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@ -534,3 +534,5 @@ mod p2269_find_the_k_beauty_of_a_number;
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mod p2012_sum_of_beauty_in_the_array;
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mod p2012_sum_of_beauty_in_the_array;
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mod p3305_count_of_substrings_containing_every_vowel_and_k_consonants_i;
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mod p3305_count_of_substrings_containing_every_vowel_and_k_consonants_i;
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mod p3306_count_of_substrings_containing_every_vowel_and_k_consonants_ii;
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@ -0,0 +1,63 @@
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/**
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* [3306] Count of Substrings Containing Every Vowel and K Consonants II
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*/
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pub struct Solution {}
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// submission codes start here
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use std::collections::{HashMap, HashSet};
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impl Solution {
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pub fn count_of_substrings(word: String, k: i32) -> i64 {
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let word: Vec<char> = word.chars().collect();
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let vowel_set = HashSet::from(['a', 'e', 'i', 'o', 'u']);
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let count = |m| -> i64 {
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let mut consonants = 0;
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let mut result = 0;
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let mut occurence = HashMap::new();
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let mut right = 0;
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for left in 0..word.len() {
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while right < word.len() && (consonants < m || occurence.len() < vowel_set.len()) {
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if vowel_set.contains(&word[right]) {
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let entry = occurence.entry(word[right]).or_insert(0i64);
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*entry += 1;
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} else {
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consonants += 1;
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}
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right += 1;
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}
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if consonants >= m && occurence.len() == vowel_set.len() {
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result += (word.len() - right + 1) as i64;
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}
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if vowel_set.contains(&word[left]) {
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let v = occurence.get_mut(&word[left]).unwrap();
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*v -= 1;
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if *v == 0 {
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occurence.remove(&word[left]);
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}
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} else {
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consonants -= 1;
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}
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}
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result
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};
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count(k) - count(k + 1)
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}
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}
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// submission codes end
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#[cfg(test)]
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mod tests {
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use super::*;
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#[test]
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fn test_3306() {
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assert_eq!(0, Solution::count_of_substrings("aeioqq".to_owned(), 1));
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}
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}
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