Most vowels in any k letters

def max_vowels(s, k):
    vowels = set("aeiou")
    count = sum(c in vowels for c in s[:k])      # the first window
    best = count
    for i in range(k, len(s)):
        count += (s[i] in vowels) - (s[i - k] in vowels)
        best = max(best, count)
    return best

#include <string>
#include <algorithm>

static int isV(char c) { return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u'; }

int solve(const std::string& s, int k) {
    int count = 0;
    for (int i = 0; i < k; ++i) count += isV(s[i]);        // the first window
    int best = count;
    for (int i = k; i < (int)s.size(); ++i) {
        count += isV(s[i]) - isV(s[i - k]);
        best = std::max(best, count);
    }
    return best;
}

class Solution {
    private int isV(char c) { return "aeiou".indexOf(c) >= 0 ? 1 : 0; }

    public int maxVowels(String s, int k) {
        int count = 0;
        for (int i = 0; i < k; i++) count += isV(s.charAt(i));   // the first window
        int best = count;
        for (int i = k; i < s.length(); i++) {
            count += isV(s.charAt(i)) - isV(s.charAt(i - k));
            best = Math.max(best, count);
        }
        return best;
    }
}

function maxVowels(s, k) {
  const isV = c => 'aeiou'.includes(c) ? 1 : 0;
  let count = 0;
  for (let i = 0; i < k; i++) count += isV(s[i]);     // the first window
  let best = count;
  for (let i = k; i < s.length; i++) {
    count += isV(s[i]) - isV(s[i - k]);
    best = Math.max(best, count);
  }
  return best;
}

fn solve(s: &str, k: usize) -> i32 {
    let b = s.as_bytes();
    let v = |c: u8| matches!(c, b'a' | b'e' | b'i' | b'o' | b'u') as i32;
    let mut count: i32 = b[..k].iter().map(|&c| v(c)).sum();   // the first window
    let mut best = count;
    for i in k..b.len() {
        count += v(b[i]) - v(b[i - k]);
        best = best.max(count);
    }
    best
}

-- Indicators of vowels, then the running window sum of width k.
solve :: String -> Int -> Int
solve s k = maximum (scanl step first (zip (drop k ind) ind))
  where
    ind = [if c `elem` "aeiou" then 1 else 0 | c <- s]
    first = sum (take k ind)
    step c (new, old) = c + new - old