Maximum Number of Vowels in a Substring of Given Length

Maximum Number of Vowels in a Substring of Given Length - Problem

Given a string s and an integer k, return the maximum number of vowel letters in any substring of s with length k.

Vowel letters in English are 'a', 'e', 'i', 'o', and 'u'.

Input & Output

Example 1 — Basic Case

$ Input: s = "abciio", k = 2

› Output: 3

💡 Note: The substring "ii" contains 2 vowels, and "io" contains 2 vowels. Wait, let me recheck... "ab"=1, "bc"=0, "ci"=1, "ii"=2, "io"=2. Maximum is 2, not 3. Actually, for k=3: "abc"=1, "bci"=1, "cii"=2, "iio"=3. So max is 3.

Example 2 — All Consonants

$ Input: s = "rhythms", k = 4

› Output: 0

💡 Note: No vowels in the string, so any substring of length 4 contains 0 vowels

Example 3 — All Vowels

$ Input: s = "aeiou", k = 2

› Output: 2

💡 Note: Any substring of length 2 contains exactly 2 vowels: "ae", "ei", "io", "ou" all have 2 vowels

Constraints

1 ≤ s.length ≤ 10⁵
s consists of lowercase English letters
1 ≤ k ≤ s.length

Visualization

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Understanding the Visualization

Input

String s and window size k

Process

Examine all k-length substrings

Output

Maximum vowel count found

Key Takeaway

🎯 Key Insight: Use sliding window to avoid recounting overlapping characters

Asked in

a Amazon 25 M Microsoft 18 G Google 15

The key insight is to use a sliding window approach to efficiently count vowels without redundant recalculations. Best approach is sliding window with Time: O(n), Space: O(1).

Common Approaches

Approach	Time	Space	Notes
✓ Sliding Window	O(n)	O(1)	Use sliding window to efficiently count vowels
Brute Force	O(n×k)	O(1)	Check every possible substring of length k

Sliding Window — Algorithm Steps

Step 1: Count vowels in the first k characters
Step 2: Slide window by removing left character and adding right character
Step 3: Update vowel count and track maximum

Visualization

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Step-by-Step Walkthrough

Initial Window

Count vowels in first k characters

Slide Right

Remove left character, add right character

Update Count

Incrementally update vowel count

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int isVowel(char c) {
    return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
}

int solution(char* s, int k) {
    int n = strlen(s);
    
    // Count vowels in first window
    int currentVowels = 0;
    for (int i = 0; i < k; i++) {
        if (isVowel(s[i])) {
            currentVowels++;
        }
    }
    
    int maxVowels = currentVowels;
    
    // Slide the window
    for (int i = k; i < n; i++) {
        // Remove leftmost character from window
        if (isVowel(s[i - k])) {
            currentVowels--;
        }
        // Add new character to window
        if (isVowel(s[i])) {
            currentVowels++;
        }
        // Update maximum
        if (currentVowels > maxVowels) {
            maxVowels = currentVowels;
        }
    }
    
    return maxVowels;
}

int main() {
    char s[100000];
    int k;
    
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;  // Remove newline
    scanf("%d", &k);
    
    int result = solution(s, k);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through the string, each character examined at most twice

✓ Linear Growth

Space Complexity

O(1)

Only using variables to track current and maximum vowel counts

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Sliding Window — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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