carloscn
commented
1 year ago Code
fn is_vowel(c:char) -> bool
{
if c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u' {
return true;
}
return false;
}
fn is_contiguous(w:&Vec<char>, start:usize, end:usize) -> bool
{
let copied_slice = &w[start..=end];
let mut a_f:bool = false;
let mut e_f:bool = false;
let mut i_f:bool = false;
let mut o_f:bool = false;
let mut u_f:bool = false;
for e in copied_slice {
if *e == 'a' {
a_f = true;
} else if *e == 'e' {
e_f = true;
} else if *e == 'i' {
i_f = true;
} else if *e == 'o' {
o_f = true;
} else if *e == 'u' {
u_f = true;
} else {
return false;
}
}
return a_f && e_f && i_f && o_f && u_f;
}
pub fn count_vowel_substrings(word: &str) -> i32
{
let mut ret:i32 = 0;
if word.len() < 5 {
return ret;
}
let w_v:Vec<char> = word.chars().collect();
for i in 0..w_v.len() - 4 {
let mut contiguous_count = 0_i32;
for j in i..w_v.len() {
if is_vowel(w_v[j]) {
contiguous_count += 1;
} else {
contiguous_count = 0;
}
if contiguous_count >= 5 {
if is_contiguous(&w_v, i, j) {
ret += 1;
}
}
}
}
return ret;
}
Description
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):
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):
Constraints:
1 <= word.length <= 100 word consists of lowercase English letters only.