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
Solution:
Iterator overs words, from left to right. Check the first and last element of the string.
Time complexity: O(|right – left + 1|)
Space complexity: O(1)
C++
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// Author: Huahua class Solution { public: int vowelStrings(vector<string>& words, int left, int right) { auto isVowel = [](char c) { return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u'; }; return count_if(begin(words) + left, begin(words) + right + 1, [&](const string& w) { return isVowel(w.front()) && isVowel(w.back()); }); } }; |