Add solution #291

Author: JoshCrozier

README.md

@@ -279,6 +279,7 @@
 288|[Unique Word Abbreviation](./solutions/0288-unique-word-abbreviation.js)|Medium|
 289|[Game of Life](./solutions/0289-game-of-life.js)|Medium|
 290|[Word Pattern](./solutions/0290-word-pattern.js)|Easy|
+291|[Word Pattern II](./solutions/0291-word-pattern-ii.js)|Medium|
 292|[Nim Game](./solutions/0292-nim-game.js)|Easy|
 295|[Find Median from Data Stream](./solutions/0295-find-median-from-data-stream.js)|Hard|
 297|[Serialize and Deserialize Binary Tree](./solutions/0297-serialize-and-deserialize-binary-tree.js)|Hard|

solutions/0291-word-pattern-ii.js

@@ -0,0 +1,51 @@
+/**
+ * 291. Word Pattern II
+ * https://leetcode.com/problems/word-pattern-ii/
+ * Difficulty: Medium
+ *
+ * Given a pattern and a string s, return true if s matches the pattern.
+ *
+ * A string s matches a pattern if there is some bijective mapping of single characters to
+ * non-empty strings such that if each character in pattern is replaced by the string it
+ * maps to, then the resulting string is s. A bijective mapping means that no two characters
+ * map to the same string, and no character maps to two different strings.
+ */
+
+/**
+ * @param {string} pattern
+ * @param {string} s
+ * @return {boolean}
+ */
+var wordPatternMatch = function(pattern, s) {
+  const charToWord = new Map();
+  const usedWords = new Set();
+
+  return backtrack(0, 0);
+
+  function backtrack(patIndex, strIndex) {
+    if (patIndex === pattern.length && strIndex === s.length) return true;
+    if (patIndex >= pattern.length || strIndex >= s.length) return false;
+
+    const char = pattern[patIndex];
+    if (charToWord.has(char)) {
+      const word = charToWord.get(char);
+      if (!s.startsWith(word, strIndex)) return false;
+      return backtrack(patIndex + 1, strIndex + word.length);
+    }
+
+    for (let i = strIndex + 1; i <= s.length; i++) {
+      const word = s.slice(strIndex, i);
+      if (usedWords.has(word)) continue;
+
+      charToWord.set(char, word);
+      usedWords.add(word);
+
+      if (backtrack(patIndex + 1, strIndex + word.length)) return true;
+
+      charToWord.delete(char);
+      usedWords.delete(word);
+    }
+
+    return false;
+  }
+};