Solution Explanation

This problem requires finding customers who spent at least $100 in both June and July 2020. The solution involves joining three tables (Customers, Orders, and Product), filtering by date, grouping by customer, and finally using a HAVING clause to check the spending condition.

Approach:

1. Join Tables: We start by joining the three tables (Customers, Orders, and Product) using appropriate join conditions (Customers.customer_id = Orders.customer_id and Orders.product_id = Product.product_id). This combines customer information, order details, and product prices into a single dataset.

2. Filter by Date: We filter this combined data to include only orders from June and July 2020 using a WHERE clause and the YEAR() and MONTH() functions. This ensures we consider only relevant orders.

3. Group by Customer: We group the filtered data by customer_id using GROUP BY. This aggregates orders for each customer.

4. Calculate Monthly Spending: Within the GROUP BY aggregation, we calculate the total spending for each customer in June and July separately using conditional aggregation with SUM(IF(...)). IF(MONTH(order_date) = 6, quantity * price, 0) calculates spending for June, and similarly for July.

5. Apply Spending Condition: Finally, the HAVING clause filters the grouped results, keeping only customers whose total spending in both June and July meets the $100 threshold. This ensures that only the qualified customers are included in the output.

Time Complexity Analysis:

The time complexity of this SQL query is dominated by the JOIN operations and the GROUP BY operation. The complexity of joining tables depends on the join algorithm used by the database system (e.g., hash join, merge join). In the worst case, a nested loop join could result in O(n*m) complexity, where n and m are the sizes of the tables being joined. However, optimized join algorithms typically achieve much better performance. The GROUP BY operation generally has a complexity of O(n log n) or O(n) depending on the implementation, where n is the number of rows after joining and filtering. The HAVING clause adds a negligible overhead compared to these operations. Therefore, the overall time complexity is roughly proportional to the time taken for the joins and the GROUP BY which can be considered O(n log n) or better under optimal conditions. The actual execution time will depend on various factors, including database indexing, data distribution, and the specific database system used.

MySQL Code:

SELECT customer_id, name
FROM
    Orders
    JOIN Product USING (product_id)
    JOIN Customers USING (customer_id)
WHERE YEAR(order_date) = 2020
GROUP BY 1
HAVING
    SUM(IF(MONTH(order_date) = 6, quantity * price, 0)) >= 100
    AND SUM(IF(MONTH(order_date) = 7, quantity * price, 0)) >= 100;

No other languages were specified, but the core logic would be similar if implemented in other SQL dialects (PostgreSQL, SQL Server, etc.). The specific syntax for date functions and conditional aggregation might vary slightly, but the overall approach remains the same.