2586 count-the-number-of-vowel-strings-in-range

题目

You 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

思路

看题半个小时，解题三分钟系列 integer 指制定数组的范围，不是元音字符串的范围

解题

class Solution:
    def vowelStrings(self, words: List[str], left: int, right: int) -> int:
        VOWEL = 'aeiou'
        res = 0
    
        for word in words[left:right+1]:
            if (word[0] in VOWEL) and (word[-1] in VOWEL):
                res += 1

        return res

复杂度分析

• 时间复杂度 O(N) / N = right - left + 1 仅遍历部分数组
• 空间复杂度 O(1) 仅用到部分变量，未用到额外存储空间