Solution Explanation for LeetCode Problem 2308: Arrange Table by Gender

This problem requires rearranging a table of user IDs and genders to alternate between 'female', 'other', and 'male', with IDs within each gender sorted ascendingly. We can achieve this efficiently using window functions in SQL.

Approach 1: Using RANK() and CASE

This approach uses two ranking functions:

1. RANK() OVER (PARTITION BY gender ORDER BY user_id): This assigns a rank to each user within their gender group, ordered by user_id. This ensures that IDs within each gender are sorted.

2. CASE WHEN gender = 'female' THEN 0 WHEN gender = 'other' THEN 1 ELSE 2 END AS rk2: This assigns a secondary rank based on gender. 'female' gets 0, 'other' gets 1, and 'male' gets 2. This dictates the alternating order.

The final ORDER BY rk1, rk2 clause sorts the results first by the rank within each gender (rk1) and then by the gender rank (rk2), producing the desired alternating pattern.

MySQL Code (Approach 1):

WITH
    t AS (
        SELECT
            *,
            RANK() OVER (
                PARTITION BY gender
                ORDER BY user_id
            ) AS rk1,
            CASE
                WHEN gender = 'female' THEN 0
                WHEN gender = 'other' THEN 1
                ELSE 2
            END AS rk2
        FROM Genders
    )
SELECT user_id, gender
FROM t
ORDER BY rk1, rk2;

Time Complexity: O(N log N) due to the RANK() window function, which typically has logarithmic time complexity. Sorting is involved implicitly in the RANK() function. Space Complexity: O(N) to store the intermediate result set t.

Approach 2: Simplified Ordering

This approach cleverly uses the RANK() function's inherent ordering to simplify the solution. It leverages the fact that the RANK() function itself already sorts users within each gender. By ordering by the rank first and then using the constant 2 in the ORDER BY clause, we effectively achieve the alternating pattern. The constant 2 acts as a placeholder, ensuring that the ordering happens by the rank and not some other default ordering column.

MySQL Code (Approach 2):

SELECT
    user_id,
    gender
FROM Genders
ORDER BY
    (
        RANK() OVER (
            PARTITION BY gender
            ORDER BY user_id
        )
    ),
    2;

Time Complexity: O(N log N), same as Approach 1 due to the RANK() function. Space Complexity: O(N), as it implicitly requires storage for calculating the rank.

Conclusion

Both approaches achieve the desired result. Approach 2 is more concise and arguably more elegant, directly leveraging the features of the RANK() function for the final ordering. The time complexity remains the same for both methods. The choice between them largely depends on personal preference and readability considerations.