[LeetCode] 2559. Count Vowel Strings in Ranges

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 ith 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

统计范围内的元音字符串数。

给你一个下标从 0 开始的字符串数组 words 以及一个二维整数数组 queries 。

每个查询 queries[i] = [li, ri] 会要求我们统计在 words 中下标在 li 到 ri 范围内（包含 这两个值）并且以元音开头和结尾的字符串的数目。

返回一个整数数组，其中数组的第 i 个元素对应第 i 个查询的答案。

注意：元音字母是 'a'、'e'、'i'、'o' 和 'u' 。

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/count-vowel-strings-in-ranges
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思路

思路是前缀和。这是前缀和的基础应用。如果 input 数组的长度是 n，那么我们需要一个长度为 n + 1 的数组记录前 n 个单词里面有多少个单词是以元音开头和结尾的。然后我们就可以以O(1)的时间找到下标在 li 到 ri 范围内（包含 这两个值）并且以元音开头和结尾的字符串的数目。

复杂度

时间O(n)
空间O(n)

代码

Java实现

class Solution {
    public int[] vowelStrings(String[] words, int[][] queries) {
        int n = words.length;
        int[] presum = new int[n + 1];
        presum[0] = 0;
        for (int i = 0; i < n; i++) {
            int cur = helper(words[i]) == true ? 1 : 0;
            presum[i + 1] = presum[i] + cur;
        }

        int len = queries.length;
        int[] res = new int[len];
        for (int i = 0; i < len; i++) {
            int left = queries[i][0];
            int right = queries[i][1];
            res[i] = presum[right + 1] - presum[left];
        }
        return res;
    }

    private boolean helper(String str) {
        char first = str.charAt(0);
        char last = str.charAt(str.length() - 1);
        return isVowel(first) && isVowel(last);
    }

    private boolean isVowel(char c) {
        return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
    }
}