Here is the full interview question "Coding - Serialize and Deserialize Dictionary Trie" asked at xAI:

Problem Statement: Implement a dictionary trie that allows for insertion and search operations. Additionally, implement methods to serialize and deserialize the trie, allowing it to be stored in a string format.

Examples:

1. Insertion:

   • Insert words: ["a", "to", "tea", "ted", "ten", "i", "in", "inn"]
   • Trie structure after insertion: A / \ T I / \ \ E N N \ S / \ T E \ D

2. Search:

   • Search for "a": True
   • Search for "b": False

3. Starts with:

   • Search for words starting with "te": ["tea", "ted", "ten"]

Constraints:

• The input words are composed of lowercase English letters.
• The number of words is at most 1000.
• The length of each word is at most 1000.

Hints:

1. Use a trie data structure to store the dictionary.
2. Implement a recursive approach to serialize and deserialize the trie.
3. Use a queue to perform a level-order traversal of the trie.

Solution: `python class TrieNode: def init(self): self.children = {} self.is_end_of_word = False

class Trie: def init(self): self.root = TrieNode()

def insert(self, word: str) -> None:
    node = self.root
    for char in word:
        if char not in node.children:
            node.children[char] = TrieNode()
        node = node.children[char]
    node.is_end_of_word = True

def search(self, word: str) -> bool:
    node = self.root
    for char in word:
        if char not in node.children:
            return False
        node = node.children[char]
    return node.is_end_of_word

def starts_with(self, prefix: str) -> bool:
    node = self.root
    for char in prefix:
        if char not in node.children:
            return False
        node = node.children[char]
    return True

def serialize(self) -> str:
    def serialize_helper(node, depth=0):
        if not node:
            return "#" * (depth + 1)
        res = ""
        for char, child in node.children.items():
            res += char + serialize_helper(child, depth + 1)
        if not res:
            res = "#" * (depth + 1)
        return res

    return serialize_helper(self.root)

def deserialize(self, data: str) -> None:
    def deserialize_helper(node, chars):
        if not chars:
            return
        char = chars[0]
        if char != "#":
            node.children[char] = TrieNode()
            deserialize_helper(node.children[char], chars[1:])
        else:
            node.is_end_of_word = True
            deserialize_helper(node, chars[1:])

    self.root = TrieNode()
    deserialize_helper(self.root, list(data))

Example usage:

trie = Trie() words = ["a", "to", "tea", "ted", "ten", "i", "in", "inn"] for word in words: trie.insert(word)

print(trie.search("a")) # True print(trie.search("b")) # False

prefix = "te" result = [word for word in words if word.startswith(prefix)] print(trie.starts_with(prefix)) # True print(result) # ["tea", "ted", "ten"]

serialized_trie = trie.serialize() print(serialized_trie)

new_trie = Trie() new_trie.deserialize(serialized_trie) print(new_trie.search("a")) # True `

Source:

• LeetCode Problem: Serialize and Deserialize Dictionary Trie