Add solution #290

JoshCrozier · JoshCrozier · commit 15e6f2a73e6d · 2023-01-01T17:06:40.000-06:00
diff --git a/0290-word-pattern.js b/0290-word-pattern.js
@@ -0,0 +1,29 @@
+/**
+ * 290. Word Pattern
+ * https://leetcode.com/problems/word-pattern/
+ * Difficulty: Easy
+ *
+ * Given a pattern and a string s, find if s follows the same pattern.
+ *
+ * Here follow means a full match, such that there is a bijection between
+ * a letter in pattern and a non-empty word in s.
+ */
+
+/**
+ * @param {string} pattern
+ * @param {string} s
+ * @return {boolean}
+ */
+var wordPattern = function(pattern, s) {
+  const words = s.split(/\s+/);
+  const map = new Map();
+
+  return words.every((word, index) => {
+    const offset = pattern.length === words.length ? index + 1 : words.length / pattern.length;
+    const sequence = pattern.slice(index, offset);
+    if (!map.has(sequence) && !Array.from(map.values()).includes(word)) {
+      map.set(sequence, word);
+    }
+    return map.get(sequence) === word && pattern.length <= words.length;
+  });
+};
diff --git a/README.md b/README.md
@@ -113,6 +113,7 @@
 273|[Integer to English Words](./0273-integer-to-english-words.js)|Hard|
 278|[First Bad Version](./0278-first-bad-version.js)|Medium|
 283|[Move Zeroes](./0283-move-zeroes.js)|Easy|
+290|[Word Pattern](./0290-word-pattern.js)|Easy|
 326|[Power of Three](./0326-power-of-three.js)|Easy|
 342|[Power of Four](./0342-power-of-four.js)|Easy|
 344|[Reverse String](./0344-reverse-string.js)|Easy|
@@ -285,4 +286,4 @@
 
 [MIT License](https://opensource.org/licenses/MIT)
 
-Copyright (c) 2019-2022 [Josh Crozier](https://joshcrozier.com)
+Copyright (c) 2019-2023 [Josh Crozier](https://joshcrozier.com)
