Mastering the Maximum Number of Vowels in a Substring Problem in Python – LeetCode 75 Explained

The Maximum Number of Vowels in a Substring of Given Length problem from LeetCode 75 is a common coding interview question that tests your understanding of sliding window techniques, string traversal and efficient counting.

The task is to find the maximum number of vowels in any substring of a given length k in a string s. This problem emphasizes efficient window-based processing instead of recalculating counts for each substring which is crucial for large strings.

In this post, we’ll walk through a clean Python solution, explain it step by step and discuss its relevance in interviews and real-world scenarios.

Problem Statement

Given a string s and an integer k, find the maximum number of vowels in any substring of length k.

Vowels are 'a', 'e', 'i', 'o', 'u'.

Understanding the Problem

A substring of length k is any contiguous sequence of k characters in the string.

Instead of checking every possible substring (which would be inefficient), we use the sliding window technique to maintain a running count of vowels as the window moves across the string.

Examples

Example 1:

Input: s = "abciiidef", k = 3  
Output: 3  
Explanation: The substring "iii" contains 3 vowels.

Example 2:

Input: s = "leetcode", k = 2  
Output: 2  
Explanation: The substring "ee" contains 2 vowels.

Python Solution – Step-by-Step Explanation

Here’s a clean and efficient Python solution using a sliding window approach:

class Solution:
    def maxVowels(self, s: str, k: int) -> int:
        vowels = set(['a','e','i','o','u'])
        max_vowels = 0
        current_vowels = 0

        # Count vowels in the initial window of size k
        for i in range(k):
            if s[i] in vowels:
                current_vowels += 1

        max_vowels = current_vowels

        # Slide the window and update the count of vowels
        for i in range(k, len(s)):
            if s[i-k] in vowels:
                current_vowels -= 1
            if s[i] in vowels:
                current_vowels += 1
            max_vowels = max(max_vowels, current_vowels)

        return max_vowels

Step 1: Initialize the Vowels Set and Counters

vowels = set(['a','e','i','o','u'])
max_vowels = 0
current_vowels = 0

• vowels helps in O(1) vowel lookups.
• current_vowels tracks vowels in the current window.
• max_vowels stores the maximum found so far.

Step 2: Count Vowels in the Initial Window

for i in range(k):
    if s[i] in vowels:
        current_vowels += 1
max_vowels = current_vowels

• Count vowels in the first k characters.
• Initialize max_vowels with this count.

Step 3: Slide the Window Across the String

for i in range(k, len(s)):
    if s[i-k] in vowels:
        current_vowels -= 1
    if s[i] in vowels:
        current_vowels += 1

• Remove the vowel leaving the window (s[i-k]).
• Add the vowel entering the window (s[i]).
• This avoids recomputing the count for each substring.

Step 4: Update Maximum Vowel Count

max_vowels = max(max_vowels, current_vowels)

• After updating the count, compare it with max_vowels and store the maximum.

Step 5: Return the Result

return max_vowels

• max_vowels now contains the highest number of vowels in any substring of length k.

Why This Solution Works

• Sliding window ensures each character is checked only once.
• Efficiently updates the vowel count without recomputation.
• Handles both lowercase vowels and large strings optimally.

Time and Space Complexity

Metric	Value
Time Complexity	O(n) – traverse the string once
Space Complexity	O(1) – only counters and a set of vowels

Edge Cases to Consider

Case	Output	Explanation
s = “aaaaa”, k = 2	2	All characters are vowels
s = “bcdfg”, k = 3	0	No vowels in the substring
s = “”, k = 1	0	Empty string
k > len(s)	0	Window size larger than string

Real-World Applications

• Text analysis: Determine regions with high vowel density in words or sentences.
• Linguistics: Study vowel distribution in language processing.
• Substring optimization: Useful in problems involving specific character constraints.
• Interview preparation: Demonstrates sliding window mastery.

Conclusion

The Maximum Number of Vowels in a Substring of Given Length problem is a classic example of sliding window optimization. By maintaining a running count and updating it as the window slides, we achieve linear time complexity and constant space usage.

Mastering this problem strengthens your skills in efficient string processing and window-based algorithms which are widely applicable in coding interviews and real-world scenarios.

Related Reads

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External Links

LeetCode Problem Page – Maximum Number of Vowels in a Substring

LeetCode Discuss – Sliding Window Solutions

GeeksforGeeks – Maximum Vowels in Substring