What is the time complexity of Vowels of All Substrings?

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

What data structures are used to solve Vowels of All Substrings?

This problem uses Math, String, Dynamic Programming, Combinatorics. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Vowels of All Substrings?

Insight 1: Instead of generating all substrings, we can focus on each vowel's contribution to the total count. A vowel at index 'i' will be present in all substrings that include index 'i'. Insight 2: The number of substrings that include index 'i' can be calculated by multiplying the number of possible start positions (from 0 to i) by the number of possible end positions (from i to n-1), where 'n' is the length of the string.

Which companies ask LeetCode 2063. Vowels of All Substrings in interviews?

This problem is frequently asked at ServiceNow.

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Vowels of All Substrings - LeetCode 2063 Solution

Vowels of All Substrings - Complete Solution Guide

Vowels of All Substrings is LeetCode problem 2063, 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 string word , return the sum of the number of vowels ( 'a' , 'e' , 'i' , 'o' , and 'u' ) in every substring of word . A substring is a contiguous (non-empty) sequence of characters within a string. Note: Due to the large constraints, the answer may not fit in a signed 32-bit integer. Please be careful during the calculations. Example 1: Input: word = "aba" Output: 6 Explanation: All possible substrings are: "a", "ab", "aba", "b", "ba", and "a". - "b" has 0 vowels in it - "a", "ab", "ba",

Detailed Explanation

The problem asks us to calculate the sum of vowels present in all possible substrings of a given string 'word'. A substring is a contiguous sequence of characters within 'word'. We need to count the number of vowels ('a', 'e', 'i', 'o', 'u') in each substring and then sum these counts up. The constraints specify that the length of 'word' can be up to 10^5, so a brute-force approach of generating all substrings and counting vowels in each might be too slow, suggesting a more efficient approach is necessary.

Solution Approach

The provided solutions efficiently count the vowels by iterating through the input string 'word'. For each character, it checks if it's a vowel. If it is, it calculates the number of substrings this vowel contributes to and adds it to the total count. The number of substrings a vowel at index 'i' contributes to is calculated as (i + 1) * (n - i), where 'n' is the length of the string. This avoids generating all substrings, resulting in a linear time complexity.

Step-by-Step Algorithm

Step 1: Initialize a variable 'total_vowels' to 0 to store the sum of vowel counts.
Step 2: Iterate through the input string 'word' using a loop with index 'i'.
Step 3: For each character at index 'i', check if it's a vowel ('a', 'e', 'i', 'o', 'u').
Step 4: If the character is a vowel, calculate the number of substrings it contributes to: (i + 1) * (n - i), where 'n' is the length of the string.
Step 5: Add this contribution to the 'total_vowels' variable.
Step 6: After iterating through all characters, return the 'total_vowels'.

Key Insights

Insight 1: Instead of generating all substrings, we can focus on each vowel's contribution to the total count. A vowel at index 'i' will be present in all substrings that include index 'i'.
Insight 2: The number of substrings that include index 'i' can be calculated by multiplying the number of possible start positions (from 0 to i) by the number of possible end positions (from i to n-1), where 'n' is the length of the string.
Insight 3: Since the answer can be very large, we need to use 'long' or 'long long' data types to avoid integer overflow (especially in Java/C++)

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Math, String, Dynamic Programming, Combinatorics.

Companies

Asked at: ServiceNow.

Vowels of All Substrings | LeetCode 2063

Medium

Math String Dynamic Programming Combinatorics

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

Solution Code

Available in:

class Solution {
    public long countVowels(String word) {
        int n = word.length();
        long total_vowels = 0;
        
        for (int i = 0; i < n; i++) {
            char charAtI = word.charAt(i);
            if ("aeiou".indexOf(charAtI) != -1) {
                // For a vowel at index i, it contributes to all substrings that contain it.
                // A substring is defined by its start and end indices.
                // To contain index i, the substring must start at an index j <= i
                // and end at an index k >= i.
                // Number of possible start indices j (from 0 to i): i + 1
                // Number of possible end indices k (from i to n-1): n - i
                // Total number of substrings containing character at index i is (i + 1) * (n - i).
                // Each such substring gets +1 vowel from this character.
                total_vowels += (long)(i + 1) * (n - i);
                
            }
        }
        
        return total_vowels;
    }
}

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Problem Info

Description

Problem 2063: Vowels of All Substrings - Detailed explanation and solution approaches.

