docs(docs): add MDX page for LeetCode 211 design problem

apps/docs/content/(string)/211-design-add-and-search-words-data-structure.mdx

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+---
+title: '211. Design Add and Search Words Data Structure'
+description: Design a data structure that supports adding new words and finding if a string matches any previously added string
+sidebar:
+  label: 'Design Add and Search Words Data Structure'
+  badge: 'Medium'
+---
+
+<Badge variant="accent">Trie</Badge>
+
+### Constraints:
+
+- `1 <= word.length <= 25`
+- `word` in addWord consists of lowercase English letters.
+- `word` in search consist of '.' or lowercase English letters.
+- There will be at most 2 dots in `word` for search queries.
+- At most 10^4 calls will be made to addWord and search.
+
+## Solution
+
+```py
+class TrieNode:
+    def __init__(self):
+        self.children = {}
+        self.endOfWord = False
+
+
+class WordDictionary:
+    def __init__(self):
+        self.root = TrieNode()
+
+    def addWord(self, word: str) -> None:
+        cur = self.root
+        for c in word:
+            if c not in cur.children:
+                cur.children[c] = TrieNode()
+            cur = cur.children[c]
+        cur.endOfWord = True
+
+    # Leverage Recursion
+    def search(self, word: str) -> bool:
+        def dfs(j, root):
+            cur = root
+            for i in range(j, len(word)):
+                c = word[i]
+
+                if c == ".":
+                    for child in cur.children.values():
+                        if dfs(i + 1, child):
+                            return True
+                    return False
+                else:
+                    if c not in cur.children:
+                        return False
+                    cur = cur.children[c]
+            return cur.endOfWord
+
+        return dfs(0, self.root)
+
+
+# Your WordDictionary object will be instantiated and called as such:
+# obj = WordDictionary()
+# obj.addWord(word)
+# param_2 = obj.search(word)
+```