Solve Problem 3306. Count Of Substrings Containing Every Vowel And K Consonants II

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Problem Overview

The problem 3306. Count of Substrings Containing Every Vowel and K Consonants II requires us to count the number of substrings in a given string that contain every vowel and K consonants. This problem is a variation of the previous problem, where we need to consider both vowels and consonants in the substrings.

Understanding the Problem

To solve this problem, we need to understand the requirements and constraints. We are given a string s and an integer k. We need to count the number of substrings in s that contain every vowel and K consonants. The vowels are 'a', 'e', 'i', 'o', and 'u', and the consonants are all the remaining alphabets.

Approach: Sliding Window Technique

To solve this problem, we can use a sliding window technique. The idea is to maintain a window of characters that contain every vowel and K consonants. We can use two pointers, left and right, to represent the start and end of the window, respectively. We can then move the right pointer to the right and expand the window, and move the left pointer to the right and shrink the window, until we have a window that contains every vowel and K consonants.

Code Implementation

Here is a Python code implementation of the sliding window technique:

def countSubstrings(s: str, k: int) -> int:
    vowels = set('aeiou')
    n = len(s)
    res = 0
    
    for left in range(n):
        vowel_count = 0
        consonant_count = 0
        for right in range(left, n):
            if s[right] in vowels:
                vowel_count += 1
            else:
                consonant_count += 1
            
            if vowel_count == 5 and consonant_count == k:
                res += 1
    
    return res

Explanation

The code implementation uses a sliding window technique to count the number of substrings that contain every vowel and K consonants. The outer loop iterates over the string s and maintains a window of characters that contain every vowel and K consonants. The inner loop expands the window to the right and counts the number of vowels and consonants in the window. If the window contains every vowel and K consonants, we increment the result counter.

Time Complexity

The time complexity of the code implementation is O(n^2), where n is the length of the string s. This is because we are using a nested loop to iterate over the string s.

Space Complexity

The space complexity of the code implementation is O(1), which means the space required does not change with the size of the input string s. This is because we are using a constant amount of space to store the variables vowels, n, res, left, right, vowel_count, and consonant_count.

Example Use Cases

Here are some example use cases of the code implementation:

  • countSubstrings("abcde", 0) returns 0, because there are no substrings that contain every vowel and 0 consonants.
  • countSubstrings("abcde", 1) returns 0, because there are no substrings that contain every vowel and 1 consonant.
  • countSubstrings("abcde", 2) returns 0, because there are no substrings that contain every vowel and 2 consonants.
  • countSubstrings("abcde", 3) returns 0, because there are no substrings that contain every vowel and 3 consonants.
  • countSubstrings("abcde", 4) returns 0, because there are no substrings that contain every vowel and 4 consonants.
  • countSubstrings("abcde", 5) returns 0, because there are no substrings that contain every vowel and 5 consonants.
  • countSubstrings("abcde", 6) returns 0, because there are no substrings that contain every vowel and 6 consonants.
  • countSubstrings("abcde", 7) returns 0, because there are no substrings that contain every vowel and 7 consonants.
  • countSubstrings("abcde", 8) returns 0, because there are no substrings that contain every vowel and 8 consonants.
  • countSubstrings("abcde", 9) returns 0, because there are no substrings that contain every vowel and 9 consonants.
  • countSubstrings("abcde", 10) returns 0, because there are no substrings that contain every vowel and 10 consonants.

Conclusion

Q: What is the problem 3306. Count of Substrings Containing Every Vowel and K Consonants II?

A: The problem 3306. Count of Substrings Containing Every Vowel and K Consonants II is a LeetCode problem that requires us to count the number of substrings in a given string that contain every vowel and K consonants.

Q: What are the requirements and constraints of the problem?

A: The requirements and constraints of the problem are as follows:

  • We are given a string s and an integer k.
  • We need to count the number of substrings in s that contain every vowel and K consonants.
  • The vowels are 'a', 'e', 'i', 'o', and 'u', and the consonants are all the remaining alphabets.

Q: How can we solve the problem using a sliding window technique?

A: We can solve the problem using a sliding window technique by maintaining a window of characters that contain every vowel and K consonants. We can use two pointers, left and right, to represent the start and end of the window, respectively. We can then move the right pointer to the right and expand the window, and move the left pointer to the right and shrink the window, until we have a window that contains every vowel and K consonants.

Q: What is the time complexity of the code implementation?

A: The time complexity of the code implementation is O(n^2), where n is the length of the string s. This is because we are using a nested loop to iterate over the string s.

Q: What is the space complexity of the code implementation?

A: The space complexity of the code implementation is O(1), which means the space required does not change with the size of the input string s. This is because we are using a constant amount of space to store the variables vowels, n, res, left, right, vowel_count, and consonant_count.

Q: Can you provide example use cases of the code implementation?

A: Here are some example use cases of the code implementation:

  • countSubstrings("abcde", 0) returns 0, because there are no substrings that contain every vowel and 0 consonants.
  • countSubstrings("abcde", 1) returns 0, because there are no substrings that contain every vowel and 1 consonant.
  • countSubstrings("abcde", 2) returns 0, because there are no substrings that contain every vowel and 2 consonants.
  • countSubstrings("abcde", 3) returns 0, because there are no substrings that contain every vowel and 3 consonants.
  • countSubstrings("abcde", 4) returns 0, because there are no substrings that contain every vowel and 4 consonants.
  • countSubstrings("abcde", 5) returns 0, because there are no substrings that contain every vowel and 5 consonants.
  • countSubstrings("abcde", 6) returns 0, because there are no substrings that contain every vowel and 6 consonants.
  • countSubstrings("abcde", 7) returns 0, because there are no substrings that contain every vowel and 7 consonants.
  • countSubstrings("abcde", 8) returns 0, because there are no substrings that contain every vowel and 8 consonants.
  • countSubstrings("abcde", 9) returns 0, because there are no substrings that contain every vowel and 9 consonants.
  • countSubstrings("abcde", 10) returns 0, because there are no substrings that contain every vowel and 10 consonants.

Q: Can you provide a Python code implementation of the sliding window technique?

A: Here is a Python code implementation of the sliding window technique:

def countSubstrings(s: str, k: int) -> int:
    vowels = set('aeiou')
    n = len(s)
    res = 0
    
    for left in range(n):
        vowel_count = 0
        consonant_count = 0
        for right in range(left, n):
            if s[right] in vowels:
                vowel_count += 1
            else:
                consonant_count += 1
            
            if vowel_count == 5 and consonant_count == k:
                res += 1
    
    return res

Q: What are the advantages and disadvantages of the sliding window technique?

A: The advantages of the sliding window technique are:

  • It is a simple and efficient technique for solving problems that require counting the number of substrings that contain certain characters.
  • It can be used to solve problems that require counting the number of substrings that contain certain characters in a string.

The disadvantages of the sliding window technique are:

  • It can be slow for large input strings, because it requires iterating over the string multiple times.
  • It can be memory-intensive, because it requires storing the variables vowels, n, res, left, right, vowel_count, and consonant_count.

Conclusion

In this article, we have discussed how to solve the problem 3306. Count of Substrings Containing Every Vowel and K Consonants II using a sliding window technique. We have implemented a Python code solution that uses a sliding window technique to count the number of substrings that contain every vowel and K consonants. We have also analyzed the time and space complexity of the code implementation and provided example use cases to demonstrate the usage of the code solution.