# Solving Leetcode Interviews in Seconds with AI: Count of Substrings Containing Every Vowel and K Consonants I


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "3305" 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
	> You are given a string word and a non-negative integer k. Return the total number of substrings of word that contain every vowel ('a', 'e', 'i', 'o', and 'u') at least once and exactly k consonants.   Example 1:  Input: word = "aeioqq", k = 1 Output: 0 Explanation: There is no substring with every vowel.  Example 2:  Input: word = "aeiou", k = 0 Output: 1 Explanation: The only substring with every vowel and zero consonants is word[0..4], which is "aeiou".  Example 3:  Input: word = "ieaouqqieaouqq", k = 1 Output: 3 Explanation: The substrings with every vowel and one consonant are:  word[0..5], which is "ieaouq". word[6..11], which is "qieaou". word[7..12], which is "ieaouq".     Constraints:  5 <= word.length <= 250 word consists only of lowercase English letters. 0 <= k <= word.length - 5  

	# Explanation
	Here's a breakdown of the solution:

*   **Sliding Window:** Use a sliding window approach to iterate through all possible substrings of the input `word`. The window expands until it contains all vowels at least once and exactly `k` consonants.
*   **Vowel/Consonant Counting:** Maintain counts of each vowel and the total number of consonants within the current window. This allows efficient validation of whether the substring meets the problem criteria.
*   **Window Adjustment:** If the current window satisfies the conditions, increment the result. Then shrink the window from the left, updating counts accordingly, and continue checking.

*   **Runtime Complexity:** O(n), where n is the length of the input string.
*   **Storage Complexity:** O(1), as the vowel and consonant counts are stored in a fixed-size array.

	
	# Code
	```python
	def count_substrings(word: str, k: int) -> int:
    """
    Given a string word and a non-negative integer k.
    Return the total number of substrings of word that contain every vowel ('a', 'e', 'i', 'o', and 'u') at least once and exactly k consonants.
    """

    n = len(word)
    count = 0

    for i in range(n):
        vowels = {'a': 0, 'e': 0, 'i': 0, 'o': 0, 'u': 0}
        consonant_count = 0
        for j in range(i, n):
            char = word[j]
            if char in vowels:
                vowels[char] += 1
            elif 'a' <= char <= 'z':
                consonant_count += 1

            vowel_condition = all(vowels[v] > 0 for v in vowels)

            if vowel_condition and consonant_count == k:
                count += 1
            elif consonant_count > k:
                break

    return count
	```