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 $O(1)$.
Python3
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
Java
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';
}
}
C++
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;
}
};
Go
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 := cnt
for i := k; i < len(s); i++ {
if isVowel(s[i-k]) {
cnt--
}
if isVowel(s[i]) {
cnt++
}
ans = max(ans, cnt)
}
return ans
}
TypeScript
function maxVowels(s: string, k: number): number {
const vowels = new Set(['a', 'e', 'i', 'o', 'u']);
let cnt = 0;
for (let i = 0; i < k; i++) {
if (vowels.has(s[i])) {
cnt++;
}
}
let ans = cnt;
for (let i = k; i < s.length; i++) {
if (vowels.has(s[i])) {
cnt++;
}
if (vowels.has(s[i - k])) {
cnt--;
}
ans = Math.max(ans, cnt);
}
return ans;
}
PHP
class Solution {
/**
* @param String $s
* @param Integer $k
* @return Integer
*/
function isVowel($c) {
return $c === 'a' || $c === 'e' || $c === 'i' || $c === 'o' || $c === 'u';
}
function maxVowels($s, $k) {
$cnt = 0;
for ($i = 0; $i < $k; $i++) {
if ($this->isVowel($s[$i])) {
$cnt++;
}
}
$ans = $cnt;
for ($j = $k; $j < strlen($s); $j++) {
if ($this->isVowel($s[$j - $k])) {
$cnt--;
}
if ($this->isVowel($s[$j])) {
$cnt++;
}
$ans = max($ans, $cnt);
}
return $ans;
}
}