LeetCode-in-Java
2062. Count Vowel Substrings of a String
Easy
A substring is a contiguous (non-empty) sequence of characters within a string.
A vowel substring is a substring that only consists of vowels ('a', 'e', 'i', 'o', and 'u') and has all five vowels present in it.
Given a string word, return the number of vowel substrings in word.
Example 1:
Input: word = “aeiouu”
Output: 2
Explanation: The vowel substrings of word are as follows (underlined):
-
“aeiouu”
-
“aeiouu”
Example 2:
Input: word = “unicornarihan”
Output: 0
Explanation: Not all 5 vowels are present, so there are no vowel substrings.
Example 3:
Input: word = “cuaieuouac”
Output: 7
Explanation: The vowel substrings of word are as follows (underlined):
-
“cuaieuouac”
-
“cuaieuouac”
-
“cuaieuouac”
-
“cuaieuouac”
-
“cuaieuouac”
-
“cuaieuouac”
-
“cuaieuouac”
Constraints:
1 <= word.length <= 100wordconsists of lowercase English letters only.
Solution
public class Solution {
public int countVowelSubstrings(String word) {
final int length = word.length();
boolean[] vows = new boolean[128];
vows['a'] = true;
vows['e'] = true;
vows['i'] = true;
vows['o'] = true;
vows['u'] = true;
int[] counts = new int[128];
int uniqVows = 0;
int originalBegin = 0;
int begin = 0;
int result = 0;
for (int i = 0; i < length; i++) {
char ch = word.charAt(i);
if (vows[ch]) {
counts[ch]++;
if (counts[ch] == 1) {
uniqVows++;
}
while (uniqVows == 5) {
uniqVows -= --counts[word.charAt(begin)] == 0 ? 1 : 0;
begin++;
}
result += begin - originalBegin;
} else {
if (uniqVows != 0) {
uniqVows = 0;
counts['a'] = 0;
counts['e'] = 0;
counts['i'] = 0;
counts['o'] = 0;
counts['u'] = 0;
}
originalBegin = begin = i + 1;
}
}
return result;
}
}