How do you solve Immediate Food Delivery II on LeetCode?

Immediate Food Delivery II (LeetCode #1174) 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: Understanding the difference between immediate and scheduled orders is crucial.

What is the time complexity of Immediate Food Delivery II?

The optimal solution for Immediate Food Delivery II runs in O(n) time and uses O(n) space. This complexity is linear because we only pass through the data a couple of times, making it efficient.

What is the brute force solution for Immediate Food Delivery II?

The brute force approach for Immediate Food Delivery II is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we will iterate through each customer's orders to find their first order and check if it is immediate. This method is straightforward but inefficient as it requires multiple passes over the data. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Immediate Food Delivery II?

Immediate Food Delivery II uses the following concepts: Database. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Immediate Food Delivery II?

Explain your thought process clearly while coding. Practice writing SQL queries as they are common in database-related problems. Be prepared to discuss the trade-offs of different approaches.

What are common mistakes when solving Immediate Food Delivery II?

Not considering edge cases where a customer has only one order. Failing to round the percentage correctly.

#1174

Immediate Food Delivery II

Medium

DatabaseHash 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)

In the optimal solution, we use a single pass to gather the first orders and check for immediacy in one go. This reduces the number of iterations and improves efficiency.

⚙️

Algorithm

4 steps

1Step 1: Use a HashMap to store the first order date for each customer.
2Step 2: Iterate through the Delivery table and populate the HashMap with the earliest order date for each customer.
3Step 3: Count the immediate orders by checking if the preferred delivery date matches the first order date.
4Step 4: Calculate the percentage of immediate orders and round to 2 decimal places.

solution.py6 lines

1# Full working Python code
2SELECT ROUND(SUM(CASE WHEN order_date = customer_pref_delivery_date THEN 1 ELSE 0 END) * 100.0 / COUNT(*), 2) AS immediate_percentage
3FROM (SELECT customer_id, MIN(order_date) AS first_order_date
4      FROM Delivery
5      GROUP BY customer_id) AS first_orders
6JOIN Delivery ON first_orders.customer_id = Delivery.customer_id AND first_orders.first_order_date = Delivery.order_date;

ℹ

Complexity note: This complexity is linear because we only pass through the data a couple of times, making it efficient.

1Understanding the difference between immediate and scheduled orders is crucial.
2Using a HashMap can significantly reduce time complexity.

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