What is the time complexity of Longest Substring Of All Vowels in Order?

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

What data structures are used to solve Longest Substring Of All Vowels in Order?

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

What are the key insights for solving Longest Substring Of All Vowels in Order?

Insight 1: The problem can be solved efficiently using a sliding window approach. We can expand the window as long as the vowels are in order and reset when we encounter a vowel that breaks the alphabetical order. Insight 2: We need to keep track of the number of distinct vowels seen in the current window to determine if all five vowels are present.

Which companies ask LeetCode 1839. Longest Substring Of All Vowels in Order in interviews?

This problem is frequently asked at Thomson Reuters.

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Longest Substring Of All Vowels in Order - LeetCode 1839 Solution

Longest Substring Of All Vowels in Order - Complete Solution Guide

Longest Substring Of All Vowels in Order is LeetCode problem 1839, 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

A string is considered beautiful if it satisfies the following conditions: Each of the 5 English vowels ( 'a' , 'e' , 'i' , 'o' , 'u' ) must appear at least once in it. The letters must be sorted in alphabetical order (i.e. all 'a' s before 'e' s, all 'e' s before 'i' s, etc.). For example, strings "aeiou" and "aaaaaaeiiiioou" are considered beautiful , but "uaeio" , "aeoiu" , and "aaaeeeooo" are not beautiful . Given a string word consisting of English vowels, return the length of the longest b

Detailed Explanation

The problem asks us to find the longest substring within a given string `word` that meets two criteria to be considered 'beautiful': 1) It must contain all five vowels ('a', 'e', 'i', 'o', 'u') at least once. 2) The vowels must appear in alphabetical order (e.g., 'a' before 'e', 'e' before 'i', and so on). The input `word` consists only of vowels. If no such beautiful substring exists, the function should return 0.

Solution Approach

The provided solution uses a sliding window technique. It iterates through the string using a right pointer `r` and maintains a left pointer `l` to define the current window. It keeps track of the count of distinct vowels encountered in the current window. If the current character `word[r]` is less than the previous character `word[r-1]`, it means the alphabetical order is broken. In that case, the window is reset by moving the left pointer `l` to the current position `r` and resetting the vowel count to 1 (as the current character represents the start of a potentially new beautiful substring). If `word[r]` is greater than `word[r-1]`, it means a new distinct vowel is encountered so we increase the count. When the count reaches 5, meaning all vowels are present, the current window length (r - l + 1) is compared with the maximum length found so far, and the maximum is updated.

Step-by-Step Algorithm

Step 1: Initialize `ans` (the maximum length), `l` (left pointer), and `count` (vowel count) to 0, 0, and 1 respectively.
Step 2: Iterate through the string `word` from index 1 (the second character) to the end using a right pointer `r`.
Step 3: Inside the loop, compare the current character `word[r]` with the previous character `word[r-1]`.
Step 4: If `word[r]` is less than `word[r-1]`, reset the left pointer `l` to `r` and the vowel count `count` to 1 (start a new potential substring).
Step 5: If `word[r]` is greater than `word[r-1]`, increment the vowel count `count` by 1 (a new vowel is encountered).
Step 6: If `count` equals 5, it means all vowels are present in the current window. Update `ans` with the maximum of its current value and the length of the current window (r - l + 1).
Step 7: After the loop finishes, return the value of `ans`.

Key Insights

Insight 1: The problem can be solved efficiently using a sliding window approach. We can expand the window as long as the vowels are in order and reset when we encounter a vowel that breaks the alphabetical order.
Insight 2: We need to keep track of the number of distinct vowels seen in the current window to determine if all five vowels are present.
Insight 3: Resetting the window start and vowel count when out-of-order vowels are encountered is crucial for finding the correct longest substring.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: String, Sliding Window.

Companies

Asked at: Thomson Reuters.

Longest Substring Of All Vowels in Order | LeetCode 1839

Medium

String Sliding Window

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int longestBeautifulSubstring(String word) {
        int ans = 0;
        int l = 0;
        int count = 1;

        for (int r = 1; r < word.length(); r++) {
            if (word.charAt(r) < word.charAt(r - 1)) {
                l = r;
                count = 1;
            } else if (word.charAt(r) > word.charAt(r - 1)) {
                count++;
            }

            if (count == 5) {
                ans = Math.max(ans, r - l + 1);
            }
        }

        return ans;
    }
}

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Problem Info

Description

Problem 1839: Longest Substring Of All Vowels in Order - Detailed explanation and solution approaches.

