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 <= 1051 <= words[i].length <= 40words[i] consists only of lowercase English letters.sum(words[i].length) <= 3 * 1051 <= queries.length <= 1050 <= li <= ri < words.lengthclass Solution {
public:
vector<char> vowels = { 'a', 'e', 'i', 'o', 'u'};
bool isVowelString(string s){
int c=0;
for(auto v:vowels){
if(v==s.front()) c++;
if(v==s.back()) c++;
}
return c==2;
}
vector<int> vowelStrings(vector<string>& words, vector<vector<int>>& queries) {
vector<int> v = {0};
for(auto w:words)
v.push_back(isVowelString(w)?1:0);
for(int i=1; i<v.size(); i++)
v[i] += v[i-1];
v.push_back(v.back());
vector<int> ans;
for(auto q: queries)
ans.push_back(v[q[1]+1]-v[q[0]]);
return ans;
}
};class Solution {
private static Set<Character> vowels = new HashSet<>(Arrays.asList('a', 'e', 'i', 'o', 'u'));
public int[] vowelStrings(String[] words, int[][] queries) {
int[] vowelFlankedWords = new int[words.length], ans = new int[queries.length];
int prefixSum = 0;
for (int i = 0; i < words.length; i++) {
String s = words[i];
vowelFlankedWords[i] = prefixSum;
if (vowels.contains(s.charAt(0))
&& vowels.contains(s.charAt(s.length() - 1)))
prefixSum = vowelFlankedWords[i] += 1;
}
for (int i = 0; i < queries.length; i++) {
int[] q = queries[i];
ans[i] = q[0] == 0 ? vowelFlankedWords[q[1]] : vowelFlankedWords[q[1]] - vowelFlankedWords[q[0] - 1];
}
return ans;
}
}