to the left until it points to a vowel.
Step 5: If `left` is still less than `right`, swap the characters at indices `left` and `right`.
Step 6: Increment `left` and decrement `right`.
Step 7: Return the modified string.

Key Insights

Insight 1: Identifying vowels efficiently: A fast way to check if a character is a vowel is crucial for performance. Using a string containing all vowels and checking for membership (e.g., using `in` in Python, `indexOf` in Java, `find` in C++, or `strchr` in C) is efficient.
Insight 2: Two-pointer approach: Using two pointers, one at the beginning and one at the end of the string, allows for in-place swapping of vowels without needing extra space proportional to the string length. This is significantly more efficient than creating a new string.
Insight 3: Handling edge cases: The solution needs to handle cases where there are no vowels, only vowels, or a mix of vowels and consonants. The code must correctly manage pointer movements and avoid index out-of-bounds errors.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Two Pointers, String.

Companies

Asked at: Accenture, Accolite, Twitch, USAA, Zoho.

Reverse Vowels of a String | LeetCode 345

Easy

Two Pointers String

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

Solution Code

Available in:

class Solution {
    public String reverseVowels(String s) {
        char[] chars = s.toCharArray();
        int left = 0;
        int right = s.length() - 1;
        while (left < right) {
            while (left < right && !isVowel(chars[left])) {
                left++;
            }
            while (left < right && !isVowel(chars[right])) {
                right--;
            }
            if (left < right) {
                char temp = chars[left];
                chars[left] = chars[right];
                chars[right] = temp;
                left++;
                right--;
            }
        }
        return new String(chars);
    }

    private boolean isVowel(char c) {
        return "AEIOUaeiou".indexOf(c) != -1;
    }
}

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Problem Info

Description

Problem 345: Reverse Vowels of a String - Detailed explanation and solution approaches.

Topics

Two Pointers String

Companies

AccentureAccoliteTwitchUSAAZoho

Stats

Difficulty:Easy

Languages:4

Companies:5

Problem #:345

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 two pointers, `left` to 0 and `right` to the last index of the string.
2
Step 2: While `left` is less than `right`:
3
Step 3: Move `left` pointer to the right until it points to a vowel.
4
Step 4: Move `right` pointer to the left until it points to a vowel.
5
Step 5: If `left` is still less than `right`, swap the characters at indices `left` and `right`.
6
Step 6: Increment `left` and decrement `right`.
7
Step 7: Return the modified 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 pointers traverse the string at most once, resulting in linear time complexity. The space used for pointers and temporary variables is constant, resulting in constant 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

Separate Vowel Array Approach

Extract all vowels into a separate array, reverse that array, and then iterate through the original string, replacing vowels with the reversed vowels from the array.

Time: O(n)Space: O(k), where k is the number of vowels (worst case O(n))

Trade-offs: This approach is less efficient in space complexity if the number of vowels is significant. The two-pointer approach is generally preferred for its better space efficiency.

✨

Key Insights

1Insight 1: Identifying vowels efficiently: A fast way to check if a character is a vowel is crucial for performance. Using a string containing all vowels and checking for membership (e.g., using `in` in Python, `indexOf` in Java, `find` in C++, or `strchr` in C) is efficient.
2Insight 2: Two-pointer approach: Using two pointers, one at the beginning and one at the end of the string, allows for in-place swapping of vowels without needing extra space proportional to the string length. This is significantly more efficient than creating a new string.
3Insight 3: Handling edge cases: The solution needs to handle cases where there are no vowels, only vowels, or a mix of vowels and consonants. The code must correctly manage pointer movements and avoid index out-of-bounds errors.

📚

Key Terms Explained

For Students

Two Pointers

Using two indices to traverse data from different positions simultaneously.

String

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

Edge Cases

Edge case 1: Empty string: The solution handles an empty string correctly, as the `while` loop condition prevents any actions from occurring.
Edge case 2: String with only consonants: The solution correctly does nothing, as the pointers will not find any vowels to swap.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrect vowel check: Failing to correctly identify both uppercase and lowercase vowels leads to incorrect results. Using a string or set of vowels simplifies this.

✗Common mistake 2: Off-by-one errors: Incorrectly handling pointer movements (e.g., forgetting to increment `left` or decrement `right` after a swap) leads to errors. Careful attention to loop conditions and pointer updates is necessary.

✅

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 processing and manipulation: This algorithm could be used in text editors or natural language processing applications to perform specific text transformations.

🏦Real-world scenario 2: Data sanitization: In certain data preprocessing tasks, reversing specific characters (vowels in this case) could be a required step, for instance, in anonymizing data.

💡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

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ 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: November 13, 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

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

@InterviewProJun 30

❓

Frequently Asked Questions

How do I avoid infinite loops with two pointers?

Always ensure your loop condition guarantees termination (typically left < right or slow != fast). Move at least one pointer in each iteration, and handle edge cases where pointers might not move (like skipping duplicates). Draw out pointer movement on paper first.

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 should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

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 "Reverse Vowels of a String"?

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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