2559

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

Companies: PayPal

Related Topics:
Array, String, Prefix Sum

Similar Questions:

• Jump Game VII (Medium)

Solution 1.

// OJ: https://leetcode.com/problems/count-vowel-strings-in-ranges
// Author: github.com/lzl124631x
// Time: O(N + Q)
// Space: O(N)
class Solution {
    bool isVowel(char c) {
        return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
    }
public:
    vector<int> vowelStrings(vector<string>& A, vector<vector<int>>& Q) {
        int N = A.size();
        vector<int> sum(N + 1), ans;
        for (int i = 0; i < N; ++i) sum[i + 1] = sum[i] + (isVowel(A[i][0]) && isVowel(A[i].back()));
        for (auto &q : Q) {
            ans.push_back(sum[q[1] + 1] - sum[q[0]]);
        }
        return ans;
    }
};