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2691. Count Vowel Strings In Ranges

Difficulty: Medium

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2691. 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 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

Solution

class Solution {
    public int[] vowelStrings(String[] words, int[][] queries) {
        int n = words.length;
        int arr[] = new int[n];
        for(int i = 0; i < n; i++) {
            String current = words[i];
            if(isvowel(current.charAt(0)) && isvowel(current.charAt(current.length() - 1))) arr[i] = 1;
            else arr[i] = 0;
        }
        int sum = 0;
        int pre[] = new int[n];
        for(int i = 0; i < n; i++) {
            sum += arr[i];
            pre[i] = sum;
        }
        int res[] = new int[queries.length];
        int k = 0;
        for(int current[] : queries) {
            int l = current[0], r = current[1];
            if(l == 0) res[k++] = pre[r];
            else res[k++] = pre[r] - pre[l - 1];
        }
        return res;
    }
    private boolean isvowel(char current) {
        if ((current == 'a') || (current == 'e') || (current == 'i' ) || (current == 'o')  || (current == 'u')) return true;
        return false;
    }
}

Complexity Analysis

  • Time Complexity: O(?)
  • Space Complexity: O(?)

Approach

Detailed explanation of the approach will be added here