What is the brute force solution for Count of Substrings Containing Every Vowel and K Consonants I?

The brute force approach for Count of Substrings Containing Every Vowel and K Consonants I is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks all possible substrings of the given string. For each substring, we count the vowels and consonants to see if it meets the criteria of containing every vowel and exactly k consonants. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count of Substrings Containing Every Vowel and K Consonants I?

Count of Substrings Containing Every Vowel and K Consonants I uses the following concepts: Hash Table, String, Sliding Window. The recommended patterns to study are: Hash Map, Sliding Window.

What are the interview tips for Count of Substrings Containing Every Vowel and K Consonants I?

Always clarify the requirements, especially regarding edge cases. Think about the efficiency of your approach and be ready to explain it. Practice explaining your thought process as you code.

What are common mistakes when solving Count of Substrings Containing Every Vowel and K Consonants I?

Not checking for all vowels correctly in the substring. Confusing consonant counts with total character counts.

#3305

Count of Substrings Containing Every Vowel and K Consonants I

Medium

Hash TableStringSliding WindowHash MapSliding Window

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal approach uses a sliding window technique to efficiently count the valid substrings. By maintaining a window that expands and contracts, we can check for the required vowels and consonants without generating all substrings explicitly.

⚙️

Algorithm

6 steps

1Step 1: Initialize a hash map to count vowels and a variable for consonants.
2Step 2: Use two pointers (left and right) to represent the current window in the string.
3Step 3: Expand the right pointer to include more characters until all vowels are found.
4Step 4: Once all vowels are found, check if the consonants in the window equal k.
5Step 5: If they do, count valid substrings by moving the left pointer to find new valid substrings.
6Step 6: Repeat until the right pointer reaches the end of the string.

solution.py27 lines

1# Full working Python code
2
3def countSubstrings(word, k):
4    vowels = set('aeiou')
5    vowel_count = {v: 0 for v in vowels}
6    left = 0
7    consonant_count = 0
8    count = 0
9    for right in range(len(word)):
10        if word[right] in vowels:
11            vowel_count[word[right]] += 1
12        elif word[right].isalpha():
13            consonant_count += 1
14        while all(vowel_count[v] > 0 for v in vowels) and consonant_count > k:
15            if word[left] in vowels:
16                vowel_count[word[left]] -= 1
17            elif word[left].isalpha():
18                consonant_count -= 1
19            left += 1
20        if all(vowel_count[v] > 0 for v in vowels) and consonant_count == k:
21            count += left + 1
22    return count
23
24# Example usage
25print(countSubstrings('aeioqq', 1))  # Output: 0
26print(countSubstrings('aeiou', 0))    # Output: 1
27print(countSubstrings('ieaouqqieaouqq', 1))  # Output: 3

ℹ

Complexity note: The time complexity is O(n) because we only traverse the string once with two pointers. The space complexity is O(1) since we only use a fixed amount of space for the vowel counts.

1Using a sliding window can significantly reduce the number of checks needed compared to brute force.
2Tracking counts of characters with a hash map allows for quick checks of conditions.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.