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

Description

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 <= 105
  • s consists of lowercase English letters.
  • 1 <= k <= s.length

Solutions

Solution 1: Sliding Window

First, we count the number of vowels in the first $k$ characters, denoted as $cnt$, and initialize the answer $ans$ as $cnt$.

Then we start traversing the string from $k$. For each iteration, we add the current character to the window. If the current character is a vowel, we increment $cnt$. We remove the first character from the window. If the removed character is a vowel, we decrement $cnt$. Then, we update the answer $ans = \max(ans, cnt)$.

After the traversal, we return the answer.

The time complexity is $O(n)$, where $n$ is the length of the string $s$. The space complexity is $O(1)$.

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class Solution:
    def maxVowels(self, s: str, k: int) -> int:
        vowels = set("aeiou")
        ans = cnt = sum(c in vowels for c in s[:k])
        for i in range(k, len(s)):
            cnt += int(s[i] in vowels) - int(s[i - k] in vowels)
            ans = max(ans, cnt)
        return ans

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class Solution {
    public int maxVowels(String s, int k) {
        int cnt = 0;
        for (int i = 0; i < k; ++i) {
            if (isVowel(s.charAt(i))) {
                ++cnt;
            }
        }
        int ans = cnt;
        for (int i = k; i < s.length(); ++i) {
            if (isVowel(s.charAt(i))) {
                ++cnt;
            }
            if (isVowel(s.charAt(i - k))) {
                --cnt;
            }
            ans = Math.max(ans, cnt);
        }
        return ans;
    }

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

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class Solution {
public:
    int maxVowels(string s, int k) {
        auto isVowel = [](char c) {
            return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
        };
        int cnt = count_if(s.begin(), s.begin() + k, isVowel);
        int ans = cnt;
        for (int i = k; i < s.size(); ++i) {
            cnt += isVowel(s[i]) - isVowel(s[i - k]);
            ans = max(ans, cnt);
        }
        return ans;
    }
};

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func maxVowels(s string, k int) int {
    isVowel := func(c byte) bool {
        return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u'
    }
    cnt := 0
    for i := 0; i < k; i++ {
        if isVowel(s[i]) {
            cnt++
        }
    }
    ans :=<