# Solving Leetcode Interviews in Seconds with AI: Count Vowel Substrings of a String


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2062" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), 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
	> A substring is a contiguous (non-empty) sequence of characters within a string. A vowel substring is a substring that only consists of vowels ('a', 'e', 'i', 'o', and 'u') and has all five vowels present in it. Given a string word, return the number of vowel substrings in word.   Example 1:  Input: word = "aeiouu" Output: 2 Explanation: The vowel substrings of word are as follows (underlined): - "aeiouu" - "aeiouu"  Example 2:  Input: word = "unicornarihan" Output: 0 Explanation: Not all 5 vowels are present, so there are no vowel substrings.  Example 3:  Input: word = "cuaieuouac" Output: 7 Explanation: The vowel substrings of word are as follows (underlined): - "cuaieuouac" - "cuaieuouac" - "cuaieuouac" - "cuaieuouac" - "cuaieuouac" - "cuaieuouac" - "cuaieuouac"    Constraints:  1 <= word.length <= 100 word consists of lowercase English letters only.  

	# Explanation
	Here's a solution to the problem, focusing on efficiency:

*   **Sliding Window:** Iterate through all possible start positions of substrings. For each start position, expand the substring (window) to the right.
*   **Vowel Tracking:** Use a set to keep track of the vowels present in the current substring.
*   **Early Exit:** If a non-vowel character is encountered, or all five vowels are not present after expansion, move on to the next start position.

*   **Runtime Complexity:** O(n^2), **Storage Complexity:** O(1) - due to the fixed size of the vowel set.

	
	# Code
	```python
	def count_vowel_substrings(word: str) -> int:
    """
    Counts the number of vowel substrings in a given string.

    Args:
        word: The input string.

    Returns:
        The number of vowel substrings in the string.
    """

    count = 0
    n = len(word)

    for i in range(n):
        for j in range(i, n):
            substring = word[i:j+1]
            vowels = set()
            is_vowel_substring = True

            for char in substring:
                if char in 'aeiou':
                    vowels.add(char)
                else:
                    is_vowel_substring = False
                    break
            
            if is_vowel_substring and len(vowels) == 5:
                count += 1

    return count
	```