Welcome to Subscribe On Youtube

Formatted question description: https://leetcode.ca/all/1456.html

1456. Maximum Number of Vowels in a Substring of Given Length (Medium)

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, 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.

Example 4:

Input: s = "rhythms", k = 4
Output: 0
Explanation: We can see that s doesn't have any vowel letters.

Example 5:

Input: s = "tryhard", k = 4
Output: 1

 

Constraints:

  • 1 <= s.length <= 10^5
  • s consists of lowercase English letters.
  • 1 <= k <= s.length

Related Topics:
String, Sliding Window

Solution 1. Sliding window

We keep a sliding window of size k. Let i be the index of the new element getting into the window. We increment cnt if s[i] is vowel.

If i >= k, we need to pop s[i - k] out of the window and decrement cnt if s[i - k] is vowel. The answer is the maximum cnt.

// OJ: https://leetcode.com/problems/maximum-number-of-vowels-in-a-substring-of-given-length/
// Time: O(N)
// Space: O(1)
class Solution {
    bool isVowel(char c) {
        return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
    }
public:
    int maxVowels(string s, int k) {
        int cnt = 0, ans = 0;
        for (int i = 0; i < s.size(); ++i) {
            if (i >= k) cnt -= isVowel(s[i - k]);
            cnt += isVowel(s[i]);
            ans = max(ans, cnt);
        }
        return ans;
    }
};

  • class Solution {
    public int maxVowels(String s, int k) {
        int count = 0;
        int length = s.length();
        if (k > length)
            return 0;
        for (int i = 0; i < k; i++) {
            if (isVowel(s.charAt(i)))
                count++;
        }
        int maxCount = count;
        for (int i = k; i < length; i++) {
            char prevC = s.charAt(i - k), currC = s.charAt(i);
            if (isVowel(prevC))
                count--;
            if (isVowel(currC))
                count++;
            maxCount = Math.max(maxCount, count);
        }
        return maxCount;
    }

    public boolean isVowel(char c) {
        return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
    }
}

  • // OJ: https://leetcode.com/problems/maximum-number-of-vowels-in-a-substring-of-given-length/
// Time: O(N)
// Space: O(1)
class Solution {
    bool isVowel(char c) {
        return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
    }
public:
    int maxVowels(string s, int k) {
        int cnt = 0, ans = 0;
        for (int i = 0; i < s.size(); ++i) {
            if (i >= k) cnt -= isVowel(s[i - k]);
            cnt += isVowel(s[i]);
            ans = max(ans, cnt);
        }
        return ans;
    }
};

  • class Solution:
    def maxVowels(self, s: str, k: int) -> int:
        vowels = set('aeiou')
        t = sum(c in vowels for c in s[:k])
        ans = t
        for i in range(k, len(s)):
            t += s[i] in vowels
            t -= s[i - k] in vowels
            ans = max(ans, t)
        return ans

All Problems

All Solutions