Introduction

In this blog post, we will explore how to solve the LeetCode problem "2586" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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]. Example 1: Input: words = ["are","amy","u"], left = 0, right = 2 Output: 2 Explanation: - "are" is a vowel string because it starts with 'a' and ends with 'e'. - "amy" is not a vowel string because it does not end with a vowel. - "u" is a vowel string because it starts with 'u' and ends with 'u'. The number of vowel strings in the mentioned range is 2. Example 2: Input: words = ["hey","aeo","mu","ooo","artro"], left = 1, right = 4 Output: 3 Explanation: - "aeo" is a vowel string because it starts with 'a' and ends with 'o'. - "mu" is not a vowel string because it does not start with a vowel. - "ooo" is a vowel string because it starts with 'o' and ends with 'o'. - "artro" is a vowel string because it starts with 'a' and ends with 'o'. The number of vowel strings in the mentioned range is 3. Constraints: 1 <= words.length <= 1000 1 <= words[i].length <= 10 words[i] consists of only lowercase English letters. 0 <= left <= right < words.length

Explanation

• Iterate through the given subarray words[left:right+1].
  • For each word, check if it starts and ends with a vowel.
  • Increment a counter if the word is a vowel string.

• Runtime Complexity: O(n*m), where n is the number of words in the range [left, right] and m is the length of the longest word. Storage Complexity: O(1)

Code

    def vowel_strings(words, left, right):
    """
    Counts the number of vowel strings in the given range.

    Args:
        words: A list of strings.
        left: The left index of the range (inclusive).
        right: The right index of the range (inclusive).

    Returns:
        The number of vowel strings in the range [left, right].
    """

    count = 0
    vowels = "aeiou"
    for i in range(left, right + 1):
        word = words[i]
        if word[0] in vowels and word[-1] in vowels:
            count += 1
    return count