What is the time complexity of Maximum Number of Vowels in a Substring of Given Length?

The optimal solution for Maximum Number of Vowels in a Substring of Given Length runs in O(n) time and uses O(1) space. The time complexity is O(n) because we only traverse the string once. The space complexity is O(1) as we are using a fixed number of variables for counting.

What is the brute force solution for Maximum Number of Vowels in a Substring of Given Length?

The brute force approach for Maximum Number of Vowels in a Substring of Given Length is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible substring of length k in the string and counting the vowels in each substring. This is straightforward but can be inefficient for long strings. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Number of Vowels in a Substring of Given Length?

Maximum Number of Vowels in a Substring of Given Length uses the following concepts: String, Sliding Window. The recommended patterns to study are: Sliding Window, Two Pointers.

What are the interview tips for Maximum Number of Vowels in a Substring of Given Length?

Always clarify the constraints and edge cases before diving into coding. Explain your thought process as you code; this shows your understanding. Practice recognizing patterns like sliding windows or two pointers in similar problems.

What are common mistakes when solving Maximum Number of Vowels in a Substring of Given Length?

Not considering edge cases where k equals the length of the string. Failing to update the count correctly when sliding the window.

#1456

Maximum Number of Vowels in a Substring of Given Length

Medium

StringSliding WindowSliding WindowTwo Pointers

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)

The optimal solution uses a sliding window approach to maintain a window of size k. As we move the window, we efficiently update the count of vowels without recounting them from scratch.

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Algorithm

3 steps

1Step 1: Initialize a variable to count vowels in the first window of size k.
2Step 2: Slide the window one character at a time, adding the next character and removing the first character of the previous window.
3Step 3: Update the maximum count of vowels found during the sliding process.

solution.py13 lines

1def maxVowels(s, k):
2    vowels = 'aeiou'
3    max_count = 0
4    current_count = 0
5    for i in range(len(s)):
6        if s[i] in vowels:
7            current_count += 1
8        if i >= k:
9            if s[i - k] in vowels:
10                current_count -= 1
11        if i >= k - 1:
12            max_count = max(max_count, current_count)
13    return max_count

ℹ

Complexity note: The time complexity is O(n) because we only traverse the string once. The space complexity is O(1) as we are using a fixed number of variables for counting.

1Using a sliding window allows us to efficiently manage the count of vowels without re-evaluating the entire substring.
2The problem can be solved in linear time, which is crucial for handling larger inputs.

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