Solution Explanation: Reporting Contiguous Dates

This problem requires generating a report summarizing contiguous periods of task success or failure based on two input tables: Failed and Succeeded. The solution leverages SQL's capabilities for data manipulation and aggregation.

Approach:

The core idea is to combine the data from both tables, identify contiguous sequences of the same status (success or failure), and then group these sequences to determine the start and end dates of each period.

1. Combining the Tables:

First, we use a UNION ALL operation to combine the Failed and Succeeded tables. This creates a single table T containing all dates, with an additional column st indicating the status ('failed' or 'succeeded'). The WHERE clause filters the dates to only include those within 2019.

2. Identifying Contiguous Periods:

The key to identifying contiguous periods is calculating a "period identifier" (pt). We achieve this using the RANK() window function. RANK() assigns a rank to each date within each status group (partitioned by st and ordered by dt). Subtracting this rank from the date yields a consistent value for dates belonging to the same contiguous period. Dates within a continuous successful period will all have the same pt value, and similarly for failed periods.

3. Grouping and Aggregating:

Finally, we group the results by the status (st) and the period identifier (pt). The MIN(dt) and MAX(dt) functions determine the start and end dates of each continuous period. The result is ordered by the start date (start_date).

Time Complexity Analysis:

The time complexity is dominated by the window function (RANK()) and the GROUP BY operations. The RANK() operation typically has a time complexity of O(N log N) where N is the number of rows in the combined table (though the exact complexity depends on database implementation). The GROUP BY operation also has a complexity related to sorting and grouping which is typically O(N log N) in the worst case. Therefore, the overall time complexity is approximately O(N log N).

Space Complexity Analysis:

The space complexity depends on the size of the intermediate tables created during the query execution. In the worst case, if all dates are in one continuous period of the same status, the space complexity is proportional to the number of rows in the input tables, which is O(N).

Code Implementation (MySQL):

WITH
    T AS (
        SELECT fail_date AS dt, 'failed' AS st
        FROM Failed
        WHERE YEAR(fail_date) = 2019
        UNION ALL
        SELECT success_date AS dt, 'succeeded' AS st
        FROM Succeeded
        WHERE YEAR(success_date) = 2019
    )
SELECT
    st AS period_state,
    MIN(dt) AS start_date,
    MAX(dt) AS end_date
FROM
    (
        SELECT
            *,
            SUBDATE(
                dt,
                RANK() OVER (
                    PARTITION BY st
                    ORDER BY dt
                )
            ) AS pt
        FROM T
    ) AS t
GROUP BY 1, pt
ORDER BY 2;
 

This MySQL code directly implements the described algorithm. Other SQL dialects (PostgreSQL, SQL Server, etc.) would have similar implementations, with minor syntax adjustments. The core logic involving UNION ALL, RANK(), GROUP BY, MIN(), and MAX() remains the same.