You are given a 0-indexed array of strings words
and a 2D array of integers queries
.
Each query queries[i] = [l<sub>i</sub>, r<sub>i</sub>]
asks us to find the number of strings present in the range l<sub>i</sub>
to r<sub>i</sub>
(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].
class Solution {
public:
vector<int> vowelStrings(vector<string>& words, vector<vector<int>>& queries) {
vector<int> prefixSum(words.size()+1,0);
vector<int> result;
function<bool(char)> isVowel = [](char ch)->bool{
if( ch == 'a' || ch == 'e' || ch =='i' || ch == 'o' || ch=='u'){
return true;
}
return false;
};
int index = 1;
for( string str:words){
if( isVowel(str.front()) && isVowel(str.back()) ){
prefixSum[index] = prefixSum[index-1]+1;
} else
prefixSum[index] = prefixSum[index-1];
index++;
}
for( auto q: queries ){
result.push_back( prefixSum[q[1]+1] - prefixSum[q[0]] );
}
return result;
}
};