What is the time complexity of Sort Vowels in a String?

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

What data structures are used to solve Sort Vowels in a String?

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

What are the key insights for solving Sort Vowels in a String?

Insight 1: Identifying vowels is a straightforward but essential part. We can use a set or similar data structure for efficient lookup. Insight 2: Counting the frequency of each vowel allows us to create a sorted sequence without needing to perform a full sorting algorithm on the entire string.

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Sort Vowels in a String - LeetCode 2785 Solution

Q: How to solve Sort Vowels in a String?

The solution iterates through the input string and counts the occurrences of each vowel (both uppercase and lowercase). It stores these counts in an array `vowel_counts` indexed by the ASCII value of the character. Then, it iterates through a pre-defined order of vowels (`ordered_vowel_chars`), constructing a sorted list of vowels (`sorted_vowels`) based on the counts. Finally, it iterates through the input string again, replacing each vowel with the next vowel from the sorted list, while leaving the consonants in their original positions.

Sort Vowels in a String - Complete Solution Guide

Sort Vowels in a String is LeetCode problem 2785, 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 0-indexed string s , permute s to get a new string t such that: All consonants remain in their original places. More formally, if there is an index i with 0 <= i < s.length such that s[i] is a consonant, then t[i] = s[i] . The vowels must be sorted in the nondecreasing order of their ASCII values. More formally, for pairs of indices i , j with 0 <= i < j < s.length such that s[i] and s[j] are vowels, then t[i] must not have a higher ASCII value than t[j] . Return the resulting string . T

Detailed Explanation

The problem asks us to sort the vowels within a given string while preserving the positions of all consonants. The vowels should be sorted in non-decreasing order based on their ASCII values. The string contains both uppercase and lowercase English letters. For example, given the string 'lEetcOde', we need to return 'lEOtcede' where 'E', 'O', and 'e' (vowels) are sorted.

Solution Approach

The solution iterates through the input string and counts the occurrences of each vowel (both uppercase and lowercase). It stores these counts in an array `vowel_counts` indexed by the ASCII value of the character. Then, it iterates through a pre-defined order of vowels (`ordered_vowel_chars`), constructing a sorted list of vowels (`sorted_vowels`) based on the counts. Finally, it iterates through the input string again, replacing each vowel with the next vowel from the sorted list, while leaving the consonants in their original positions.

Step-by-Step Algorithm

Step 1: Initialize a set `vowels_set` to efficiently check if a character is a vowel.
Step 2: Initialize an array `vowel_counts` of size 128 (to accommodate all ASCII characters) to store the count of each vowel.
Step 3: Iterate through the input string `s`. If a character is a vowel, increment its corresponding count in `vowel_counts`.
Step 4: Create a list `sorted_vowels` by iterating through a pre-defined ordered list of vowels `ordered_vowel_chars`. For each vowel, add it to `sorted_vowels` as many times as it appears in `vowel_counts`.
Step 5: Create an iterator `sorted_vowels_iter` for the `sorted_vowels` list.
Step 6: Initialize an empty list `result_chars` to store the characters of the result string.
Step 7: Iterate through the original string `s` again. If a character is a vowel, append the next vowel from `sorted_vowels_iter` to `result_chars`. Otherwise, append the original character to `result_chars`.
Step 8: Join the characters in `result_chars` to form the final sorted string.

Key Insights

Insight 1: Identifying vowels is a straightforward but essential part. We can use a set or similar data structure for efficient lookup.
Insight 2: Counting the frequency of each vowel allows us to create a sorted sequence without needing to perform a full sorting algorithm on the entire string.
Insight 3: Creating an ordered list of vowels (A, E, I, O, U, a, e, i, o, u) to iterate through simplifies the sorting and guarantees correct ASCII order.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: String, Sorting.

