[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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思路是前缀和。首先我们需要一个与 words 数组等长的数组 goods 记录哪些单词满足以元音开头和结尾这个条件，遍历 words 数组的时候，判断每一个单词 words[i] 是否满足条件，如果是，则在 goods[i] 处标记为 1。我们还需要另外一个数组 presums 记录 goods 数组的前缀和。最后我们通过 presums 快速地找到每一个 query 的答案。

时间O(n)

空间O(n)

Java实现

class Solution {
    public int[] vowelStrings(String[] words, int[][] queries) {
        int[] good = new int[words.length];
        for (int i = 0; i < words.length; i++) {
            if (helper(words[i])) {
                good[i] = 1;
            }
        }

        int[] presum = new int[words.length + 1];
        presum[0] = good[0];
        for (int i = 0; i < words.length; i++) {
            presum[i + 1] = presum[i] + good[i];
        }

        int n = queries.length;
        int[] res = new int[n];
        for (int i = 0; i < n; i++) {
            int start = queries[i][0];
            int end = queries[i][1];
            res[i] = presum[end + 1] - presum[start];
        }
        return res;
    }

    private boolean helper(String word) {
        Set<Character> set = new HashSet<>();
        set.add('a');
        set.add('e');
        set.add('i');
        set.add('o');
        set.add('u');
        if (set.contains(word.charAt(0)) && set.contains(word.charAt(word.length() - 1))) {
            return true;
        }
        return false;
    }
}

 

LeetCode 题目总结