Solution Explanation and Code

This problem requires calculating a 7-day moving average of customer spending at a restaurant. The solution involves using SQL window functions or self-joins to achieve this.

Approach 1: Using Window Functions (MySQL)

This approach leverages MySQL's window functions for efficient calculation.

1. Subquery tt: This subquery groups the Customer table by visited_on and sums the amount for each day. This is necessary because multiple customers might visit on the same day.

2. CTE t: This Common Table Expression (CTE) does the following:

• SUM(amount) OVER (ORDER BY visited_on ROWS 6 PRECEDING): This is the core of the moving average calculation. It calculates a running sum of amount for the current day and the six preceding days. The ROWS 6 PRECEDING clause specifies the window frame.
• RANK() OVER (ORDER BY visited_on ROWS 6 PRECEDING): This assigns a rank to each row based on the visited_on date within the 7-day window. This helps filter out rows where the 7-day window is not fully populated (for the first 6 days).

3. Final SELECT Statement: This selects visited_on, the calculated amount (sum of the 7-day window), and the average amount, rounded to two decimal places using ROUND(amount / 7, 2). The WHERE rk > 6 clause filters out the initial rows where the 7-day window isn't complete.

MySQL Code:

WITH
    t AS (
        SELECT
            visited_on,
            SUM(amount) OVER (
                ORDER BY visited_on
                ROWS 6 PRECEDING
            ) AS amount,
            RANK() OVER (
                ORDER BY visited_on
                ROWS 6 PRECEDING
            ) AS rk
        FROM
            (
                SELECT visited_on, SUM(amount) AS amount
                FROM Customer
                GROUP BY visited_on
            ) AS tt
    )
SELECT visited_on, amount, ROUND(amount / 7, 2) AS average_amount
FROM t
WHERE rk > 6;

Time Complexity: O(N log N) due to the RANK() function, where N is the number of rows in the Customer table after grouping by visited_on. The sorting implicit in window functions contributes to this complexity. If RANK() wasn't used, it would be closer to O(N).

Space Complexity: O(N) for the CTE.

Approach 2: Using Self-Join (MySQL)

This approach uses a self-join to find the sum of amounts within a 7-day window for each date.

1. Subquery a: This selects distinct visited_on dates from the Customer table. This generates a list of all dates.

2. JOIN: The a subquery is joined with the Customer table (b) using DATEDIFF. This ensures that only rows within a 7-day window (including the current day) are included in the sum.

3. WHERE clause: Filters out dates that don't have a full 7-day window available (at least the first six days).

4. GROUP BY and ORDER BY: Groups the results by visited_on to sum the amounts for each 7-day period and orders by visited_on.

MySQL Code:

SELECT
    a.visited_on,
    SUM(b.amount) AS amount,
    ROUND(SUM(b.amount) / 7, 2) AS average_amount
FROM
    (SELECT DISTINCT visited_on FROM customer) AS a
    JOIN customer AS b ON DATEDIFF(a.visited_on, b.visited_on) BETWEEN 0 AND 6
WHERE a.visited_on >= (SELECT MIN(visited_on) FROM customer) + 6
GROUP BY 1
ORDER BY 1;
 

Time Complexity: O(N*M), where N is the number of distinct dates and M is the average number of rows per date. This is because of the self-join. In the worst case (many customers each day), it could be O(N^2).

Space Complexity: O(N) to store the distinct dates.

In Summary:

Both approaches achieve the desired result. The window function approach is generally preferred for its conciseness and potentially better performance, especially for large datasets, although the self-join is more easily understood for those unfamiliar with window functions. The time complexity analysis highlights the potential performance differences between these two methods. Choosing the best approach often depends on database engine specific optimizations and the size of the dataset.