Solution Explanation for LeetCode 2175: The Change in Global Rankings

This problem requires calculating the change in global rankings of national teams after updating their points based on a PointsChange table. The solution involves several steps:

1. Updating Team Points: We need to add the points_change from the PointsChange table to the points in the TeamPoints table for each team.

2. Calculating Initial and Final Ranks: We determine the initial rank before the points update and the final rank after the update. The ranking is based on points (descending) and then name (lexicographical order) as tie-breaker.

3. Calculating Rank Difference: Finally, we subtract the initial rank from the final rank to find the change in ranking for each team.

SQL Solution (MySQL)

The provided MySQL solution efficiently performs these steps using common table expressions (CTEs) and window functions:

WITH
    P AS (
        SELECT team_id, SUM(points_change) AS delta
        FROM PointsChange
        GROUP BY team_id
    )
SELECT
    team_id,
    name,
    CAST(RANK() OVER (ORDER BY points DESC, name) AS SIGNED) - CAST(
        RANK() OVER (ORDER BY (points + delta) DESC, name) AS SIGNED
    ) AS 'rank_diff'
FROM
    TeamPoints
    JOIN P USING (team_id);

Explanation:

• CTE P: This CTE calculates the total points_change for each team_id from the PointsChange table. SUM(points_change) aggregates changes for a team if they have multiple entries. GROUP BY team_id ensures that we have one row per team.

• Main Query:

  • JOIN P USING (team_id): This joins TeamPoints with P based on team_id, combining the team's initial points with their total point change.
  • RANK() OVER (ORDER BY points DESC, name): This calculates the initial rank of each team based on their initial points (descending) and name (ascending) as a tie-breaker. RANK() assigns the same rank to teams with equal points.
  • RANK() OVER (ORDER BY (points + delta) DESC, name): This calculates the final rank using the updated points (points + delta).
  • CAST(... AS SIGNED): The RANK() function returns integers, but we need to allow for negative differences, so we cast them to SIGNED integers.
  • ... - ...: Subtracting the initial rank from the final rank gives the rank_diff.

Time Complexity Analysis

The time complexity of this SQL query is dominated by the RANK() window function. Window functions typically have a time complexity of O(N log N) where N is the number of rows in the TeamPoints table, due to the sorting required for ranking. The JOIN operation is also O(N) on average (assuming appropriate indexing). Therefore, the overall time complexity of the solution is O(N log N).

Space Complexity Analysis

The space complexity depends on the size of the intermediate CTE P and the result table. In the worst case, P could have the same size as TeamPoints if each team has only one entry in PointsChange. The result table also has the same size as TeamPoints. Therefore, the space complexity is O(N).