Count the Number of Vowel Strings in Range - Problem

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].

Input & Output

Example 1 — Basic Range

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

› Output: 2

💡 Note: "are" starts with 'a' and ends with 'e' (both vowels). "u" starts and ends with 'u' (vowel). "amy" starts with 'a' but ends with 'y' (not a vowel).

Example 2 — Partial Range

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

› Output: 3

💡 Note: In range [1,4]: "aeo" (a-o), "ooo" (o-o), "artro" (a-o) are vowel strings. "mu" (m-u) is not because it starts with 'm'.

Example 3 — Single Element

$ Input: words = ["a","b","c"], left = 1, right = 1

› Output: 0

💡 Note: Only checking words[1] = "b", which starts and ends with 'b' (not a vowel).

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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Asked in

a Amazon 25 M Microsoft 18

The key insight is to iterate through only the specified range and check if each string starts and ends with a vowel. Best approach is direct iteration with O(n) time and O(1) space complexity.

Common Approaches

✓ Direct Range Iteration

⏱️ Time: O(n) Space: O(1)

Go through each index from left to right, check if the string at that index starts and ends with a vowel, and count valid strings.

Direct Range Iteration — Algorithm Steps

Iterate through indices from left to right
For each string, check first and last character
Count strings that start and end with vowels

Visualization

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Step-by-Step Walkthrough

Range Selection

Select indices from left to right

Vowel Check

Check first and last character of each string

Count Result

Count valid vowel strings

Code -

solution.c — C

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

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

int solution(char words[][101], int wordsCount, int left, int right) {
    int count = 0;
    
    for (int i = left; i <= right; i++) {
        char* word = words[i];
        int len = strlen(word);
        if (isVowel(word[0]) && isVowel(word[len - 1])) {
            count++;
        }
    }
    
    return count;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

int main() {
    char line[10001];
    fgets(line, sizeof(line), stdin);
    
    // Parse array manually
    char words[1001][101];
    int wordsCount = 0;
    
    char* token = strtok(line, "[],\"\n");
    while (token != NULL) {
        if (strlen(token) > 0 && token[0] != ' ') {
            strcpy(words[wordsCount], token);
            wordsCount++;
        }
        token = strtok(NULL, "[],\"\n");
    }
    
    int left, right;
    scanf("%d", &left);
    scanf("%d", &right);
    
    int result = solution(words, wordsCount, left, right);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

We iterate through at most n strings in the range, checking first and last character is O(1)

✓ Linear Growth

Space Complexity

O(1)

Only using a counter variable and vowel set, no extra data structures

✓ Linear Space

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Count the Number of Vowel Strings in Range - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Direct Range Iteration — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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