Add solution #3306

Author: JoshCrozier

README.md

@@ -836,6 +836,7 @@
 3174|[Clear Digits](./3174-clear-digits.js)|Easy|
 3208|[Alternating Groups II](./3208-alternating-groups-ii.js)|Medium|
 3223|[Minimum Length of String After Operations](./3223-minimum-length-of-string-after-operations.js)|Medium|
+3306|[Count of Substrings Containing Every Vowel and K Consonants II](./3306-count-of-substrings-containing-every-vowel-and-k-consonants-ii.js)|Medium|
 3392|[Count Subarrays of Length Three With a Condition](./3392-count-subarrays-of-length-three-with-a-condition.js)|Easy|
 3396|[Minimum Number of Operations to Make Elements in Array Distinct](./3396-minimum-number-of-operations-to-make-elements-in-array-distinct.js)|Easy|
 3397|[Maximum Number of Distinct Elements After Operations](./3397-maximum-number-of-distinct-elements-after-operations.js)|Medium|

solutions/3306-count-of-substrings-containing-every-vowel-and-k-consonants-ii.js

@@ -0,0 +1,48 @@
+/**
+ * 3306. Count of Substrings Containing Every Vowel and K Consonants II
+ * https://leetcode.com/problems/count-of-substrings-containing-every-vowel-and-k-consonants-ii/
+ * Difficulty: Medium
+ *
+ * You are given a string word and a non-negative integer k.
+ *
+ * Return the total number of substrings of word that contain every vowel ('a', 'e', 'i', 'o', and
+ * 'u') at least once and exactly k consonants.
+ */
+
+/**
+ * @param {string} word
+ * @param {number} k
+ * @return {number}
+ */
+function countOfSubstrings(word, k) {
+  return helper(word, k) - helper(word, k - 1);
+
+  function helper(word, k) {
+    const vowels = { a: 0, e: 0, i: 0, o: 0, u: 0 };
+    let count = 0;
+    let left = 0;
+    let consonants = k;
+
+    for (let right = 0; right < word.length; right++) {
+      const character = word[right];
+      if (character in vowels) {
+        vowels[character]++;
+      } else {
+        consonants--;
+      }
+
+      while (vowels.a > 0 && vowels.e > 0 && vowels.i > 0 && vowels.o > 0
+             && vowels.u > 0 && consonants < 0) {
+        const leftCharacter = word[left];
+        if (leftCharacter in vowels) {
+          vowels[leftCharacter]--;
+        } else {
+          consonants++;
+        }
+        left++;
+      }
+      count += right - left + 1;
+    }
+    return count;
+  }
+}