Count Vowel Strings in Ranges

Problem Description

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'.

Examples

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 <= 10^5
• 1 <= words[i].length <= 40
• sum(words[i].length) <= 3 * 10^5

Approach

We can preprocess all the indices of the strings that start and end with a vowel, and record them in order in the array $nums$.

Next, we iterate through each query $(l, r)$, and use binary search to find the first index $i$ in $nums$ that is greater than or equal to $l$, and the first index $j$ that is greater than $r$. Therefore, the answer to the current query is $j - i$.

The time complexity is $O(n + m \times \log n)$, and the space complexity is $O(n)$. Where $n$ and $m$ are the lengths of the arrays $words$ and $queries$, respectively.

Python3

class Solution:
    def vowelStrings(self, words: List[str], queries: List[List[int]]) -> List[int]:
        vowels = set("aeiou")
        nums = [i for i, w in enumerate(words) if w[0] in vowels and w[-1] in vowels]
        return [bisect_right(nums, r) - bisect_left(nums, l) for l, r in queries]

Java

class Solution {
    private List<Integer> nums = new ArrayList<>();

    public int[] vowelStrings(String[] words, int[][] queries) {
        Set<Character> vowels = Set.of('a', 'e', 'i', 'o', 'u');
        for (int i = 0; i < words.length; ++i) {
            char a = words[i].charAt(0), b = words[i].charAt(words[i].length() - 1);
            if (vowels.contains(a) && vowels.contains(b)) {
                nums.add(i);
            }
        }
        int m = queries.length;
        int[] ans = new int[m];
        for (int i = 0; i < m; ++i) {
            int l = queries[i][0], r = queries[i][1];
            ans[i] = search(r + 1) - search(l);
        }
        return ans;
    }

    private int search(int x) {
        int l = 0, r = nums.size();
        while (l < r) {
            int mid = (l + r) >> 1;
            if (nums.get(mid) >= x) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }
}

C++

class Solution {
public:
    vector<int> vowelStrings(vector<string>& words, vector<vector<int>>& queries) {
        unordered_set<char> vowels = {'a', 'e', 'i', 'o', 'u'};
        vector<int> nums;
        for (int i = 0; i < words.size(); ++i) {
            char a = words[i][0], b = words[i].back();
            if (vowels.count(a) && vowels.count(b)) {
                nums.push_back(i);
            }
        }
        vector<int> ans;
        for (auto& q : queries) {
            int l = q[0], r = q[1];
            int cnt = upper_bound(nums.begin(), nums.end(), r) - lower_bound(nums.begin(), nums.end(), l);
            ans.push_back(cnt);
        }
        return ans;
    }
};