What is the time complexity of Count Vowel Strings in Ranges?

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

What data structures are used to solve Count Vowel Strings in Ranges?

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

What are the key insights for solving Count Vowel Strings in Ranges?

Insight 1: The core of the problem lies in efficiently determining if a string starts and ends with a vowel. Insight 2: Using prefix sums can significantly optimize the query processing time, avoiding repeated iterations over the `words` array for each query.

Which companies ask LeetCode 2559. Count Vowel Strings in Ranges in interviews?

This problem is frequently asked at Atlassian, IBM, PayPal.

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Count Vowel Strings in Ranges - LeetCode 2559 Solution

Count Vowel Strings in Ranges - Complete Solution Guide

Count Vowel Strings in Ranges is LeetCode problem 2559, 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

You are given a 0-indexed array of strings words and a 2D array of integers queries . Each query queries[i] = [l i , r i ] asks us to find the number of strings present at the indices ranging from l i to r i (both inclusive ) of words that start and end with a vowel. Return an array ans of size queries.length , where ans[i] is the answer to the i th query . Note that the vowel letters are 'a' , 'e' , 'i' , 'o' , and 'u' . Example 1: Input: words = ["aba","bcb","ece","aa","e"], queries = [[0,2],[

Detailed Explanation

The problem asks us to process a list of strings (`words`) and a list of queries (`queries`). Each query specifies a range of indices within the `words` list. For each query, we need to count how many strings within the specified range start and end with a vowel ('a', 'e', 'i', 'o', 'u'). The final result should be an array where each element represents the answer to the corresponding query.

Solution Approach

The provided solutions utilize the prefix sum technique to efficiently answer the queries. First, an array `prefix_sums` is created. `prefix_sums[i]` stores the cumulative count of vowel strings from the beginning of the `words` array up to index `i-1`. To answer a query `[l, r]`, we can simply calculate `prefix_sums[r+1] - prefix_sums[l]`. This gives us the count of vowel strings within the range `[l, r]` in O(1) time.

Step-by-Step Algorithm

Step 1: Initialize a set (or similar data structure) containing the vowels: 'a', 'e', 'i', 'o', 'u'.
Step 2: Create a `prefix_sums` array of size `len(words) + 1`, initialized with all zeros. `prefix_sums[0]` is always 0.
Step 3: Iterate through the `words` array. For each word, check if it starts and ends with a vowel using the vowel set.
Step 4: If a word is a vowel string, increment the corresponding `prefix_sums` value: `prefix_sums[i+1] = prefix_sums[i] + 1`. Otherwise, `prefix_sums[i+1] = prefix_sums[i]`.
Step 5: Iterate through the `queries` array. For each query `[l, r]`, calculate the answer as `prefix_sums[r+1] - prefix_sums[l]` and store it in the result array.
Step 6: Return the result array.

Key Insights

Insight 1: The core of the problem lies in efficiently determining if a string starts and ends with a vowel.
Insight 2: Using prefix sums can significantly optimize the query processing time, avoiding repeated iterations over the `words` array for each query.
Insight 3: Pre-calculating and storing whether each word is a vowel string allows for O(1) access during prefix sum calculation.

Complexity Analysis

Time Complexity: O(n+q)

Space Complexity: O(n)

Topics

This problem involves: Array, String, Prefix Sum.

Companies

Asked at: Atlassian, IBM, PayPal.

Count Vowel Strings in Ranges | LeetCode 2559

Medium

Array String Prefix Sum

Time: O(n+q)Space: O(n)

Solution Code

Available in:

class Solution {
    public int[] vowelStrings(String[] words, int[][] queries) {
        java.util.Set<Character> vowels = new java.util.HashSet<>(java.util.Arrays.asList('a', 'e', 'i', 'o', 'u'));
        
        int[] prefix_sums = new int[words.length + 1];
        
        for (int i = 0; i < words.length; i++) {
            String word = words[i];
            boolean is_vowel_string = vowels.contains(word.charAt(0)) && vowels.contains(word.charAt(word.length() - 1));
            prefix_sums[i + 1] = prefix_sums[i] + (is_vowel_string ? 1 : 0);
        }
        
        int[] result = new int[queries.length];
        for (int i = 0; i < queries.length; i++) {
            int li = queries[i][0];
            int ri = queries[i][1];
            result[i] = prefix_sums[ri + 1] - prefix_sums[li];
        }
        
        return result;
    }
}

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Problem Info

Description

Problem 2559: Count Vowel Strings in Ranges - Detailed explanation and solution approaches.

