How do you solve Search Suggestions System on LeetCode?

Search Suggestions System (LeetCode #1268) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * m) time complexity and O(n * m) space complexity. Key insight: Sorting the products is crucial for lexicographical order.

What is the brute force solution for Search Suggestions System?

The brute force approach for Search Suggestions System is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking all products for each prefix of the searchWord. This is straightforward but inefficient as it requires repeated scanning of the product list. The optimal Optimal Solution approach improves this to O(n * m).

What are the interview tips for Search Suggestions System?

Explain your thought process clearly. Discuss the trade-offs of different approaches. Be ready to optimize your solution if asked.

What are common mistakes when solving Search Suggestions System?

Not sorting the products before searching. Failing to limit suggestions to three products.

#1268

Search Suggestions System

Medium

ArrayStringBinary SearchTrieSortingHeap (Priority Queue)Hash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n * m)Space O(n * m)

The optimal solution uses a Trie (prefix tree) to efficiently store and retrieve suggestions based on prefixes. This allows for quick lookups as we build the searchWord character by character.

⚙️

Algorithm

3 steps

1Step 1: Build a Trie from the sorted products list.
2Step 2: For each character in searchWord, traverse the Trie to find matching products.
3Step 3: Collect at most three suggestions for each prefix and store them in a result list.

solution.py33 lines

1# Full working Python code
2class TrieNode:
3    def __init__(self):
4        self.children = {}
5        self.products = []
6
7class Trie:
8    def __init__(self):
9        self.root = TrieNode()
10
11    def insert(self, product):
12        node = self.root
13        for char in product:
14            if char not in node.children:
15                node.children[char] = TrieNode()
16            node = node.children[char]
17            if len(node.products) < 3:
18                node.products.append(product)
19
20def suggestedProducts(products, searchWord):
21    products.sort()
22    trie = Trie()
23    for product in products:
24        trie.insert(product)
25    result = []
26    node = trie.root
27    for char in searchWord:
28        if char in node.children:
29            node = node.children[char]
30        else:
31            node = TrieNode()
32        result.append(node.products)
33    return result

ℹ

Complexity note: The time complexity is O(n * m) where n is the number of products and m is the maximum length of a product. This is due to inserting each product into the Trie. The space complexity is O(n * m) for storing the Trie structure.

1Sorting the products is crucial for lexicographical order.
2Using a Trie allows for efficient prefix searching.

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