Topics

Math String Dynamic Programming Combinatorics

Companies

ServiceNow

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2063

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 variable 'total_vowels' to 0 to store the sum of vowel counts.
2
Step 2: Iterate through the input string 'word' using a loop with index 'i'.
3
Step 3: For each character at index 'i', check if it's a vowel ('a', 'e', 'i', 'o', 'u').
4
Step 4: If the character is a vowel, calculate the number of substrings it contributes to: (i + 1) * (n - i), where 'n' is the length of the string.
5
Step 5: Add this contribution to the 'total_vowels' variable.
6
Step 6: After iterating through all characters, return the 'total_vowels'.

💡

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 we perform a single pass through the input string. The space complexity is constant because the amount of memory used does not grow with the input size; it's independent of the input string's length.

📖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 Approach

Generate all possible substrings of 'word' and then, for each substring, count the number of vowels. Sum these counts up.

Time: O(n^3) in worst case. O(n^2) to generate substrings, and O(n) to count vowels in worst case for each substring. However, on average the vowel counting may be O(1), so the total might be closer to O(n^2).Space: O(1). Dominant space usage is for the substring itself which is bounded by length 'n'.

Trade-offs: The brute-force approach is conceptually simpler but much less efficient. It's suitable for very small input sizes but becomes impractical for larger strings due to its higher time complexity. The provided solution is significantly better for typical problem constraints.

✨

Key Insights

1Insight 1: Instead of generating all substrings, we can focus on each vowel's contribution to the total count. A vowel at index 'i' will be present in all substrings that include index 'i'.
2Insight 2: The number of substrings that include index 'i' can be calculated by multiplying the number of possible start positions (from 0 to i) by the number of possible end positions (from i to n-1), where 'n' is the length of the string.
3Insight 3: Since the answer can be very large, we need to use 'long' or 'long long' data types to avoid integer overflow (especially in Java/C++)

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

String

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

Edge Cases

Edge case 1: Empty string. The solution correctly handles this because the loop won't execute, and 0 will be returned.
Edge case 2: String with no vowels. The solution correctly handles this because the conditional check for vowels will always be false, so 'total_vowels' will remain 0.
Edge case 3: Large string with many vowels. The use of `long` (or `long long`) for `total_vowels` ensures that potential integer overflows are avoided, particularly for strings close to the maximum length allowed by the constraints.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Using 'int' instead of 'long' or 'long long' for 'total_vowels'. This can lead to integer overflow when the number of vowels and string length is large, resulting in incorrect answers.

✗Common mistake 2: Incorrectly calculating the contribution of a vowel at index 'i'. Forgetting to add 1 to 'i' or incorrectly calculating the number of substrings ending at or after 'i' will lead to wrong results.

✗Common mistake 3: Trying to generate all the substrings explicitly can lead to time limit exceeded errors. The key is to count the number of substrings that contain a given vowel at index i.

✅

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 large text datasets to quantify vowel usage in different contexts or languages. The core algorithm can be adapted to count other character sets or patterns.

🏦Real-world scenario 2: Bioinformatics. Finding frequently occurring DNA or RNA subsequences (akin to substrings) that contain certain nucleotides (analogous to vowels) within genetic sequences.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

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 23, 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

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

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

@TechLeadJul 6

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

@DevExpert92Jun 29

❓

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 "Vowels of All Substrings"?

For this medium-level Math 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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