Add solution #639

Author: JoshCrozier

README.md

@@ -481,6 +481,7 @@
 636|[Exclusive Time of Functions](./0636-exclusive-time-of-functions.js)|Medium|
 637|[Average of Levels in Binary Tree](./0637-average-of-levels-in-binary-tree.js)|Easy|
 638|[Shopping Offers](./0638-shopping-offers.js)|Medium|
+639|[Decode Ways II](./0639-decode-ways-ii.js)|Hard|
 643|[Maximum Average Subarray I](./0643-maximum-average-subarray-i.js)|Easy|
 645|[Set Mismatch](./0645-set-mismatch.js)|Medium|
 648|[Replace Words](./0648-replace-words.js)|Medium|

solutions/0639-decode-ways-ii.js

@@ -0,0 +1,52 @@
+/**
+ * 639. Decode Ways II
+ * https://leetcode.com/problems/decode-ways-ii/
+ * Difficulty: Hard
+ *
+ * A message containing letters from A-Z can be encoded into numbers using the following mapping:
+ * - 'A' -> "1"
+ * - 'B' -> "2"
+ * - 'Z' -> "26"
+ *
+ * To decode an encoded message, all the digits must be grouped then mapped back into letters using
+ * the reverse of the mapping above (there may be multiple ways). For example, "11106" can be mapped
+ * into:
+ * - "AAJF" with the grouping (1 1 10 6)
+ * - "KJF" with the grouping (11 10 6)
+ *
+ * Note that the grouping (1 11 06) is invalid because "06" cannot be mapped into 'F' since "6" is
+ * different from "06".
+ *
+ * In addition to the mapping above, an encoded message may contain the '*' character, which can
+ * represent any digit from '1' to '9' ('0' is excluded). For example, the encoded message "1*"
+ * may represent any of the encoded messages "11", "12", "13", "14", "15", "16", "17", "18", or
+ * "19". Decoding "1*" is equivalent to decoding any of the encoded messages it can represent.
+ *
+ * Given a string s consisting of digits and '*' characters, return the number of ways to decode it.
+ *
+ * Since the answer may be very large, return it modulo 109 + 7.
+ */
+
+/**
+ * @param {string} s
+ * @return {number}
+ */
+var numDecodings = function(s) {
+  const mod = 1e9 + 7;
+  let [i0, i1, i2, i3] = [1, 0, 0, 0];
+
+  for (const c of s) {
+    if (c == '*') {
+      i3 = 9 * i0 + 9 * i1 + 6 * i2;
+      i1 = i0;
+      i2 = i0;
+    } else {
+      i3 = (c > '0') * i0 + i1 + (c <= '6') * i2;
+      i1 = (c == '1') * i0;
+      i2 = (c == '2') * i0;
+    }
+    i0 = i3 % mod;
+  }
+
+  return i0;
+};