Topics

String Sliding Window

Companies

Thomson Reuters

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1839

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 `ans` (the maximum length), `l` (left pointer), and `count` (vowel count) to 0, 0, and 1 respectively.
2
Step 2: Iterate through the string `word` from index 1 (the second character) to the end using a right pointer `r`.
3
Step 3: Inside the loop, compare the current character `word[r]` with the previous character `word[r-1]`.
4
Step 4: If `word[r]` is less than `word[r-1]`, reset the left pointer `l` to `r` and the vowel count `count` to 1 (start a new potential substring).
5
Step 5: If `word[r]` is greater than `word[r-1]`, increment the vowel count `count` by 1 (a new vowel is encountered).
6
Step 6: If `count` equals 5, it means all vowels are present in the current window. Update `ans` with the maximum of its current value and the length of the current window (r - l + 1).
7
Step 7: After the loop finishes, return the value of `ans`.

💡

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(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is linear because the algorithm iterates through the input string once. The space complexity is constant because the algorithm uses a fixed number of variables that do not depend on the size of the input.

📖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

Brute Force

Generate all possible substrings and check if each substring is beautiful. This involves iterating through all possible start and end indices of substrings.

Time: O(n^3) where n is the length of the string. O(n^2) for generating substrings and O(n) for checking if each substring is beautiful.Space: O(n) in the worst case, to store the substring.

Trade-offs: This approach is simple to understand but very inefficient for larger input strings. It's not recommended due to its high time complexity. The sliding window approach is much better.

✨

Key Insights

1Insight 1: The problem can be solved efficiently using a sliding window approach. We can expand the window as long as the vowels are in order and reset when we encounter a vowel that breaks the alphabetical order.
2Insight 2: We need to keep track of the number of distinct vowels seen in the current window to determine if all five vowels are present.
3Insight 3: Resetting the window start and vowel count when out-of-order vowels are encountered is crucial for finding the correct longest substring.

📚

Key Terms Explained

For Students

Sliding Window

A moving "window" over data to process subsets efficiently.

String

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

Edge Cases

Edge case 1: Empty string. The loop doesn't execute, and the initial value of `ans` (0) is returned, which is the correct behavior.
Edge case 2: String with only one character. The loop doesn't execute, and 0 is returned.
Edge case 3: String where no beautiful substring exists (e.g., "aaaaa"). The `count` never reaches 5, so `ans` remains 0, which is correct.
Edge case 4: String containing just one beautiful substring at the beginning (e.g., "aeiouxyz"). The algorithm correctly identifies and returns the length of "aeiou".
Edge case 5: String containing consecutive identical vowels after 'u' e.g., "aeiouuuuu". This doesn't cause issues because l pointer will not be reset after reaching vowel 'u', and the length of longest substring is calculated correctly when count equals 5.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not resetting the left pointer and count when a vowel out of order is encountered. This leads to incorrect substring lengths being considered.

✗Common mistake 2: Incorrectly checking for the presence of all five vowels. It's important to check for the *distinct* vowel count and not just the total number of vowel characters.

✗Common mistake 3: Assuming that a beautiful substring must start from index 0. The sliding window handles different starting positions.

✅

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: Data validation: This algorithm can be adapted to validate strings based on predefined sequential patterns. For example, verifying that a user input follows a specific sequence of characters.

🏦Real-world scenario 2: Bioinformatics: Finding specific patterns of nucleotides (A, C, G, T) in DNA sequences. You might need to find the longest sequence containing a specific set of nucleotides in a particular order.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

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

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: July 10, 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

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 14

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

@DevExpert92Jul 7

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

@AlgoMasterJun 30

❓

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.

What data structures complement sliding window?

Hash maps track character frequencies within the window. Deques enable O(1) access to both window boundaries. Heaps help find min/max within the window. Choose based on what property you need to maintain as the window slides.

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 "Longest Substring Of All Vowels in Order"?

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

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