2559. Count Vowel Strings in Ranges

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 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

 

Solutions

Solution: Prefix Sum

• Time complexity: O(n+queries.length)
• Space complexity: O(n)

 

JavaScript

/**
 * @param {string[]} words
 * @param {number[][]} queries
 * @return {number[]}
 */
const vowelStrings = function (words, queries) {
  const VOWEL = 'aeiou';
  const n = words.length;
  const prefixVowel = Array.from({ length: n + 1 }, () => 0);

  for (let index = 1; index <= n; index++) {
    const word = words[index - 1];
    const prefix = prefixVowel[index - 1];
    const isVowelWord = VOWEL.includes(word[0]) && VOWEL.includes(word.at(-1));

    prefixVowel[index] = prefix + (isVowelWord ? 1 : 0);
  }

  return queries.map(([l, r]) => {
    return prefixVowel[r + 1] - prefixVowel[l];
  });
};