LeetCode //C - 1456. Maximum Number of Vowels in a Substring of Given Length

1456. Maximum Number of Vowels in a Substring of Given Length

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’.
?

Example 1:

Input: s = “abciiidef”, k = 3
Output: 3
Explanation: The substring “iii” contains 3 vowel letters.

Example 2:

Input: s = “aeiou”, k = 2
Output: 2
Explanation: Any substring of length 2 contains 2 vowels.

Example 3:

Input: s = “leetcode”, k = 3
Output: 2
Explanation: “lee”, “eet” and “ode” contain 2 vowels.

Constraints:

• $1<=s.length<=10^5$
• s consists of lowercase English letters.
• 1 <= k <= s.length

From: LeetCode
Link: 1456. Maximum Number of Vowels in a Substring of Given Length

Solution:

Ideas：

• s is the input string.
• k is the length of the substring.
• isVowel is a helper function to check if a character is a vowel.
• The function first counts the number of vowels in the first window of size k.
• Then it iterates through the string, sliding the window one character at a time, updating the vowel count, and keeping track of the maximum vowel count found.

Code:

// Helper function to check if a character is a vowel
bool isVowel(char c) {
    return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
}

int maxVowels(char *s, int k) {
    int maxVowelCount = 0, currentVowelCount = 0;

    // Initial count of vowels in the first window
    for (int i = 0; i < k; ++i) {
        if (isVowel(s[i])) {
            currentVowelCount++;
        }
    }
    maxVowelCount = currentVowelCount;

    // Slide the window and update counts
    for (int i = k; s[i] != '\0'; ++i) {
        if (isVowel(s[i])) {
            currentVowelCount++;
        }
        if (isVowel(s[i - k])) {
            currentVowelCount--;
        }
        if (currentVowelCount > maxVowelCount) {
            maxVowelCount = currentVowelCount;
        }
    }

    return maxVowelCount;
}