Solution Explanations for LeetCode 626: Exchange Seats

This problem requires swapping the seats of consecutive students in a database table. Several SQL solutions are provided, each with different approaches and complexities. Let's analyze them.

Solution 1: Using LEFT JOIN and Bitwise XOR

This solution cleverly uses a LEFT JOIN and bitwise XOR (^) operation to achieve the swap.

SQL (MySQL):

SELECT s1.id, COALESCE(s2.student, s1.student) AS student
FROM
    Seat AS s1
    LEFT JOIN Seat AS s2 ON (s1.id + 1) ^ 1 - 1 = s2.id
ORDER BY 1;

Explanation:

• (s1.id + 1) ^ 1 - 1: This expression is the key. Let's break it down:
  • s1.id + 1: Gets the next ID.
  • ^ 1: XORing with 1 flips the least significant bit. This effectively toggles between even and odd numbers.
  • - 1: Subtracts 1 to get the correct ID for swapping.
• LEFT JOIN Seat AS s2 ON ...: This joins s1 (current row) with s2 (next row) based on the calculated ID. If it's the last row, s2 will be NULL.
• COALESCE(s2.student, s1.student): If a matching s2 is found (even ID), it uses s2.student; otherwise (odd ID or last row), it keeps s1.student.
• ORDER BY 1: Orders the result by ID.

Time Complexity: O(N log N) due to the sorting (though the join itself is likely faster than a nested loop). The complexity is dominated by the ORDER BY clause. The JOIN and the rest of the operations are relatively efficient.

Space Complexity: O(N) due to the creation of the result set.

Solution 2: Conditional Logic with CASE

This solution uses conditional logic with CASE statements to determine whether to increment or decrement the ID.

SQL (MySQL):

SELECT
    id + (
        CASE
            WHEN id % 2 = 1
            AND id != (SELECT MAX(id) FROM Seat) THEN 1
            WHEN id % 2 = 0 THEN -1
            ELSE 0
        END
    ) AS id,
    student
FROM Seat
ORDER BY 1;

Explanation:

• id % 2 = 1: Checks if the ID is odd.
• id != (SELECT MAX(id) FROM Seat): Checks if it's not the last row.
• CASE statement: Adds 1 to odd IDs (except the last one) and subtracts 1 from even IDs.
• ORDER BY 1: Orders the result by the modified ID.

Time Complexity: O(N log N) due to the sorting. The CASE statement and subquery are performed for each row, leading to a linear time complexity in this part. However, sorting dominates the complexity.

Space Complexity: O(N) due to the result set.

Solution 3: Using RANK() Window Function

This is a more concise solution using the RANK() window function.

SQL (MySQL):

SELECT
    RANK() OVER (ORDER BY (id - 1) ^ 1) AS id,
    student
FROM Seat;

Explanation:

• (id - 1) ^ 1: Similar to Solution 1, this manipulates the ID to create the desired ordering.
• RANK() OVER (ORDER BY ...): Assigns ranks based on the modified ID, effectively swapping the order of consecutive students.

Time Complexity: O(N log N). Window functions often have logarithmic time complexity in the order of the number of rows due to sorting internally.

Space Complexity: O(N) for the result set.

Solution 4: Using ROW_NUMBER() and CASE

This solution combines ROW_NUMBER() with CASE for more explicit control.

SQL (MySQL):

SELECT
    CASE
        WHEN id & 1 = 0 THEN id - 1
        WHEN ROW_NUMBER() OVER (ORDER BY id) != COUNT(id) OVER () THEN id + 1
        ELSE id
    END AS id,
    student
FROM Seat
ORDER BY 1;

Explanation:

• id & 1 = 0: Checks if the ID is even (using bitwise AND).
• ROW_NUMBER() OVER (ORDER BY id): Assigns a unique rank to each row.
• COUNT(id) OVER (): Gets the total number of rows.
• CASE statement: Swaps even and odd IDs, handling the last row appropriately.

Time Complexity: O(N log N) dominated by the window functions and sorting in ORDER BY.

Space Complexity: O(N) for the result set.

Summary:

All four solutions achieve the same outcome but with varying approaches and complexities. Solutions 1 and 3 are generally more efficient because they leverage bitwise operations and window functions more effectively. The choice of the best solution might depend on the specific database system and performance considerations. However, for practical purposes, the differences in performance will often be negligible unless dealing with very large datasets.