LeetCode #1456 — MEDIUM

Maximum Number of Vowels in a Substring of Given Length

Move from brute-force thinking to an efficient approach using sliding window strategy.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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'.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Sliding Window

Example 1

"abciiidef"
3

Example 2

"aeiou"
2

Example 3

"leetcode"
3

Core Insight

What unlocks the optimal approach

Keep a window of size k and maintain the number of vowels in it.
Keep moving the window and update the number of vowels while moving. Answer is max number of vowels of any window.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1456: Maximum Number of Vowels in a Substring of Given Length
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';
    }
}

// Accepted solution for LeetCode #1456: Maximum Number of Vowels in a Substring of Given Length
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
}

# Accepted solution for LeetCode #1456: Maximum Number of Vowels in a Substring of Given Length
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

// Accepted solution for LeetCode #1456: Maximum Number of Vowels in a Substring of Given Length
struct Solution;

impl Solution {
    fn max_vowels(s: String, k: i32) -> i32 {
        let n = s.len();
        let s: Vec<char> = s.chars().collect();
        let k = k as usize;
        let mut cur = 0;
        let mut res = 0;
        for i in 0..n {
            if Self::is_vowel(s[i]) {
                cur += 1;
            }
            if i >= k {
                if Self::is_vowel(s[i - k]) {
                    cur -= 1;
                }
            }
            res = res.max(cur);
        }
        res
    }

    fn is_vowel(c: char) -> bool {
        matches!(c, 'a' | 'e' | 'i' | 'o' | 'u')
    }
}

#[test]
fn test() {
    let s = "abciiidef".to_string();
    let k = 3;
    let res = 3;
    assert_eq!(Solution::max_vowels(s, k), res);
    let s = "aeiou".to_string();
    let k = 2;
    let res = 2;
    assert_eq!(Solution::max_vowels(s, k), res);
    let s = "leetcode".to_string();
    let k = 3;
    let res = 2;
    assert_eq!(Solution::max_vowels(s, k), res);
    let s = "rhythms".to_string();
    let k = 4;
    let res = 0;
    assert_eq!(Solution::max_vowels(s, k), res);
    let s = "tryhard".to_string();
    let k = 4;
    let res = 1;
    assert_eq!(Solution::max_vowels(s, k), res);
}

// Accepted solution for LeetCode #1456: Maximum Number of Vowels in a Substring of Given Length
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(k)

Approach Breakdown

BRUTE FORCE

O(n × k) time

O(1) space

For each starting index, scan the next k elements to compute the window aggregate. There are n−k+1 starting positions, each requiring O(k) work, giving O(n × k) total. No extra space since we recompute from scratch each time.

SLIDING WINDOW

O(n) time

O(k) space

The window expands and contracts as we scan left to right. Each element enters the window at most once and leaves at most once, giving 2n total operations = O(n). Space depends on what we track inside the window (a hash map of at most k distinct elements, or O(1) for a fixed-size window).

Shortcut: Each element enters and exits the window once → O(n) amortized, regardless of window size.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Shrinking the window only once

Wrong move: Using `if` instead of `while` leaves the window invalid for multiple iterations.

Usually fails on: Over-limit windows stay invalid and produce wrong lengths/counts.

Fix: Shrink in a `while` loop until the invariant is valid again.