How do you solve Design Add and Search Words Data Structure on LeetCode?

Design Add and Search Words Data Structure (LeetCode #211) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a Trie allows efficient prefix searching.

What is the brute force solution for Design Add and Search Words Data Structure?

The brute force approach for Design Add and Search Words Data Structure is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking each word in the data structure against the search query. If the query contains dots, we replace them with every possible character and check for matches. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Design Add and Search Words Data Structure?

Design Add and Search Words Data Structure uses the following concepts: String, Depth-First Search, Design, Trie. The recommended patterns to study are: Trie, Depth-First Search.

What are the interview tips for Design Add and Search Words Data Structure?

Explain your thought process clearly while coding. Consider edge cases and discuss them with the interviewer. Practice implementing a Trie before the interview.

What are common mistakes when solving Design Add and Search Words Data Structure?

Not considering edge cases with dots. Failing to correctly manage the Trie structure.

#211

Design Add and Search Words Data Structure

Medium

StringDepth-First SearchDesignTrieTrieDepth-First Search

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Using a Trie (prefix tree) allows us to efficiently store words and perform searches, especially with wildcards. The Trie structure enables us to traverse through characters, making it easier to handle dots as any character.

⚙️

Algorithm

3 steps

1Step 1: Create a TrieNode class to represent each node in the Trie.
2Step 2: Implement the WordDictionary class that uses the Trie for adding and searching words.
3Step 3: For searching, recursively traverse the Trie, handling dots by exploring all possible child nodes.

solution.py32 lines

1class TrieNode:
2    def __init__(self):
3        self.children = {}
4        self.is_end_of_word = False
5
6class WordDictionary:
7    def __init__(self):
8        self.root = TrieNode()
9
10    def addWord(self, word: str) -> None:
11        node = self.root
12        for char in word:
13            if char not in node.children:
14                node.children[char] = TrieNode()
15            node = node.children[char]
16        node.is_end_of_word = True
17
18    def search(self, word: str) -> bool:
19        return self._search_in_node(word, self.root)
20
21    def _search_in_node(self, word: str, node: TrieNode) -> bool:
22        for i, char in enumerate(word):
23            if char == '.':
24                for child in node.children.values():
25                    if self._search_in_node(word[i + 1:], child):
26                        return True
27                return False
28            else:
29                if char not in node.children:
30                    return False
31                node = node.children[char]
32        return node.is_end_of_word

ℹ

Complexity note: The time complexity is O(n) for both adding and searching words, where n is the length of the word. The space complexity is O(n) because we store each character of the words in the Trie.

1Using a Trie allows efficient prefix searching.
2Handling wildcards with recursion simplifies the search process.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.