LeetCode 2586. Count the Number of Vowel Strings in Range Solution in Java, C++, Python & More | Explanation + Code

Description

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

Solutions

Solution 1: Simulation

We just need to traverse the string in the interval [left,.. right], and check if it starts and ends with a vowel. If so, the answer plus one.

After the traversal, return the answer.

The time complexity is O(m), and the space complexity is O(1). Where m = right - left + 1.

PythonJavaC++GoTypeScriptRust

class Solution: def vowelStrings(self, words: List[str], left: int, right: int) -> int: return sum( w[0] in 'aeiou' and w[-1] in 'aeiou' for w in words[left : right + 1] )(code-box)

class Solution { public int vowelStrings(String[] words, int left, int right) { int ans = 0; for (int i = left; i <= right; ++i) { var w = words[i]; if (check(w.charAt(0)) && check(w.charAt(w.length() - 1))) { ++ans; } } return ans; } private boolean check(char c) { return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u'; } }(code-box)

class Solution { public: int vowelStrings(vector<string>& words, int left, int right) { auto check = [](char c) -> bool { return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u'; }; int ans = 0; for (int i = left; i <= right; ++i) { auto w = words[i]; ans += check(w[0]) && check(w[w.size() - 1]); } return ans; } };(code-box)

func vowelStrings(words []string, left int, right int) (ans int) { check := func(c byte) bool { return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u' } for _, w := range words[left : right+1] { if check(w[0]) && check(w[len(w)-1]) { ans++ } } return }(code-box)

function vowelStrings(words: string[], left: number, right: number): number { let ans = 0; const check: string[] = ['a', 'e', 'i', 'o', 'u']; for (let i = left; i <= right; ++i) { const w = words[i]; if (check.includes(w[0]) && check.includes(w.at(-1))) { ++ans; } } return ans; }(code-box)

impl Solution { pub fn vowel_strings(words: Vec<String>, left: i32, right: i32) -> i32 { let check = |c: u8| -> bool { c == b'a' || c == b'e' || c == b'i' || c == b'o' || c == b'u' }; let mut ans = 0; for i in left..=right { let w = words[i as usize].as_bytes(); if check(w[0]) && check(w[w.len() - 1]) { ans += 1; } } ans } }(code-box)