Maximum Number of Vowels in a Substring of Given Length

Medium Arrays & Hashing Sliding Window · Fixed Original on LeetCode

Maximum Number of Vowels in a Substring of Given Length is a medium arrays & hashing problem solved with the sliding window · fixed pattern. The best approach, optimal (fixed sliding window), runs in O(n) time and O(1) space. Below are 2 approaches in Java, from brute force up.

Problem

Given a lowercase string s and an integer k, return the largest number of vowels in any substring of s of length exactly k.

Examples

Example 1

Input: s = "leetcode", k = 3
Output: 2
Why: "lee", "eet" and "ode" each contain 2 vowels.

Example 2

Input: s = "rhythm", k = 2
Output: 0

Constraints

1 <= k <= s.length <= 10^5
s is lowercase English letters; vowels are a, e, i, o, u.

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Solutions

Try it yourself first. Then compare: each approach lists its idea, the steps, its time and space, and the Java code.

Approach	Time	Space
Brute force	`O(n · k)`	`O(1)`
Optimal (fixed sliding window)	`O(n)`	`O(1)`

1Brute force

TimeO(n · k)

SpaceO(1)

Count vowels in each window of size k from scratch.

For each start i, count vowels in s[i..i+k-1].

Java

class Solution {
    public int maxVowels(String s, int k) {
        int best = 0;
        for (int i = 0; i + k <= s.length(); i++) {
            int c = 0;
            for (int j = i; j < i + k; j++) if (isVowel(s.charAt(j))) c++;
            best = Math.max(best, c);
        }
        return best;
    }

    private boolean isVowel(char ch) { return "aeiou".indexOf(ch) >= 0; }
}

2Optimal (fixed sliding window)

TimeO(n)

SpaceO(1)

Keep the vowel count of the current window. Add 1 when a vowel enters on the right and subtract 1 when a vowel leaves on the left.

Add each character; once i >= k, remove s[i - k].
After the window reaches size k, update best.

Java

class Solution {
    public int maxVowels(String s, int k) {
        int count = 0, best = 0;
        for (int i = 0; i < s.length(); i++) {
            if (isVowel(s.charAt(i))) count++;
            if (i >= k && isVowel(s.charAt(i - k))) count--;
            if (i >= k - 1) best = Math.max(best, count);
            if (best == k) break;
        }
        return best;
    }

    private boolean isVowel(char ch) { return "aeiou".indexOf(ch) >= 0; }
}

Edge cases to test

No vowels at all
The answer reaches k, so you can stop early

Hints

Hint 1

Count vowels in the first window, then update the count by one character in and one out.

FAQ

What is the best time complexity for Maximum Number of Vowels in a Substring of Given Length?

Optimal (fixed sliding window) runs in O(n) time and O(1) extra space.

Which pattern does Maximum Number of Vowels in a Substring of Given Length use?

It is a arrays & hashing problem that uses the sliding window · fixed pattern. Other problems with the same pattern: Maximum Average Subarray I, K Radius Subarray Averages.

Is there a brute force solution for Maximum Number of Vowels in a Substring of Given Length?

Yes. Brute force takes O(n · k) time and O(1) space. Count vowels in each window of size k from scratch.

Which edge cases should I test for Maximum Number of Vowels in a Substring of Given Length?

No vowels at all; The answer reaches k, so you can stop early.