Solution Explanation for LeetCode 2372: Calculate the Influence of Each Salesperson

This problem requires calculating the total sales for each salesperson. The solution involves joining three tables: Salesperson, Customer, and Sales.

Approach:

We employ a LEFT JOIN strategy to ensure that all salespeople are included in the result, even those without associated sales. The process unfolds as follows:

1. LEFT JOIN Salesperson and Customer: This join combines the Salesperson table with the Customer table using salesperson_id as the common key. A left join ensures that all rows from the Salesperson table are included, regardless of whether a match exists in the Customer table. If a salesperson has no customers, the corresponding customer columns will be NULL.

2. LEFT JOIN Customer and Sales: This join links the combined result with the Sales table using customer_id. Again, a left join retains all rows from the previous join (Salesperson and Customer). If a customer has no sales, price will be NULL.

3. GROUP BY and SUM: We group the results by salesperson_id to aggregate sales for each salesperson. The SUM(price) function calculates the total sales for each group. IFNULL(SUM(price),0) handles cases where a salesperson has no sales, replacing NULL with 0.

4. Result: The final query returns the salesperson_id, name, and the calculated total sales for each salesperson.

MySQL Code:

SELECT sp.salesperson_id, name, IFNULL(SUM(price), 0) AS total
FROM
    Salesperson AS sp
    LEFT JOIN Customer AS c ON sp.salesperson_id = c.salesperson_id
    LEFT JOIN Sales AS s ON s.customer_id = c.customer_id
GROUP BY 1;

Time Complexity Analysis:

The time complexity is dominated by the JOIN operations. In the worst case, the number of operations is proportional to the product of the sizes of the tables involved. Assuming n, m, and k are the number of rows in Salesperson, Customer, and Sales tables respectively, the time complexity is approximately O(nm + mk). The GROUP BY and SUM operations have a complexity less than the join operation, thus not significantly affecting the overall time complexity. The IFNULL operation adds negligible overhead.

Space Complexity Analysis:

The space complexity is determined by the size of the intermediate and final result sets. The intermediate result after the joins can be as large as the product of the number of rows in the three tables in the worst case (nmk). The space for the final result set is proportional to the number of salespeople (n), as each salesperson occupies a single row in the output. Hence, the space complexity is O(nmk) for the intermediate results and O(n) for the final result. However, in practice, database management systems optimize these operations, so the actual space used is usually much less than the theoretical worst-case complexity.