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

###### tags: `Medium`,`String`,`Sliding Window`

[1456. Maximum Number of Vowels in a Substring of Given Length](https://leetcode.com/problems/maximum-number-of-vowels-in-a-substring-of-given-length/)

### 題目描述

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'`.

### 範例

**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`

### 解答
#### Python

class Solution:
    def maxVowels(self, s: str, k: int) -> int:
        vowels = 'aeiou'
        count = 0
        for i in range(k):
            count += s[i] in vowels
        ans = count
        for i in range(k, len(s)):
            count -= s[i - k] in vowels
            count += s[i] in vowels
            ans = max(ans, count)
        return ans

> [name=Yen-Chi Chen][time=Fri, May 5, 2023]

class Solution:
    def maxVowels(self, s: str, k: int) -> int:
        i, j = 0, 0
        vowels = 'aeiou'
        max_length = 0
        count_vowel = 0

        while i < len(s) and j < len(s):
            if j - i < k:
                if s[j] in vowels:
                    count_vowel += 1
                    max_length = max(max_length, count_vowel)
                j += 1
            else:
                if s[i] in vowels:
                    count_vowel -= 1
                i += 1

        return max_length

> [name=Ron Chen][time=Fri, May 5, 2023]

#### TypeScript

function maxVowels(s: string, k: number): number {
  let max = 0;
  let currentMax = 0;
  const vowels = ['a', 'e', 'i', 'o', 'u'];
  const queue: string[] = [];

  for (let i = 0; i < s.length; i++) {
    if (queue.length < k) {
      if (vowels.includes(s[i])) currentMax++;
      queue.push(s[i]);
    } else {
      if (vowels.includes(queue.shift()!)) currentMax--;
      if (vowels.includes(s[i])) currentMax++;
      queue.push(s[i]);
    }

    if (currentMax > max) max = currentMax;
  }

  return max;
}

學習了

function maxVowels(s: string, k: number): number {
  let max = 0;
  let currentMax = 0;
  const vowels = ['a', 'e', 'i', 'o', 'u'];

  for (let i = 0; i < s.length; i++) {
    if (i < k && vowels.includes(s[i])) {
      currentMax++;
    } else {
      if (vowels.includes(s[i])) currentMax++;
      if (vowels.includes(s[i - k])) currentMax--;
    }

    max = Math.max(max, currentMax);
  }

  return max;
}

> [name=Sheep][time=Fri, May 5, 2023]

#### C#

public int MaxVowels(string s, int k) {
    string vowels = "aeiou";
    int count = s.Take(k).Count(c => vowels.Contains(c));
    int max = 0;

    for(int i = k; i < s.Length; i++) {
        if (vowels.Contains(s[i])) count++;
        if (vowels.Contains(s[i-k])) count--;
        max = Math.Max(max, count);
    }

    return max;
}

> [name=Jim][time=Fri, May 5, 2023]

#### Javascript

function maxVowels(s, k) {
  const vowels = ['a', 'e', 'i', 'o', 'u'];
  let count = 0;
  for (let i = 0; i < k; i++) {
    if (vowels.includes(s[i])) count++;
  }

  let max = count;

  for (let i = k; i < s.length; i++) {
    if (vowels.includes(s[i])) count++;
    if (vowels.includes(s[i - k])) count--;

    max = Math.max(max, count);
  }

  return max;
}

> [name=Marsgoat][time=Fri, May 5, 2023]
### Reference

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