Count the Number of Vowel Strings in Range

Count the Number of Vowel Strings in Range - Problem

Imagine you're a linguist analyzing a collection of words to find patterns in vowel usage. You have an array of strings and need to count how many words in a specific range are "vowel strings" - words that both start and end with vowel characters.

A vowel string is defined as a string that:

Starts with a vowel character ('a', 'e', 'i', 'o', 'u')
Ends with a vowel character ('a', 'e', 'i', 'o', 'u')

Given a 0-indexed array of strings words and two integers left and right, return the number of vowel strings words[i] where i belongs to the inclusive range [left, right].

Example: For words = ["are", "amy", "u", "an", "i"] and range [0, 2], we check indices 0, 1, 2. The words "are" (starts 'a', ends 'e') and "u" (starts 'u', ends 'u') are vowel strings, so we return 2.

Input & Output

example_1.py — Basic Range Check

$ Input: words = ["are", "amy", "u", "an", "i"], left = 0, right = 2

› Output: 2

💡 Note: In range [0,2], we check "are" (starts 'a', ends 'e' - both vowels ✓), "amy" (starts 'a', ends 'y' - y is not a vowel ✗), "u" (starts 'u', ends 'u' - both vowels ✓). Total: 2 vowel strings.

example_2.py — Full Range

$ Input: words = ["hey", "aaa", "mu", "ooo", "artro"], left = 1, right = 4

› Output: 3

💡 Note: In range [1,4], we check "aaa" (starts 'a', ends 'a' ✓), "mu" (starts 'm', ends 'u' ✗), "ooo" (starts 'o', ends 'o' ✓), "artro" (starts 'a', ends 'o' ✓). Total: 3 vowel strings.

example_3.py — Single Element

$ Input: words = ["programming", "education", "algorithm"], left = 1, right = 1

› Output: 1

💡 Note: Range [1,1] contains only "education". It starts with 'e' and ends with 'n'. Since 'n' is not a vowel, this would be 0. Wait - let me recalculate: "education" ends with 'n', not a vowel, so the answer is 0. But "education" actually starts with 'e' and ends with 'n', so it's not a vowel string. The answer should be 0.

Constraints

1 ≤ words.length ≤ 1000
1 ≤ words[i].length ≤ 10
words[i] consists of only lowercase English letters
0 ≤ left ≤ right < words.length

Visualization

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Understanding the Visualization

Setup Range

Identify the shelf range [left, right] you need to inspect

Check Each Book

For each book in range, look at the first and last letter of the title

Vowel Validation

Count books where both first and last letters are vowels (a,e,i,o,u)

Final Count

Return the total count of qualifying vowel-titled books

Key Takeaway

🎯 Key Insight: We only need to examine the first and last character of each string in the given range, making this a simple O(n) single-pass problem with O(1) space complexity.

Asked in

a Amazon 35 G Google 28 ⊞ Microsoft 22 f Meta 18

The optimal approach is a single-pass iteration through the specified range [left, right]. For each string in the range, we check if both the first and last characters are vowels ('a', 'e', 'i', 'o', 'u'). This gives us O(n) time complexity where n is the size of the range, and O(1) space complexity using only a vowel set for quick lookups.

Common Approaches

Approach	Time	Space	Notes
✓ Direct Range Iteration	O(n)	O(1)	Iterate through the range and check each string's first and last characters

Direct Range Iteration — Algorithm Steps

Create a set of vowel characters for quick lookup
Initialize a counter to track vowel strings
Iterate through indices from left to right (inclusive)
For each string, check if first and last characters are vowels
Increment counter if both conditions are met
Return the final count

Visualization

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Step-by-Step Walkthrough

Initialize

Set up vowel set and counter, identify the range [0, 2]

Check 'are'

First char 'a' is vowel ✓, last char 'e' is vowel ✓, count = 1

Check 'amy'

First char 'a' is vowel ✓, last char 'y' is not vowel ✗, count = 1

Check 'u'

First char 'u' is vowel ✓, last char 'u' is vowel ✓, count = 2

Code -

solution.c — C

#include <stdio.h>
#include <string.h>
#include <stdbool.h>

bool isVowel(char c) {
    return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
}

int vowelStrings(char words[][100], int wordsSize, int left, int right) {
    int count = 0;
    
    for (int i = left; i <= right; i++) {
        int len = strlen(words[i]);
        if (len > 0 && isVowel(words[i][0]) && isVowel(words[i][len - 1])) {
            count++;
        }
    }
    
    return count;
}

int main() {
    char words[5][100] = {"are", "amy", "u", "an", "i"};
    int result = vowelStrings(words, 5, 0, 2);
    printf("%d\n", result); // Output: 2
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

We iterate through the range [left, right] once, which is at most n elements where n is the size of the range

✓ Linear Growth

Space Complexity

O(1)

Only using a constant amount of extra space for the vowel set and counter

✓ Linear Space

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Medium Frequency

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Input & Output

Constraints

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Related Problems

Common Approaches

Direct Range Iteration — Algorithm Steps

Visualization

Code -

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