Add solution #686

Author: JoshCrozier

README.md

@@ -18,6 +18,7 @@
 606|[Construct String from Binary Tree](./0606-construct-string-from-binary-tree.js)|Easy|
 617|[Merge Two Binary Trees](./0617-merge-two-binary-trees.js)|Easy|
 648|[Replace Words](./0648-replace-words.js)|Medium|
+686|[Repeated String Match](./0686-repeated-string-match.js)|Easy|
 739|[Daily Temperatures](./0739-daily-temperatures.js)|Medium|
 791|[Custom Sort String](./0791-custom-sort-string.js)|Medium|
 804|[Unique Morse Code Words](./0804-unique-morse-code-words.js)|Easy|

solutions/0686-repeated-string-match.js

@@ -0,0 +1,25 @@
+/**
+ * 686. Repeated String Match
+ * https://leetcode.com/problems/repeated-string-match/
+ * Difficulty: Easy
+ *
+ * Given two strings A and B, find the minimum number of times A has to be
+ * repeated such that B is a substring of it. If no such solution, return -1.
+ *
+ * For example, with A = "abcd" and B = "cdabcdab".
+ *
+ * Return 3, because by repeating A three times (“abcdabcdabcd”), B is a
+ * substring of it; and B is not a substring of A repeated two times ("abcdabcd").
+ */
+
+/**
+ * @param {string} A
+ * @param {string} B
+ * @return {number}
+ */
+var repeatedStringMatch = function(A, B) {
+  const length = Math.ceil(B.length / A.length);
+  if (A.repeat(length).includes(B)) return length;
+  if ((A.repeat(length) + A).includes(B)) return length + 1;
+  return -1;
+};