LeetCode 2586: Count the Number of Vowel Strings in Range Solution

Master LeetCode problem 2586 (Count the Number of Vowel Strings in Range), a easy challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2586. Count the Number of Vowel Strings in Range

Easy

Array String Counting

Problem Explanation

Explanation:

To solve this problem, we need to iterate through the given range [left, right] and for each word at index i, check if it is a vowel string (starts and ends with a vowel character). To determine if a word is a vowel string, we can simply check if the first and last characters of the word are vowels.

Algorithm:

Initialize a variable count to store the count of vowel strings.
Iterate through the range [left, right] inclusive.
For each word at index i, check if it is a vowel string by verifying if the first and last characters are vowels.
If the word is a vowel string, increment the count.
Finally, return the count.

Time Complexity:

The time complexity of this algorithm is O(n), where n is the number of words in the given range [left, right].

Space Complexity:

The space complexity of this algorithm is O(1) as we only use a constant amount of extra space.

Solution Code

class Solution {
    public int countVowelStringsInRange(String[] words, int left, int right) {
        int count = 0;
        for (int i = left; i <= right; i++) {
            if (isVowelString(words[i])) {
                count++;
            }
        }
        return count;
    }
    
    private boolean isVowelString(String word) {
        char firstChar = word.charAt(0);
        char lastChar = word.charAt(word.length() - 1);
        return isVowel(firstChar) && isVowel(lastChar);
    }
    
    private boolean isVowel(char c) {
        return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
    }
}

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What is the time complexity of LeetCode 2586 (Count the Number of Vowel Strings in Range)?

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