1456 Maximum Number of Vowels in a Substring of Given Length 1

Joel Castillo Espinosa 2

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’.

Examples

  • 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

SOLUTION 3

max_vowels <- function(s, k){
  s_split <- strsplit(s, "")[[1]]
  # define the first max for the k-first elements in s
  max_vo <- sum(grepl(pattern = "[aeiou]", s_split[1:k]))
  # case if the k == length of s
  if (k == length(s)) {
    return(max_vo)
  } else {
    # in each iteration, calculate the number of vowels in the subarray s[i:i+k-1]
    max_ext <- length(s_split) - k + 1 
    for (i in 2:max_ext) {  # max_ext: max value to get a k-length subarray 
      ext <- i+k-1
      # calculate the number of vowels
      val <- sum(grepl(pattern = "[aeiou]", s_split[i:ext]))
      # and compare with the max
      max_vo <- ifelse(max_vo > val, max_vo, val)
    }
    return(max_vo)
  }
}

Examples using the function

We can use the examples presented before.

max_vowels("abciiidef", 3)

## [1] 3

max_vowels("aeiou", 2)

## [1] 2

max_vowels("leetcode", 3)

## [1] 2

  1. This problem is originally from LeetCode, you can find it in Leetcode. ↩

  2. Email: jocastillo@colmex.mx. For more content visit my website: https://joelcastillo.netlify.app
    If you have any questions or suggestions, I’d be grateful to hear from you. ↩

  3. This solution is entirely my own work. It was developed using R version 4.4.1 (2024-06-14 ucrt). ↩