Solution Explanation: Finding Active Users

This problem requires identifying users who have logged in for at least five consecutive days. The solution leverages SQL's window functions for efficient processing. Let's break down the approach and the code.

Approach:

The solution employs a multi-step approach using Common Table Expressions (CTEs) in SQL:

1. Data Consolidation: A CTE called T joins the Accounts and Logins tables to combine user information with their login dates. DISTINCT ensures that we only consider unique login entries for each user on each day.

2. Consecutive Day Grouping: A crucial CTE, P, calculates a grouping variable g. This variable represents the starting date of a consecutive login sequence for each user. It's calculated using ROW_NUMBER() partitioned by user ID and ordered by login date. Subtracting the row number (in days) from the login date effectively groups consecutive login days together. If a user logs in for five consecutive days, all those logins will have the same g value.

3. Active User Identification: The final SELECT statement groups the results by id (user ID) and g (consecutive login group). The HAVING clause filters out groups with fewer than 5 logins, effectively identifying active users. DISTINCT ensures we only return each active user once. The results are ordered by id.

Time Complexity Analysis:

The time complexity is dominated by the window function operations (ROW_NUMBER()) and the grouping and aggregation (GROUP BY, HAVING). Window functions generally have a time complexity of O(N log N) where N is the number of rows in the Logins table (after joining with Accounts). The GROUP BY and HAVING operations also have a time complexity that is dependent on the data, but in the worst case, could also be O(N log N). Therefore, the overall time complexity of this solution is approximately O(N log N).

Space Complexity Analysis:

The space complexity is primarily determined by the size of the intermediate CTEs (T and P). These CTEs store temporary data during the processing. In the worst case, the space required by the CTEs is proportional to the size of the input data. So, the space complexity is approximately O(N), where N is the number of rows in the Logins table after the join.

MySQL Code:

WITH
    T AS (
        SELECT DISTINCT *
        FROM
            Logins
            JOIN Accounts USING (id)
    ),
    P AS (
        SELECT
            *,
            DATE_SUB(
                login_date,
                INTERVAL ROW_NUMBER() OVER (
                    PARTITION BY id
                    ORDER BY login_date
                ) DAY
            ) g
        FROM T
    )
SELECT DISTINCT id, name
FROM P
GROUP BY id, g
HAVING COUNT(*) >= 5
ORDER BY 1;

Note: This solution directly addresses the problem statement. Adapting it for a general "n" consecutive days simply requires changing the HAVING COUNT(*) >= 5 condition to HAVING COUNT(*) >= n. No significant algorithmic changes are needed.