LeetCode-in-Java

2559. Count Vowel Strings in Ranges

Medium

You are given a 0-indexed array of strings words and a 2D array of integers queries.

Each query queries[i] = [li, ri] asks us to find the number of strings present in the range li to ri (both inclusive) of words that start and end with a vowel.

Return an array ans of size queries.length, where ans[i] is the answer to the i^th query.

Note that the vowel letters are 'a', 'e', 'i', 'o', and 'u'.

Example 1:

Input: words = [“aba”,”bcb”,”ece”,”aa”,”e”], queries = [[0,2],[1,4],[1,1]]

Output: [2,3,0]

Explanation: The strings starting and ending with a vowel are “aba”, “ece”, “aa” and “e”.

The answer to the query [0,2] is 2 (strings “aba” and “ece”).

to query [1,4] is 3 (strings “ece”, “aa”, “e”).

to query [1,1] is 0.

We return [2,3,0].

Example 2:

Input: words = [“a”,”e”,”i”], queries = [[0,2],[0,1],[2,2]]

Output: [3,2,1]

Explanation: Every string satisfies the conditions, so we return [3,2,1].

Constraints:

• 1 <= words.length <= 105
• 1 <= words[i].length <= 40
• words[i] consists only of lowercase English letters.
• sum(words[i].length) <= 3 * 105
• 1 <= queries.length <= 105
• 0 <= li <= ri < words.length

Solution

public class Solution {
    private boolean validWord(String s) {
        char cStart = s.charAt(0);
        char cEnd = s.charAt(s.length() - 1);
        boolean flag1 =
                cStart == 'a' || cStart == 'e' || cStart == 'i' || cStart == 'o' || cStart == 'u';
        boolean flag2 = cEnd == 'a' || cEnd == 'e' || cEnd == 'i' || cEnd == 'o' || cEnd == 'u';
        return flag1 && flag2;
    }

    public int[] vowelStrings(String[] words, int[][] queries) {
        int[] prefixArr = new int[words.length];
        prefixArr[0] = validWord(words[0]) ? 1 : 0;
        for (int i = 1; i < words.length; ++i) {
            if (validWord(words[i])) {
                prefixArr[i] = prefixArr[i - 1] + 1;
            } else {
                prefixArr[i] = prefixArr[i - 1];
            }
        }
        int[] res = new int[queries.length];
        for (int i = 0; i < queries.length; ++i) {
            int upperBound = queries[i][1];
            int lowerBound = queries[i][0];
            int val =
                    (lowerBound == 0)
                            ? prefixArr[upperBound]
                            : prefixArr[upperBound] - prefixArr[lowerBound - 1];
            res[i] = val;
        }
        return res;
    }
}

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