What is the time complexity of Count the Number of Vowel Strings in Range?

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

What data structures are used to solve Count the Number of Vowel Strings in Range?

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

What are the key insights for solving Count the Number of Vowel Strings in Range?

Insight 1: The problem requires iterating through a subset of the input array. Focusing on only the specified range (`left` to `right`) is crucial for efficiency. Insight 2: Efficiently checking if a character is a vowel is important. Using a string containing all vowels ('aeiou') and checking for membership using `in` (Python), `contains` (Java), `find` (C++), or manual comparison (C) is efficient.

Which companies ask LeetCode 2586. Count the Number of Vowel Strings in Range in interviews?

This problem is frequently asked at PayPal.

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Count the Number of Vowel Strings in Range - LeetCode 2586 Solution

Count the Number of Vowel Strings in Range - Complete Solution Guide

Count the Number of Vowel Strings in Range is LeetCode problem 2586, 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

You are given a 0-indexed array of string words and two integers left and right . A string is called a vowel string if it starts with a vowel character and ends with a vowel character where vowel characters are 'a' , 'e' , 'i' , 'o' , and 'u' . Return the number of vowel strings words[i] where i belongs to the inclusive range [left, right] . Example 1: Input: words = ["are","amy","u"], left = 0, right = 2 Output: 2 Explanation: - "are" is a vowel string because it starts with 'a' and ends with '

Detailed Explanation

The problem asks you to count the number of strings within a given range of a larger array that meet a specific criterion. The criterion is that a string is considered a 'vowel string' if and only if it begins and ends with a vowel (a, e, i, o, u). The input consists of an array of strings (`words`), and two integers (`left`, `right`) representing the inclusive range of indices to consider within the `words` array. The output is a single integer representing the count of vowel strings within the specified range.

Solution Approach

The solution iterates through the `words` array from index `left` to `right` (inclusive). For each word, it checks if the first and last characters are vowels. If both are vowels, the `count` is incremented. Finally, the function returns the total `count` of vowel strings found within the specified range.

Step-by-Step Algorithm

Step 1: Initialize a `count` variable to 0.
Step 2: Iterate through the `words` array from index `left` to `right`.
Step 3: For each word, extract the first and last characters.
Step 4: Check if both the first and last characters are vowels (using a helper string or manual comparison).
Step 5: If both are vowels, increment the `count`.
Step 6: After iterating through all words in the range, return the final `count`.

Key Insights

Insight 1: The problem requires iterating through a subset of the input array. Focusing on only the specified range (`left` to `right`) is crucial for efficiency.
Insight 2: Efficiently checking if a character is a vowel is important. Using a string containing all vowels ('aeiou') and checking for membership using `in` (Python), `contains` (Java), `find` (C++), or manual comparison (C) is efficient.
Insight 3: The problem only requires examining the first and last characters of each word, making the process linear in the length of the examined words within the range.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, String, Counting.

Companies

Asked at: PayPal.

Count the Number of Vowel Strings in Range | LeetCode 2586

Easy

Array String Counting

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

Solution Code

Available in:

class Solution {
    public int vowelStrings(String[] words, int left, int right) {
        int count = 0;
        for (int i = left; i <= right; i++) {
            String word = words[i];
            char first = word.charAt(0);
            char last = word.charAt(word.length() - 1);
            if ((first == 'a' || first == 'e' || first == 'i' || first == 'o' || first == 'u') &&
                    (last == 'a' || last == 'e' || last == 'i' || last == 'o' || last == 'u')) {
                count++;
            }
        }
        return count;
    }
}

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Problem Info

Description

Problem 2586: Count the Number of Vowel Strings in Range - Detailed explanation and solution approaches.

Topics

Array String Counting

Companies

PayPal

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:2586

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

6 steps

1
Step 1: Initialize a `count` variable to 0.
2
Step 2: Iterate through the `words` array from index `left` to `right`.
3
Step 3: For each word, extract the first and last characters.
4
Step 4: Check if both the first and last characters are vowels (using a helper string or manual comparison).
5
Step 5: If both are vowels, increment the `count`.
6
Step 6: After iterating through all words in the range, return the final `count`.

💡

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 O(n*m) where n is the number of words in the range (right - left + 1) and m is the maximum length of a word. This is because we iterate through the specified range of words (n) and in the worst case, we iterate through each character of a word to check if it is a vowel (m). The space complexity is O(1) because we only use a constant amount of extra space regardless of the input size. The vowel string is stored in constant space. The `count` variable also takes constant space.

📖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 Regular Expression

A regular expression could be used to check if a word starts and ends with a vowel in a single step. This would be slightly more concise but might be less efficient for very large datasets depending on regex engine implementation.

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

Trade-offs: Regular expressions offer conciseness but may have slightly higher overhead than the iterative approach for this specific problem. It can be beneficial if you already heavily rely on regular expressions in your code.

✨

Key Insights

1Insight 1: The problem requires iterating through a subset of the input array. Focusing on only the specified range (`left` to `right`) is crucial for efficiency.
2Insight 2: Efficiently checking if a character is a vowel is important. Using a string containing all vowels ('aeiou') and checking for membership using `in` (Python), `contains` (Java), `find` (C++), or manual comparison (C) is efficient.
3Insight 3: The problem only requires examining the first and last characters of each word, making the process linear in the length of the examined words within the range.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

String

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

Edge Cases

Edge case 1: Empty input array (`words`). The solution handles this correctly; the loop will not execute, and 0 will be returned.
Edge case 2: `left` and `right` are the same. The solution correctly processes a single word in this scenario.
Edge case 3: A word with only one character. The solution correctly checks if that single character is a vowel.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to check both the first and last character. Some solutions might only check one end.

✗Common mistake 2: Incorrectly handling the range of indices. Off-by-one errors are common when using `left` and `right` to define the iteration boundaries.

✅

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 analysis. This type of algorithm is useful for identifying words with specific phonetic properties in natural language processing tasks.

🏦Real-world scenario 2: Data validation. You could use this to verify that entries in a database comply with a specific pattern (e.g., strings representing codes must start and end with a vowel).

💡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

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

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

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

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 6, 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

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

@AlgoMasterJul 13

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

@InterviewProJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

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 "Count the Number of Vowel Strings in Range"?

For this easy-level Array 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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