Add solution #467

Author: JoshCrozier

README.md

@@ -374,6 +374,7 @@
 463|[Island Perimeter](./0463-island-perimeter.js)|Medium|
 464|[Can I Win](./0464-can-i-win.js)|Medium|
 466|[Count The Repetitions](./0466-count-the-repetitions.js)|Hard|
+467|[Unique Substrings in Wraparound String](./0467-unique-substrings-in-wraparound-string.js)|Medium|
 472|[Concatenated Words](./0472-concatenated-words.js)|Hard|
 476|[Number Complement](./0476-number-complement.js)|Easy|
 482|[License Key Formatting](./0482-license-key-formatting.js)|Easy|

solutions/0467-unique-substrings-in-wraparound-string.js

@@ -0,0 +1,28 @@
+/**
+ * 467. Unique Substrings in Wraparound String
+ * https://leetcode.com/problems/unique-substrings-in-wraparound-string/
+ * Difficulty: Medium
+ *
+ * We define the string base to be the infinite wraparound string of "abcdefghijklmnopqrstuvwxyz",
+ * so base will look like this:
+ * - "...zabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcd....".
+ *
+ *  Given a string s, return the number of unique non-empty substrings of s are present in base.
+ */
+
+/**
+ * @param {string} s
+ * @return {number}
+ */
+var findSubstringInWraproundString = function(s) {
+  const maxLength = new Array(26).fill(0);
+  let count = 0;
+
+  for (let i = 0; i < s.length; i++) {
+    count = i > 0 && (s.charCodeAt(i) - s.charCodeAt(i - 1) - 1) % 26 === 0 ? count + 1 : 1;
+    const index = s.charCodeAt(i) - 97;
+    maxLength[index] = Math.max(maxLength[index], count);
+  }
+
+  return maxLength.reduce((sum, n) => sum + n, 0);
+};