Maximum Vowels In Substring Length K

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In Maximum Vowels In Substring Length K, you are given a lowercase string s and an integer k. Your job is to find the largest number of vowels in any contiguous substring of length k. Think of it like sliding a fixed window across the string and tracking the best vowel count you ever see.

The efficient solution is a classic fixed-size sliding window. Start by counting vowels in the first window, then move one character at a time: add one when the incoming character is a vowel, subtract one when the outgoing character is a vowel, and update the maximum. This gives linear performance and matches judge expectations for mixed strings, all-vowel strings, and strings with zero vowels.

The judge includes edge behavior such as very short inputs and windows where counts change quickly between positions. Your function should return one integer, not a substring. The result is purely the maximum vowel count found among all valid windows of size k.

Examples

Example 1

Input

s = "abciiidef", k = 3

Output

Explanation

Substring "iii" contains 3 vowels.

Example 2

Input

s = "aeiou", k = 2

Output

Explanation

Any length-2 window has 2 vowels.

Example 3

Input

s = "leetcode", k = 3

Output

Explanation

Best window has 2 vowels.

Algorithm Flow

Recommendation Algorithm Flow for Maximum Vowels In Substring Length K - Budibadu

Best Answers

java

class Solution {
    private boolean isVowel(char c) {
        return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
    }
    public int maximum_vowels_in_substring_length_k(String s, int k) {
        int n = s.length();
        if (k <= 0 || k > n) return 0;
        int window = 0;
        for (int i = 0; i < k; i++) if (isVowel(s.charAt(i))) window++;
        int best = window;
        for (int i = k; i < n; i++) {
            if (isVowel(s.charAt(i))) window++;
            if (isVowel(s.charAt(i-k))) window--;
            if (window > best) best = window;
        }
        return best;
    }
}

javascript

function maximum_vowels_in_substring_length_k(s, k) {
    var n = s.length;
    if (k <= 0 || k > n) return 0;
    function isVowel(c){ return "aeiou".indexOf(c) !== -1; }
    var window = 0;
    for (var i = 0; i < k; i++) if (isVowel(s.charAt(i))) window++;
    var best = window;
    for (var j = k; j < n; j++) {
        if (isVowel(s.charAt(j))) window++;
        if (isVowel(s.charAt(j-k))) window--;
        if (window > best) best = window;
    }
    return best;
}

php

<?php
function maximum_vowels_in_substring_length_k($s, $k) {
    $n = strlen($s);
    if ($k <= 0 || $k > $n) return 0;
    $isVowel = function($c) { return strpos("aeiou", $c) !== false; };
    $window = 0;
    for ($i = 0; $i < $k; $i++) if ($isVowel($s[$i])) $window++;
    $best = $window;
    for ($i = $k; $i < $n; $i++) {
        if ($isVowel($s[$i])) $window++;
        if ($isVowel($s[$i-$k])) $window--;
        if ($window > $best) $best = $window;
    }
    return $best;
}
?>

python

def maximum_vowels_in_substring_length_k(s: str, k: int) -> int:
    n = len(s)
    if k <= 0 or k > n:
        return 0
    vowels = set(["a","e","i","o","u"])
    window = 0
    for i in range(k):
        if s[i] in vowels:
            window += 1
    best = window
    for i in range(k, n):
        if s[i] in vowels:
            window += 1
        if s[i-k] in vowels:
            window -= 1
        if window > best:
            best = window
    return best

rust

fn maximum_vowels_in_substring_length_k(s: String, k: i32) -> i32 {
    let chars: Vec<char> = s.chars().collect();
    let n = chars.len();
    if k <= 0 { return 0; }
    let ku = k as usize;
    if ku > n { return 0; }
    let is_vowel = |c: char| -> bool { c=='a' || c=='e' || c=='i' || c=='o' || c=='u' };
    let mut window = 0;
    for i in 0..ku { if is_vowel(chars[i]) { window += 1; } }
    let mut best = window;
    for i in ku..n {
        if is_vowel(chars[i]) { window += 1; }
        if is_vowel(chars[i-ku]) { window -= 1; }
        if window > best { best = window; }
    }
    best
}

typescript

function maximum_vowels_in_substring_length_k(s: string, k: number): number {
    var n = s.length;
    if (k <= 0 || k > n) return 0;
    function isVowel(c: string): boolean { return "aeiou".indexOf(c) !== -1; }
    var window = 0;
    for (var i = 0; i < k; i++) if (isVowel(s.charAt(i))) window++;
    var best = window;
    for (var j = k; j < n; j++) {
        if (isVowel(s.charAt(j))) window++;
        if (isVowel(s.charAt(j-k))) window--;
        if (window > best) best = window;
    }
    return best;
}

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Maximum Vowels In Substring Length K

Examples

Algorithm Flow

Best Answers

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