What algorithm or data structure is used to solve Count Vowel Substrings of a String?

Count Vowel Substrings of a String uses the following concepts: Hash Table, String. The recommended patterns to study are: Sliding Window, Hash Map.

What are the interview tips for Count Vowel Substrings of a String?

Explain your thought process clearly. Consider edge cases, such as strings without vowels. Optimize your solution after discussing the brute force approach.

What are common mistakes when solving Count Vowel Substrings of a String?

Failing to check for consonants and generating unnecessary substrings. Not using a set to track unique vowels.

Count Vowel Substrings of a String — LeetCode #2062 (Easy)

Tags: Hash Table, String

Related patterns: Sliding Window, Hash Map

Brute Force approach

Time complexity: O(n²). Space complexity: O(1).

The brute force approach involves generating all possible substrings of the input string and checking if they contain all five vowels. This method is straightforward but can be inefficient due to the large number of substrings.

The time complexity is O(n²) because for each character, we may generate up to n substrings, leading to a quadratic number of checks. The space complexity is O(1) since we are using a fixed amount of extra space.

Step 1: Iterate through each character in the string as a starting point for substrings.
Step 2: For each starting point, generate all possible substrings until a consonant is encountered.
Step 3: Check if the substring contains all five vowels. If it does, increment the count.

For word = 'aeiouu': 1. Start at index 0 ('a'), generate 'a', 'ae', 'aei', 'aeio', 'aeiou' (count = 1) 2. Generate 'aeiouu' (count = 2) 3. Start at index 1 ('e'), generate 'e', 'ei', 'eio', 'eiou', 'eiouu' (no new count) 4. Start at index 2 ('i'), generate 'i', 'io', 'iou', 'iouu' (no new count) 5. Start at index 3 ('o'), generate 'o', 'ou', 'ouu' (no new count) 6. Start at index 4 ('u'), generate 'u', 'uu' (no new count) Final count = 2.

Common Mistakes

Failing to check for consonants and generating unnecessary substrings.
Not using a set to track unique vowels.

Interview Tips

Explain your thought process clearly.
Consider edge cases, such as strings without vowels.
Optimize your solution after discussing the brute force approach.

#2062

Count Vowel Substrings of a String

Easy

Hash Table↗String↗Sliding Window↗Hash Map↗

LeetCode ↗

Approaches

💡

Intuition

Time O(n²)Space O(1)

⚙️

Algorithm

3 steps

1Step 1: Iterate through each character in the string as a starting point for substrings.
2Step 2: For each starting point, generate all possible substrings until a consonant is encountered.
3Step 3: Check if the substring contains all five vowels. If it does, increment the count.

solution.py12 lines

1def countVowelSubstrings(word):
2    vowels = set('aeiou')
3    count = 0
4    n = len(word)
5    for i in range(n):
6        if word[i] in vowels:
7            for j in range(i, n):
8                if word[j] not in vowels:
9                    break
10                if vowels.issubset(set(word[i:j+1])):
11                    count += 1
12    return count

ℹ

Complexity note: The time complexity is O(n²) because for each character, we may generate up to n substrings, leading to a quadratic number of checks. The space complexity is O(1) since we are using a fixed amount of extra space.

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