How do you solve Vowels of All Substrings on LeetCode?

Vowels of All Substrings (LeetCode #2063) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Each vowel's contribution can be calculated based on its position, avoiding the need to generate all substrings.

What is the brute force solution for Vowels of All Substrings?

The brute force approach for Vowels of All Substrings is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach generates all possible substrings and counts the vowels in each. While this is straightforward, it becomes inefficient for longer strings due to the exponential growth of substrings. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Vowels of All Substrings?

Vowels of All Substrings uses the following concepts: Math, String, Dynamic Programming, Combinatorics. The recommended patterns to study are: Combinatorics, Dynamic Programming.

What are the interview tips for Vowels of All Substrings?

Always consider the constraints of the problem and whether a brute-force solution is feasible. Think about how to optimize your solution by analyzing the problem from different angles. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Vowels of All Substrings?

Students often try to generate all substrings without realizing the inefficiency of this approach. Failing to account for the contribution of each vowel based on its position can lead to incorrect results.

#2063

Vowels of All Substrings

Medium

MathStringDynamic ProgrammingCombinatoricsCombinatoricsDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

Instead of generating all substrings, we can calculate how many substrings each vowel contributes to. Each vowel at position i contributes to substrings that start from any index before or at i and end at any index after or at i.

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Algorithm

3 steps

1Step 1: Initialize a variable to keep track of the total vowel count.
2Step 2: Iterate through each character in the string. For each vowel found, calculate how many substrings it contributes to based on its position.
3Step 3: For a vowel at index i, it contributes to (i + 1) * (n - i) substrings, where n is the length of the string.

solution.py8 lines

1# Full working Python code
2word = 'aba'
3total_vowels = 0
4n = len(word)
5for i, char in enumerate(word):
6    if char in 'aeiou':
7        total_vowels += (i + 1) * (n - i)
8print(total_vowels)

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Complexity note: The time complexity is O(n) because we only make a single pass through the string. The space complexity is O(1) since we are using a constant amount of extra space.

1Each vowel's contribution can be calculated based on its position, avoiding the need to generate all substrings.
2Understanding how many substrings a character contributes to is key to solving similar problems efficiently.

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