Implement Magic Dictionary
<p>Design a data structure that is initialized with a list of <strong>different</strong> words. Provided a string, you should determine if you can change exactly one character in this string to match any word in the data structure.</p>

<p>Implement the&nbsp;<code>MagicDictionary</code>&nbsp;class:</p>

<ul>
	<li><code>MagicDictionary()</code>&nbsp;Initializes the object.</li>
	<li><code>void buildDict(String[]&nbsp;dictionary)</code>&nbsp;Sets the data structure&nbsp;with an array of distinct strings <code>dictionary</code>.</li>
	<li><code>bool search(String searchWord)</code> Returns <code>true</code> if you can change <strong>exactly one character</strong> in <code>searchWord</code> to match any string in the data structure, otherwise returns <code>false</code>.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input</strong>
[&quot;MagicDictionary&quot;, &quot;buildDict&quot;, &quot;search&quot;, &quot;search&quot;, &quot;search&quot;, &quot;search&quot;]
[[], [[&quot;hello&quot;, &quot;leetcode&quot;]], [&quot;hello&quot;], [&quot;hhllo&quot;], [&quot;hell&quot;], [&quot;leetcoded&quot;]]
<strong>Output</strong>
[null, null, false, true, false, false]

<strong>Explanation</strong>
MagicDictionary magicDictionary = new MagicDictionary();
magicDictionary.buildDict([&quot;hello&quot;, &quot;leetcode&quot;]);
magicDictionary.search(&quot;hello&quot;); // return False
magicDictionary.search(&quot;hhllo&quot;); // We can change the second &#39;h&#39; to &#39;e&#39; to match &quot;hello&quot; so we return True
magicDictionary.search(&quot;hell&quot;); // return False
magicDictionary.search(&quot;leetcoded&quot;); // return False
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;=&nbsp;dictionary.length &lt;= 100</code></li>
	<li><code>1 &lt;=&nbsp;dictionary[i].length &lt;= 100</code></li>
	<li><code>dictionary[i]</code> consists of only lower-case English letters.</li>
	<li>All the strings in&nbsp;<code>dictionary</code>&nbsp;are <strong>distinct</strong>.</li>
	<li><code>1 &lt;=&nbsp;searchWord.length &lt;= 100</code></li>
	<li><code>searchWord</code>&nbsp;consists of only lower-case English letters.</li>
	<li><code>buildDict</code>&nbsp;will be called only once before <code>search</code>.</li>
	<li>At most <code>100</code> calls will be made to <code>search</code>.</li>
</ul>
Depth-First Search; Design; Trie; Hash Table; String
class Trie:
    __slots__ = ["children", "is_end"]

    def __init__(self):
        self.children = {}
        self.is_end = False

    def insert(self, w: str) -> None:
        node = self
        for c in w:
            if c not in node.children:
                node.children[c] = Trie()
            node = node.children[c]
        node.is_end = True

    def search(self, w: str) -> bool:
        def dfs(i: int, node: Trie, diff: int) -> bool:
            if i == len(w):
                return diff == 1 and node.is_end
            if w[i] in node.children and dfs(i + 1, node.children[w[i]], diff):
                return True
            return diff == 0 and any(
                dfs(i + 1, node.children[c], 1) for c in node.children if c != w[i]
            )

        return dfs(0, self, 0)


class MagicDictionary:
    def __init__(self):
        self.trie = Trie()

    def buildDict(self, dictionary: List[str]) -> None:
        for w in dictionary:
            self.trie.insert(w)

    def search(self, searchWord: str) -> bool:
        return self.trie.search(searchWord)


# Your MagicDictionary object will be instantiated and called as such:
# obj = MagicDictionary()
# obj.buildDict(dictionary)
# param_2 = obj.search(searchWord)
