Solution Explanation

This problem requires querying three tables: Customers, Contacts, and Invoices to generate a result set containing invoice details and contact information. The solution uses SQL joins and aggregate functions to achieve this.

Approach

The core idea is to perform a series of joins to link related information from the three tables and then use aggregate functions (COUNT) to determine the number of contacts and trusted contacts for each invoice.

1. Join Invoices and Customers: We start by joining the Invoices table with the Customers table on user_id to get the customer's name (customer_name) associated with each invoice. This is done using a LEFT JOIN to include all invoices, even if a customer isn't found (though this shouldn't happen given the problem description).

2. Join with Contacts: Next, we join the result with the Contacts table on user_id to retrieve contact information for each customer. This is also a LEFT JOIN to ensure that all invoices and customers are included regardless of whether they have contacts.

3. Count Contacts: We use COUNT(t3.user_id) to count the total number of contacts for each customer associated with an invoice. The t3 alias references the Contacts table.

4. Join with Customers (again) to count trusted contacts: To identify trusted contacts (contacts who are also customers), we perform another LEFT JOIN from the Contacts table to the Customers table on contact_email = email. This join links contacts to customer records. We then use COUNT(t4.email) to count the number of contacts whose emails exist in the Customers table (i.e., trusted contacts). The t4 alias references the second instance of the Customers table.

5. Grouping and Ordering: Finally, the results are grouped by invoice_id using GROUP BY to aggregate results for each invoice. The results are ordered by invoice_id using ORDER BY.

MySQL Code Explanation

SELECT
    invoice_id,
    t2.customer_name,
    price,
    COUNT(t3.user_id) AS contacts_cnt,
    COUNT(t4.email) AS trusted_contacts_cnt
FROM
    Invoices AS t1
    LEFT JOIN Customers AS t2 ON t1.user_id = t2.customer_id
    LEFT JOIN Contacts AS t3 ON t1.user_id = t3.user_id
    LEFT JOIN Customers AS t4 ON t3.contact_email = t4.email
GROUP BY invoice_id
ORDER BY invoice_id;

• SELECT: This clause specifies the columns to be included in the output. COUNT(t3.user_id) and COUNT(t4.email) are aggregate functions counting contacts and trusted contacts, respectively.
• FROM, LEFT JOIN: These clauses specify the tables to be joined. LEFT JOIN ensures that all rows from the left table (the one before LEFT JOIN) are included in the result, even if there's no match in the right table.
• ON: This clause specifies the join conditions. It determines how rows from different tables are matched.
• GROUP BY: This clause groups the rows based on the invoice_id, so aggregate functions calculate values for each invoice.
• ORDER BY: This clause orders the results by invoice_id.

Time Complexity Analysis

The time complexity of this SQL query depends on the size of the tables involved. The joins have a complexity that's roughly proportional to the product of the sizes of the tables being joined (though optimized database systems usually do much better than this naive analysis). The COUNT aggregate functions add a linear complexity related to the size of the grouped data. Therefore, the overall time complexity is approximately O(N * M * K), where N is the number of rows in Invoices, M is the number of rows in Customers, and K is the number of rows in Contacts. The complexity could be lower with optimized database indexing.

Space Complexity Analysis

The space complexity is largely determined by the size of the intermediate result set generated during the joins and the final result set. In the worst case, the space complexity is proportional to the size of the largest table or the combined size of the joined tables, approximately O(N + M + K). Again, this is a simplified analysis and database optimization techniques can influence the actual space used.