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 . l e n g t h < = 1 0 5 1 <= s.length <= 10^5 1<=s.length<=105
- 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 windowfor (int i = 0; i < k; ++i) {if (isVowel(s[i])) {currentVowelCount++;}}maxVowelCount = currentVowelCount;// Slide the window and update countsfor (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;
}