Add solution #211

Author: JoshCrozier

README.md

@@ -184,6 +184,7 @@
 205|[Isomorphic Strings](./0205-isomorphic-strings.js)|Easy|
 206|[Reverse Linked List](./0206-reverse-linked-list.js)|Easy|
 207|[Course Schedule](./0207-course-schedule.js)|Medium|
+211|[Design Add and Search Words Data Structure](./0211-design-add-and-search-words-data-structure.js)|Medium|
 213|[House Robber II](./0213-house-robber-ii.js)|Medium|
 214|[Shortest Palindrome](./0214-shortest-palindrome.js)|Hard|
 215|[Kth Largest Element in an Array](./0215-kth-largest-element-in-an-array.js)|Medium|

solutions/0211-design-add-and-search-words-data-structure.js

@@ -0,0 +1,60 @@
+/**
+ * 211. Design Add and Search Words Data Structure
+ * https://leetcode.com/problems/design-add-and-search-words-data-structure/
+ * Difficulty: Medium
+ *
+ * Design a data structure that supports adding new words and finding if a string matches
+ * any previously added string.
+ *
+ * Implement the WordDictionary class:
+ * - WordDictionary() Initializes the object.
+ * - void addWord(word) Adds word to the data structure, it can be matched later.
+ * - bool search(word) Returns true if there is any string in the data structure that
+ *   matches word or false otherwise. word may contain dots '.' where dots can be
+ *   matched with any letter.
+ */
+
+
+var WordDictionary = function() {
+  this.root = {};
+};
+
+/**
+ * @param {string} word
+ * @return {void}
+ */
+WordDictionary.prototype.addWord = function(word) {
+  let node = this.root;
+
+  for (const character of word) {
+    node[character] = node[character] || {};
+    node = node[character];
+  }
+
+  node.isTail = true;
+};
+
+/**
+ * @param {string} word
+ * @return {boolean}
+ */
+WordDictionary.prototype.search = function(word) {
+  return dfs(this.root, 0);
+
+  function dfs(node, i) {
+    if (word.length === i) {
+      return node.isTail;
+    }
+    if (word[i] === '.') {
+      for (const key in node) {
+        if (dfs(node[key], i + 1)) {
+          return true;
+        }
+      }
+    } else if (node[word[i]] && dfs(node[word[i]], i + 1)) {
+      return true;
+    }
+
+    return false;
+  }
+};