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Count the Number of Vowel Strings in Range — Easy Problem & Solution

You are given a 0-indexed array of strings words and two integers left and right.

  • Difficulty: Easy
  • Topics: Arrays, Strings
  • Asked at: TCS, Wipro, Cognizant
  • Time limit: 2 s
  • Memory limit: 256 MB
  • Languages: JavaScript, TypeScript, Python, Java, C++, C, C#, Go, Kotlin, Swift, Rust, PHP and Ruby

Problem statement

You are given a 0-indexed array of strings words and two integers left and right.

A string is a vowel string if it both starts and ends with a vowel (a, e, i, o or u). Return how many indices i with left <= i <= right hold a vowel string.

Example 1

Input: words = ["are","amy","u"], left = 0, right = 2
Output: 2
Explanation: "are" starts with a and ends with e; "u" is a single vowel, which counts for both ends; "amy" ends in y.

Example 2

Input: words = ["hey","aeo","mu","ooo","artro"], left = 1, right = 4
Output: 3
Explanation: "aeo", "ooo" and "artro" qualify.

Example 3

Input: words = ["kata","echo"], left = 1, right = 1
Output: 1

Constraints

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

How to solve Count the Number of Vowel Strings in Range

The predicate touches two characters per word, so the whole problem is a bounded loop with a two-character test.

Approach

  1. Loop i from left to right inclusive.
  2. Take w = words[i] and test whether w[0] and w[w.length - 1] are both vowels.
  3. Count the words that pass.

Why it works

A word of length 1 has w[0] == w[w.length - 1], so the same test correctly requires that single character to be a vowel — no special case needed.

Complexity

  • Time — O(right - left)
  • Space — O(1)

Pitfalls

  • right is inclusive; i < right drops the last word.
  • Treating 'y' as a vowel — the statement lists exactly five.

Reference solution

Python

from typing import List

def vowelStrings(words: List[str], left: int, right: int) -> int:
    vowels = set("aeiou")
    total = 0
    for i in range(left, right + 1):
        w = words[i]
        if w[0] in vowels and w[-1] in vowels:
            total += 1
    return total

JavaScript

var vowelStrings = function(words, left, right) {
    var total = 0;
    for (var i = left; i <= right; i++) {
        var w = words[i];
        if ("aeiou".indexOf(w.charAt(0)) >= 0 && "aeiou".indexOf(w.charAt(w.length - 1)) >= 0) total++;
    }
    return total;
};

Also on the editorial tab: C, C#, C++, Go, Java, Kotlin, PHP, Ruby, Rust, Swift, TypeScript.

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