Solution Explanation for LeetCode Problem 1831: Maximum Transaction Each Day

This problem requires finding the transaction IDs with the maximum amount for each day in the Transactions table. The solution uses a window function for efficient and concise querying.

Approach: Using Window Functions

The core idea is to leverage the power of window functions to rank transactions within each day based on their amounts. We use RANK() to handle ties (multiple transactions with the same maximum amount on a day). RANK() assigns the same rank to all transactions with the same amount within a partition (day), unlike ROW_NUMBER() which would arbitrarily assign unique ranks.

Steps:

1. Partitioning: We partition the data by the DAY() function applied to the day column. This groups transactions by the day of the transaction, ignoring the time component.

2. Ordering: Within each day's partition, we order transactions by amount in descending order (ORDER BY amount DESC).

3. Ranking: The RANK() function assigns ranks to each transaction within its day partition based on the amount. Transactions with the highest amount receive rank 1; transactions with the same highest amount all get rank 1.

4. Filtering: We filter the results to keep only the transactions with rank 1 (WHERE rk = 1).

5. Ordering Output: Finally, we order the results by transaction_id in ascending order as requested.

MySQL Code:

WITH
    T AS (
        SELECT
            transaction_id,
            RANK() OVER (
                PARTITION BY DAY(day)
                ORDER BY amount DESC
            ) AS rk
        FROM Transactions
    )
SELECT transaction_id
FROM T
WHERE rk = 1
ORDER BY 1;

Time Complexity Analysis:

The time complexity is dominated by the window function operation. Window functions generally have a time complexity that is linear to the number of rows in the table (O(N)), where N is the number of rows in the Transactions table. The sorting within each partition is also typically done efficiently. The filtering and ordering of the final result are also linear in the number of rows after ranking (which will be less than or equal to N). Therefore, the overall time complexity is O(N).

Space Complexity Analysis:

The space complexity depends on the size of the intermediate result (the CTE T). In the worst case (all transactions have the same maximum amount on each day), the size of T would be the same as the input table. However, in most cases, it will be smaller. The space used for the final result is also linear in the number of transactions returned. Therefore, the space complexity is O(N) in the worst case, but could be better in practice.

This solution elegantly handles the requirement to return all transactions with the maximum amount on a given day using standard SQL functionalities and achieves optimal time and space complexity. The use of window functions is a key element to a concise and efficient solution.