Leetcode #1268: Search Suggestions System
In this guide, we solve Leetcode #1268 Search Suggestions System in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given an array of strings products and a string searchWord. Design a system that suggests at most three product names from products after each character of searchWord is typed.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Trie, Array, String, Binary Search, Sorting, Heap (Priority Queue)
Intuition
The problem structure suggests a monotonic decision, which makes binary search a natural fit.
By halving the search space each step, we reach the answer efficiently.
Approach
Search either directly on a sorted array or on the answer space using a check function.
Each check is fast, and the logarithmic search keeps the overall runtime low.
Steps:
- Define the search bounds.
- Check the mid point condition.
- Narrow the bounds until convergence.
Example
Input: products = ["mobile","mouse","moneypot","monitor","mousepad"], searchWord = "mouse"
Output: [["mobile","moneypot","monitor"],["mobile","moneypot","monitor"],["mouse","mousepad"],["mouse","mousepad"],["mouse","mousepad"]]
Explanation: products sorted lexicographically = ["mobile","moneypot","monitor","mouse","mousepad"].
After typing m and mo all products match and we show user ["mobile","moneypot","monitor"].
After typing mou, mous and mouse the system suggests ["mouse","mousepad"].
Python Solution
class Trie:
def __init__(self):
self.children: List[Union[Trie, None]] = [None] * 26
self.v: List[int] = []
def insert(self, w, i):
node = self
for c in w:
idx = ord(c) - ord('a')
if node.children[idx] is None:
node.children[idx] = Trie()
node = node.children[idx]
if len(node.v) < 3:
node.v.append(i)
def search(self, w):
node = self
ans = [[] for _ in range(len(w))]
for i, c in enumerate(w):
idx = ord(c) - ord('a')
if node.children[idx] is None:
break
node = node.children[idx]
ans[i] = node.v
return ans
class Solution:
def suggestedProducts(
self, products: List[str], searchWord: str
) -> List[List[str]]:
products.sort()
trie = Trie()
for i, w in enumerate(products):
trie.insert(w, i)
return [[products[i] for i in v] for v in trie.search(searchWord)]
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.