How do you solve Find Golden Hour Customers on LeetCode?

Find Golden Hour Customers (LeetCode #3705) 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: Focus on aggregating data efficiently.

What is the time complexity of Find Golden Hour Customers?

The optimal solution for Find Golden Hour Customers runs in O(n) time and uses O(n) space. Single pass through data for aggregation leads to linear time complexity.

What is the brute force solution for Find Golden Hour Customers?

The brute force approach for Find Golden Hour Customers is Brute Force, with O(n²) time complexity and O(1) space complexity. Check each customer individually to see if they meet all criteria. This involves counting orders and ratings, leading to a lot of repeated work. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Find Golden Hour Customers?

Clarify requirements before coding. Explain your thought process while coding. Test edge cases.

What are common mistakes when solving Find Golden Hour Customers?

Neglecting to check for NULL ratings. Incorrectly calculating percentages.

#3705

Find Golden Hour Customers

Medium

Hash MapArray

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)

Use aggregation to compute counts and averages in a single pass, reducing repeated calculations.

⚙️

Algorithm

3 steps

1Step 1: Aggregate orders by customer to count total orders, rated orders, and sum ratings.
2Step 2: Calculate peak hour orders in the same aggregation.
3Step 3: Filter customers based on the criteria using HAVING clause.

solution.py5 lines

1# Full working Python code
2SELECT customer_id
3FROM restaurant_orders
4GROUP BY customer_id
5HAVING COUNT(order_id) >= 3 AND AVG(order_rating) >= 4.0 AND SUM(CASE WHEN (HOUR(order_timestamp) BETWEEN 11 AND 14 OR HOUR(order_timestamp) BETWEEN 18 AND 21) THEN 1 ELSE 0 END) / COUNT(order_id) >= 0.6 AND COUNT(order_rating) / COUNT(order_id) >= 0.5;

ℹ

Complexity note: Single pass through data for aggregation leads to linear time complexity.

1Focus on aggregating data efficiently.
2Understand the importance of filtering based on multiple criteria.

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