How do you solve List the Products Ordered in a Period on LeetCode?

List the Products Ordered in a Period (LeetCode #1327) 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 how to filter and aggregate data is crucial.

What is the time complexity of List the Products Ordered in a Period?

The optimal solution for List the Products Ordered in a Period runs in O(n) time and uses O(n) space. This complexity is due to a single pass through the Orders table to aggregate units, leading to linear time complexity.

What is the brute force solution for List the Products Ordered in a Period?

The brute force approach for List the Products Ordered in a Period is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we will check each order and sum the units for each product within the specified date range. This is straightforward but inefficient as it involves multiple scans of the data. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve List the Products Ordered in a Period?

List the Products Ordered in a Period uses the following concepts: Database. The recommended patterns to study are: Hash Map, Aggregation, Join.

What are the interview tips for List the Products Ordered in a Period?

Always clarify the requirements before jumping to code. Think about edge cases, like no orders in the given period. Practice writing SQL queries to become familiar with syntax.

What are common mistakes when solving List the Products Ordered in a Period?

Not filtering the date range correctly. Forgetting to group by product_id before summing.

#1327

List the Products Ordered in a Period

Easy

DatabaseHash MapAggregationJoin

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 approach, we will use a single pass to aggregate the units ordered for each product in February 2020, which reduces the number of scans and improves efficiency.

⚙️

Algorithm

4 steps

1Step 1: Filter the Orders table for February 2020.
2Step 2: Use a HashMap to aggregate the total units ordered for each product_id.
3Step 3: Join the aggregated results with the Products table to get product names.
4Step 4: Filter the results to include only those products with a total unit count of at least 100.

solution.py7 lines

1SELECT p.product_name, total_units
2FROM (SELECT product_id, SUM(unit) AS total_units
3      FROM Orders
4      WHERE order_date >= '2020-02-01' AND order_date < '2020-03-01'
5      GROUP BY product_id
6      HAVING SUM(unit) >= 100) AS aggregated
7JOIN Products p ON aggregated.product_id = p.product_id;

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Complexity note: This complexity is due to a single pass through the Orders table to aggregate units, leading to linear time complexity.

1Understanding how to filter and aggregate data is crucial.
2Using joins effectively can simplify data retrieval.

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