Solution Explanation for Ordering Two Columns Independently

This problem requires ordering two columns of a table independently: first_col in ascending order and second_col in descending order. A simple ORDER BY clause won't work because it orders rows based on the entire row, not individual columns. The solution uses window functions to assign ranks to each column separately and then joins the results.

Approach

The solution employs the following steps:

1. Rank first_col: A common table expression (CTE) called S is created to rank the first_col values in ascending order using ROW_NUMBER() OVER (ORDER BY first_col). This assigns a unique rank to each distinct value of first_col, or if there are duplicates, the same rank is given to each, preserving the order.

2. Rank second_col: Another CTE, T, is created to similarly rank the second_col values, but this time in descending order using ROW_NUMBER() OVER (ORDER BY second_col DESC). This ensures the highest second_col values receive the lowest rank.

3. Join the Ranked Data: Finally, S and T are joined using the rk column (the rank). Because each CTE assigns the same rank for values at the same position originally, the join effectively combines the independently ordered columns. The result is a table where first_col is in ascending order and second_col is in descending order, maintaining the original pairing of first_col and second_col as much as possible based on their ranks.

MySQL Code

WITH
    S AS (
        SELECT
            first_col,
            ROW_NUMBER() OVER (ORDER BY first_col) AS rk
        FROM Data
    ),
    T AS (
        SELECT
            second_col,
            ROW_NUMBER() OVER (ORDER BY second_col DESC) AS rk
        FROM Data
    )
SELECT first_col, second_col
FROM
    S
    JOIN T USING (rk);

Time Complexity Analysis

The time complexity of this solution is dominated by the ROW_NUMBER() function within the CTEs. ROW_NUMBER() has a time complexity of O(N log N) on average, where N is the number of rows in the Data table, due to the sorting involved. The join operation has a time complexity of O(N) in this case because it's based on a unique key (rk). Therefore, the overall time complexity of the query is O(N log N).

Space Complexity Analysis

The space complexity is O(N) because the CTEs S and T store intermediate results with the same number of rows as the input table. The space used by the final SELECT statement is also proportional to the number of rows in the input. Hence, the space complexity is O(N).