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Count the Number of Vowel Strings in Range

Tags : leetcode, array, string, cpp, easy

Problem Statement - link #

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].

Examples #

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 #

Solutions #


class Solution {
    bool isVowel(char c) {
        return (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u');
    }
public:
    int vowelStrings(vector<string>& words, int left, int right) {
        int n = words.size();
        vector<int> pre(n , 0);
        for(int i = 0; i < n; i++) {
            if(i > 0) {
                pre[i] += pre[i - 1];
            }
            if(isVowel(words[i].front()) && isVowel(words[i].back())) {
                pre[i]++;
            }
        }
        if(left == 0) {
            return pre[right];
        }
        return pre[right] - pre[left - 1];
    }
};