Sort Vowels in a String | LeetCode 2785

Medium

String Sorting

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

Solution Code

Available in:

import java.util.Arrays;
import java.util.HashSet;
import java.util.Iterator;
import java.util.Set;
import java.util.ArrayList;
import java.util.List;

class Solution {
    public String sortVowels(String s) {
        Set<Character> vowels_set = new HashSet<>(Arrays.asList('a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'));
        
        int[] vowel_counts = new int[128];
        for (char char_ : s.toCharArray()) {
            if (vowels_set.contains(char_)) {
                vowel_counts[(int) char_]++;
            }
        }
        
        List<Character> sorted_vowels = new ArrayList<>();
        char[] ordered_vowel_chars = {'A', 'E', 'I', 'O', 'U', 'a', 'e', 'i', 'o', 'u'};
        for (char v_char : ordered_vowel_chars) {
            int count = vowel_counts[(int) v_char];
            for (int i = 0; i < count; i++) {
                sorted_vowels.add(v_char);
            }
        }
        
        Iterator<Character> sorted_vowels_iter = sorted_vowels.iterator();
        
        List<Character> result_chars = new ArrayList<>();
        for (char char_ : s.toCharArray()) {
            if (vowels_set.contains(char_)) {
                result_chars.add(sorted_vowels_iter.next());
            } else {
                result_chars.add(char_);
            }
        }
        
        StringBuilder sb = new StringBuilder();
        for (char c : result_chars) {
            sb.append(c);
        }
        
        return sb.toString();
    }
}

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Problem Info

Description

Problem 2785: Sort Vowels in a String - Detailed explanation and solution approaches.

Topics

String Sorting

Stats

Difficulty:Medium

Languages:4

Problem #:2785

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

8 steps

1
Step 1: Initialize a set `vowels_set` to efficiently check if a character is a vowel.
2
Step 2: Initialize an array `vowel_counts` of size 128 (to accommodate all ASCII characters) to store the count of each vowel.
3
Step 3: Iterate through the input string `s`. If a character is a vowel, increment its corresponding count in `vowel_counts`.
4
Step 4: Create a list `sorted_vowels` by iterating through a pre-defined ordered list of vowels `ordered_vowel_chars`. For each vowel, add it to `sorted_vowels` as many times as it appears in `vowel_counts`.
5
Step 5: Create an iterator `sorted_vowels_iter` for the `sorted_vowels` list.
6
Step 6: Initialize an empty list `result_chars` to store the characters of the result string.
7
Step 7: Iterate through the original string `s` again. If a character is a vowel, append the next vowel from `sorted_vowels_iter` to `result_chars`. Otherwise, append the original character to `result_chars`.
8
Step 8: Join the characters in `result_chars` to form the final sorted 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 dominant operations are iterating through the string and building lists of vowels. The frequency counting and small vowel lists contribute constant or fixed space, so the complexity depends on the input string's size.

📖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

Sorting and Two Pointers

Extract all vowels into a separate list, sort that list, and then use two pointers (one for the original string and one for the sorted vowels) to reconstruct the string.

Time: O(n log n) due to the sorting step.Space: O(n) to store the extracted vowels.

Trade-offs: This approach may be slightly less efficient than the frequency counting approach due to the O(n log n) sorting step, but it's a reasonable alternative. It would be preferable if the vowel character set was very large and using a counting array was not feasible.

✨

Key Insights

1Insight 1: Identifying vowels is a straightforward but essential part. We can use a set or similar data structure for efficient lookup.
2Insight 2: Counting the frequency of each vowel allows us to create a sorted sequence without needing to perform a full sorting algorithm on the entire string.
3Insight 3: Creating an ordered list of vowels (A, E, I, O, U, a, e, i, o, u) to iterate through simplifies the sorting and guarantees correct ASCII order.

📚

Key Terms Explained

For Students

String

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

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Empty string. The algorithm will handle this correctly as the loops will not execute, and an empty string will be returned.
Edge case 2: String with no vowels. The `sorted_vowels` list will be empty, and the algorithm will simply return the original string.
Edge case 3: String with all vowels. The `sorted_vowels` list will contain all characters of the string, and they will be sorted correctly.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle both uppercase and lowercase vowels. The vowel check must be case-insensitive.

✗Common mistake 2: Incorrectly sorting the vowels. The problem specifies sorting based on ASCII value. Building an ordered list of vowels is a good way to achieve this.

✗Common mistake 3: Modifying the original string directly, which can lead to errors if not done carefully, and is often less readable than building a new string.

✅

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 applications where specific character sets need to be manipulated and ordered, such as lexical analysis or data cleaning.

🏦Real-world scenario 2: String manipulation in user interface development, where certain characters need to be ordered based on specific criteria for display or validation purposes.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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: August 25, 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.

@DevExpert92May 29

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

@AlgoMasterMay 22

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

@InterviewProMay 15

❓

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.

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 "Sort Vowels in a 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 sorting techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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