Topics

Array String Prefix Sum

Companies

AtlassianIBMPayPal

Stats

Difficulty:Medium

Languages:4

Companies:3

Problem #:2559

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 set (or similar data structure) containing the vowels: 'a', 'e', 'i', 'o', 'u'.
2
Step 2: Create a `prefix_sums` array of size `len(words) + 1`, initialized with all zeros. `prefix_sums[0]` is always 0.
3
Step 3: Iterate through the `words` array. For each word, check if it starts and ends with a vowel using the vowel set.
4
Step 4: If a word is a vowel string, increment the corresponding `prefix_sums` value: `prefix_sums[i+1] = prefix_sums[i] + 1`. Otherwise, `prefix_sums[i+1] = prefix_sums[i]`.
5
Step 5: Iterate through the `queries` array. For each query `[l, r]`, calculate the answer as `prefix_sums[r+1] - prefix_sums[l]` and store it in the result array.
6
Step 6: Return the result array.

💡

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+q)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The prefix sum array stores the count of vowel strings encountered up to a given index. Constructing the prefix sum array requires iterating through all the strings in the input array `words`, hence O(n). Each query then involves a constant time lookup and subtraction operation from the precomputed prefix sums, leading to O(1) per query or O(q) overall for all queries.

📖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

For each query, iterate through the `words` array within the given range and count the number of vowel strings.

Time: O(n*q), where n is the length of the `words` array and q is the number of queries.Space: O(1)

Trade-offs: This approach is simpler to implement but significantly less efficient than the prefix sum approach, especially when the number of queries is large. It avoids the O(n) space complexity of prefix sums, making it favorable only if memory usage is an extreme constraint and the number of queries is very small.

✨

Key Insights

1Insight 1: The core of the problem lies in efficiently determining if a string starts and ends with a vowel.
2Insight 2: Using prefix sums can significantly optimize the query processing time, avoiding repeated iterations over the `words` array for each query.
3Insight 3: Pre-calculating and storing whether each word is a vowel string allows for O(1) access during prefix sum calculation.

📚

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 string in `words`. The provided solutions correctly handle this case because `word[0]` and `word[-1]` checks are only performed if the string has a length greater than 0, preventing index out of bounds errors.
Edge case 2: Query where `l` and `r` are the same (e.g., `[1, 1]`). The prefix sum approach correctly calculates the result for this case as `prefix_sums[2] - prefix_sums[1]`, which is the count of vowel strings at index 1.
Edge case 3: A word with only one character. The code handles this case correctly because it still checks if the first and last (which are the same) character is a vowel.
Edge case 4: Very large input arrays (10^5). The prefix sum approach handles these efficiently because each query is resolved in O(1) time after initial setup.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where a word is empty, potentially leading to out-of-bounds access.

✗Common mistake 2: Incorrectly calculating the prefix sum, such as not including the current word in the cumulative count or offsetting the indices.

✗Common mistake 3: Forgetting that the range in the query is inclusive, leading to incorrect indices when calculating the prefix sum difference.

✗Common mistake 4: Using a less efficient data structure (e.g., list instead of set) to check for vowels. Though still functionally correct, this increases execution time slightly.

✅

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: Analyzing text data for patterns. This approach can be used to efficiently count the number of sentences (represented as strings) within a document that meet certain criteria, like starting and ending with specific keywords.

🏦Real-world scenario 2: Processing log files. When analyzing server logs or application logs, we might need to quickly count events within specific time ranges. The strings in the log can be processed and queried using prefix sums to determine the frequency of certain events within certain time windows.

💡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

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

⚠️ 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: 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

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

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 "Count Vowel Strings in Ranges"?

For this medium-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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