2586. Count the Number of Vowel Strings in Range

EasyLeetCode

AYou are given a 0-indexed array of string words and two integers left and right.

A string is called a vowel string if it starts with a vowel character and ends with a vowel character where vowel characters are 'a', 'e', 'i', 'o', and 'u'.

Return the number of vowel strings words[i] where i belongs to the inclusive range [left, right].

Example 1:

Input: words = ["are","amy","u"], left = 0, right = 2

Output: 2

Explanation:

• "are" is a vowel string because it starts with 'a' and ends with 'e'.
• "amy" is not a vowel string because it does not end with a vowel.
• "u" is a vowel string because it starts with 'u' and ends with 'u'. The number of vowel strings in the mentioned range is 2.

Example 2:

Input: words = ["hey","aeo","mu","ooo","artro"], left = 1, right = 4

Output: 3

Explanation:

• "aeo" is a vowel string because it starts with 'a' and ends with 'o'.
• "mu" is not a vowel string because it does not start with a vowel.
• "ooo" is a vowel string because it starts with 'o' and ends with 'o'.
• "artro" is a vowel string because it starts with 'a' and ends with 'o'. The number of vowel strings in the mentioned range is 3.

Constraints:

• 1 <= words.length <= 1000
• 1 <= words[i].length <= 10
• words[i] consists of only lowercase English letters
• 0 <= left <= right < words.length

How to solve the problem

• Using words[i][0] and words[i][-1]to get the first and last character.

Code

class Solution:
    def vowelStrings(self, words: List[str], left: int, right: int) -> int:
        vowel = {'a', 'e', 'i', 'o', 'u'}
        counter = 0
        for i in range(left, right + 1):
            if words[i][0] in vowel and words[i][-1] in vowel:
                counter += 1
        return counter

Complexity

Time complexity: O(n)

• One loop.

Space complexity: O(1)

• No extra space