Maximum Vowels In Substring Length K

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You are given a lowercase string s and an integer k. Among all substrings of length k, return the maximum number of vowels found in any one substring.

This is a fixed-size sliding window counting problem. Start by counting vowels in the first k characters. Then move the window one character at a time: add 1 if the incoming character is a vowel, subtract 1 if the outgoing character is a vowel, and track the best count seen so far.

Because each character is processed a constant number of times, the solution runs in O(n) time.

Example 1:

Input: s = "abciiidef", k = 3

Output: 3

Explanation: Substring "iii" has 3 vowels, which is the maximum.

Example 2:

Input: s = "aeiou", k = 2

Output: 2

Explanation: Every window of size 2 contains exactly 2 vowels.

Example 3:

Input: s = "leetcode", k = 3

Output: 2

Explanation: Best windows like "lee" or "eet" contain 2 vowels.

Constraints: 1 <= s.length <= 10^5, 1 <= k <= s.length

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

Example 1:

Example 2:

Example 3:

Algorithm Flow

Best Answers

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