How do you solve Implement Trie (Prefix Tree) on LeetCode?

Implement Trie (Prefix Tree) (LeetCode #208) 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: Tries are efficient for prefix-based searches.

What is the brute force solution for Implement Trie (Prefix Tree)?

The brute force approach for Implement Trie (Prefix Tree) is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute-force approach involves storing all words in a list and checking for existence or prefixes by iterating through the list. This is simple but inefficient for larger datasets. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Implement Trie (Prefix Tree)?

Implement Trie (Prefix Tree) uses the following concepts: Hash Table, String, Design, Trie. The recommended patterns to study are: Trie, Prefix Tree.

What are the interview tips for Implement Trie (Prefix Tree)?

Explain your thought process clearly. Consider edge cases during implementation. Practice writing clean and efficient code.

What are common mistakes when solving Implement Trie (Prefix Tree)?

Not handling edge cases like empty strings. Confusing search and startsWith logic.

#208

Implement Trie (Prefix Tree)

Medium

Hash TableStringDesignTrieTriePrefix Tree

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a trie structure, where each node represents a character of a word. This allows for efficient insertion and search operations, reducing the time complexity significantly.

⚙️

Algorithm

5 steps

1Step 1: Create a TrieNode class that contains a dictionary for children and a boolean to mark the end of a word.
2Step 2: In the Trie class, initialize a root node.
3Step 3: For 'insert', traverse the trie character by character, creating new nodes as needed, and mark the end of the word.
4Step 4: For 'search', traverse the trie and check if the entire word exists.
5Step 5: For 'startsWith', traverse the trie to check if any word starts with the given prefix.

solution.py28 lines

1class TrieNode:
2    def __init__(self):
3        self.children = {}
4        self.is_end_of_word = False
5class Trie:
6    def __init__(self):
7        self.root = TrieNode()
8    def insert(self, word):
9        node = self.root
10        for char in word:
11            if char not in node.children:
12                node.children[char] = TrieNode()
13            node = node.children[char]
14        node.is_end_of_word = True
15    def search(self, word):
16        node = self.root
17        for char in word:
18            if char not in node.children:
19                return False
20            node = node.children[char]
21        return node.is_end_of_word
22    def startsWith(self, prefix):
23        node = self.root
24        for char in prefix:
25            if char not in node.children:
26                return False
27            node = node.children[char]
28        return True

ℹ

Complexity note: The time complexity is O(n) for insert, search, and startsWith, where n is the length of the word or prefix. This is because we only traverse the length of the word/prefix. Space complexity is O(n) as we store nodes for each character in the trie.

1Tries are efficient for prefix-based searches.
2Each node represents a character, allowing for quick traversal.

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