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Solving Leetcode Interviews in Seconds with AI: Maximum Number of Vowels in a Substring of Given Length

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2 min read

Introduction

In this blog post, we will explore how to solve the LeetCode problem "1456" 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

Given a string s and an integer k, return the maximum number of vowel letters in any substring of s with length k. Vowel letters in English are 'a', 'e', 'i', 'o', and 'u'. Example 1: Input: s = "abciiidef", k = 3 Output: 3 Explanation: The substring "iii" contains 3 vowel letters. Example 2: Input: s = "aeiou", k = 2 Output: 2 Explanation: Any substring of length 2 contains 2 vowels. Example 3: Input: s = "leetcode", k = 3 Output: 2 Explanation: "lee", "eet" and "ode" contain 2 vowels. Constraints: 1 <= s.length <= 105 s consists of lowercase English letters. 1 <= k <= s.length

Explanation

Here's the solution approach, complexity analysis, and Python code:

  • Sliding Window: Use a sliding window of size k to traverse the string s.
  • Vowel Counting: Maintain a count of vowels within the current window.
  • Maximum Tracking: Update the maximum vowel count encountered so far as the window slides.

  • Runtime Complexity: O(n), where n is the length of the string s. Storage Complexity: O(1).

Code

    def maxVowels(s: str, k: int) -> int:
    """
    Finds the maximum number of vowel letters in any substring of s with length k.

    Args:
        s: The input string.
        k: The length of the substring.

    Returns:
        The maximum number of vowel letters in any substring of s with length k.
    """

    vowels = "aeiou"
    max_vowel_count = 0
    current_vowel_count = 0

    # Initialize the window
    for i in range(k):
        if s[i] in vowels:
            current_vowel_count += 1

    max_vowel_count = current_vowel_count

    # Slide the window
    for i in range(k, len(s)):
        if s[i] in vowels:
            current_vowel_count += 1
        if s[i - k] in vowels:
            current_vowel_count -= 1
        max_vowel_count = max(max_vowel_count, current_vowel_count)

    return max_vowel_count

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