Solution Explanation:

The problem requires retrieving the number of unique orders and unique customers per month, considering only orders with invoices exceeding $20. The solution uses a combination of filtering and aggregation.

Approach:

1. Filtering: The initial step involves filtering the Orders table to include only those entries where the invoice amount is greater than 20. This is crucial because the problem's constraints specify that only these orders should be considered.

2. Grouping: After filtering, the data is grouped by month. The month needs to be extracted from the order_date column. The specific method for extracting the month depends on the database system used (e.g., DATE_FORMAT in MySQL, .dt.to_period("M") in Pandas).

3. Aggregation: For each month group, we perform the following aggregations:

   • order_count: The total number of orders (using COUNT(order_id)).
   • customer_count: The number of unique customers (using COUNT(DISTINCT customer_id)).

4. Result: Finally, the aggregated results are returned in a table format, with columns representing the month, order count, and customer count.

Time Complexity Analysis:

The time complexity of this solution is dominated by the grouping and aggregation operations. The exact complexity depends on the database system's implementation, but it generally scales linearly with the number of rows in the Orders table (O(N)), where N is the number of rows. Filtering is also a linear operation. Therefore, the overall time complexity is O(N).

Space Complexity Analysis:

The space complexity is determined by the size of the intermediate result set created during the grouping and aggregation process. In the worst-case scenario, if every month has a unique entry, the space complexity will be proportional to the number of unique months. However, this is generally much smaller than the size of the original Orders table. Therefore, space complexity is considered to be relatively low and can be approximated as O(M), where M is the number of unique months.

Code in Different Languages:

MySQL:

SELECT
    DATE_FORMAT(order_date, '%Y-%m') AS month,
    COUNT(order_id) AS order_count,
    COUNT(DISTINCT customer_id) AS customer_count
FROM Orders
WHERE invoice > 20
GROUP BY 1;

This MySQL query directly performs the filtering, grouping, and aggregation in a single SQL statement. DATE_FORMAT extracts the year and month, COUNT counts orders, and COUNT(DISTINCT) counts unique customers. The GROUP BY 1 clause groups the results by the first column (month).

Pandas (Python):

import pandas as pd
 
def unique_orders_and_customers(orders: pd.DataFrame) -> pd.DataFrame:
    filtered_orders = orders[orders["invoice"] > 20]
    filtered_orders["month"] = filtered_orders["order_date"].dt.to_period("M").astype(str)
    result = (
        filtered_orders.groupby("month")
        .agg(
            order_count=("order_id", "count"), customer_count=("customer_id", "nunique")
        )
        .reset_index()
    )
    return result
 

The Pandas solution mirrors the approach. It first filters the DataFrame, then extracts the month using the .dt.to_period("M") accessor. The .groupby() method groups the data, and the .agg() method performs the aggregation using count and nunique functions. reset_index() converts the grouped result back to a DataFrame.

This detailed explanation provides a comprehensive understanding of the solution to the problem, its complexities, and its implementation in different programming languages.