1456. Maximum Number of Vowels in a Substring of Given Length

Description

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

My Solution

 
// using sliding window
/**
 * @param {string} s
 * @param {number} k
 * @return {number}
 */
var maxVowels = function(s, k) {
    let vowels = ['a', 'e', 'i', 'o', 'u'];
    let count = 0;
    let maxCount = 0;
    for(let i = 0; i < k; i++) {
        if(vowels.includes(s[i])) count++;
    }
    maxCount = count;
    for(let i = k; i < s.length; i++) {
        if(vowels.includes(s[i-k])) count--;
        if(vowels.includes(s[i])) count++;
        maxCount = Math.max(maxCount, count);
    }
    return maxCount;
};
 

Explanation

1. Purpose: The goal is to find the maximum number of vowels in any substring of length 'k' from the given string 's'.
2. Vowel List: First, make a list of vowels ('a', 'e', 'i', 'o', 'u').
3. Start Counting: Initialize two counters: 'count' for the current number of vowels, and 'maxCount' for the highest number of vowels found.
4. First Window: Look at the first substring of length 'k'. For each character, if it's a vowel (in our list), increase 'count' by 1.
5. Set Initial Max: After checking the first 'k' characters, set 'maxCount' to the value of 'count'.
6. Slide the Window: Now, start sliding the window one character at a time, from the kth character to the end of the string 's'.

• Subtract: If the character leaving the window (on the left) is a vowel, reduce 'count' by 1.
• Add: If the new character entering the window (on the right) is a vowel, increase 'count' by 1.

7. Update Max: After each slide, update 'maxCount' if 'count' is higher than the current 'maxCount'.
8. Result: Once slid through the entire string, 'maxCount' will hold the maximum number of vowels we could find in any substring of length 'k'. Return 'maxCount'.