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
When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:
The brute force method finds the maximum number of vowels by checking every possible short section of the word and counting the vowels in each. Think of it as sliding a window along the word, looking at every possible piece of the right size. It then picks the window with the most vowels.
Here's how the algorithm would work step-by-step:
def max_vowels_brute_force(word, substring_length):
maximum_number_of_vowels = 0
# Iterate through all possible substrings
for index in range(len(word) - substring_length + 1):
substring = word[index:index + substring_length]
number_of_vowels = 0
# Count vowels in the substring
for character in substring:
if character in 'aeiou':
number_of_vowels += 1
# Update the maximum vowel count if needed
if number_of_vowels > maximum_number_of_vowels:
maximum_number_of_vowels = number_of_vowels
return maximum_number_of_vowels
The most efficient way to solve this is by focusing on the current window of letters. We'll slide this window across the entire string of letters, keeping track of the maximum number of vowels we've seen at any point.
Here's how the algorithm would work step-by-step:
def find_max_vowels(input_string, substring_length):
vowels = "aeiou"
maximum_vowel_count = 0
current_vowel_count = 0
# Count vowels in the initial substring.
for i in range(substring_length):
if input_string[i] in vowels:
current_vowel_count += 1
maximum_vowel_count = current_vowel_count
# Slide the window across the string.
for i in range(substring_length, len(input_string)):
# Decrement if the leftmost character was a vowel.
if input_string[i - substring_length] in vowels:
current_vowel_count -= 1
# Increment if the new character is a vowel.
if input_string[i] in vowels:
current_vowel_count += 1
# Update the maximum vowel count if needed.
if current_vowel_count > maximum_vowel_count:
maximum_vowel_count = current_vowel_count
return maximum_vowel_count
Case | How to Handle |
---|---|
Null or empty string input | Return 0 immediately as there are no substrings to evaluate, preventing NullPointerException or incorrect calculations. |
k is zero or negative | Return 0 immediately because a substring of non-positive length is not valid. |
k is greater than the length of the string | Return 0 immediately because a substring longer than the string itself cannot exist. |
String contains only consonants | The sliding window approach will correctly count zero vowels for all windows, resulting in a final answer of 0. |
String contains only vowels | The sliding window approach will correctly count k vowels for the first window and the rest will be also k, resulting in a final answer of k. |
k is equal to the length of the string and string contains both vowels and consonants | The sliding window approach correctly counts all of the vowels to get the maximum. |
Integer overflow if k is very large | No risk of integer overflow in calculating vowel counts as they are bounded by the string length and k. |
Large string input | The sliding window approach provides an O(n) solution that scales linearly with the input string size. |