LeetCode in Kotlin

2559. 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 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

Solution

class Solution {
    fun vowelStrings(words: Array<String>, queries: Array<IntArray>): IntArray {
        val vowels = HashSet(listOf('a', 'e', 'i', 'o', 'u'))
        val n = words.size
        val validWords = IntArray(n)
        for (i in 0 until n) {
            val startChar = words[i][0]
            val endChar = words[i][words[i].length - 1]
            validWords[i] = if (vowels.contains(startChar) && vowels.contains(endChar)) 1 else 0
        }
        val prefix = IntArray(n)
        prefix[0] = validWords[0]
        for (i in 1 until n) {
            prefix[i] = prefix[i - 1] + validWords[i]
        }
        val output = IntArray(queries.size)
        for (i in queries.indices) {
            val start = queries[i][0]
            val end = queries[i][1]
            output[i] = if (start == 0) prefix[end] else prefix[end] - prefix[start - 1]
        }
        return output
    }
}

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