Solution Explanation for LeetCode 1082: Sales Analysis I

This problem requires finding the seller(s) with the highest total sales price. We'll explore two SQL solutions, analyzing their efficiency.

Solution 1: Using HAVING and ALL

This solution uses a subquery to find the maximum total sales price for any seller and then filters the main query to select sellers whose total sales match that maximum.

MySQL Code:

SELECT seller_id
FROM Sales
GROUP BY seller_id
HAVING SUM(price) >= ALL (
    SELECT SUM(price)
    FROM Sales
    GROUP BY seller_id
);

Explanation:

1. SELECT seller_id FROM Sales GROUP BY seller_id: This groups the Sales table by seller_id, preparing to aggregate sales for each seller.

2. HAVING SUM(price) >= ALL (...): This filters the grouped results. SUM(price) calculates the total sales for each seller. >= ALL (...) checks if the total sales are greater than or equal to every other seller's total sales. This elegantly handles ties.

3. SELECT SUM(price) FROM Sales GROUP BY seller_id: This subquery calculates the total sales for each seller, providing the comparison values for the ALL operator.

Time Complexity Analysis:

• The subquery has a time complexity of O(N log N) due to the GROUP BY and sorting implied by ALL. N is the number of rows in the Sales table.
• The main query also has a time complexity of O(N log N) for GROUP BY and the HAVING clause comparison.

Therefore, the overall time complexity is O(N log N).

Solution 2: Using RANK() Window Function

This solution utilizes a window function (RANK()) to assign a rank to each seller based on their total sales. This is generally more efficient than the ALL approach.

MySQL Code:

WITH T AS (
    SELECT
        seller_id,
        SUM(price) AS tot,
        RANK() OVER (ORDER BY SUM(price) DESC) AS rk
    FROM Sales
    GROUP BY seller_id
)
SELECT seller_id
FROM T
WHERE rk = 1;

Explanation:

1. WITH T AS (...): This creates a Common Table Expression (CTE) called T.

2. SELECT seller_id, SUM(price) AS tot, RANK() OVER (ORDER BY SUM(price) DESC) AS rk FROM Sales GROUP BY seller_id: This part does the following:

   • Groups the sales by seller_id.
   • Calculates the sum of price for each seller (tot).
   • Assigns a rank (rk) using the RANK() window function. ORDER BY SUM(price) DESC ranks sellers in descending order of their total sales. Sellers with the same total sales receive the same rank.

3. SELECT seller_id FROM T WHERE rk = 1: This selects the seller_id from the CTE T where the rank is 1 (the highest-ranked sellers).

Time Complexity Analysis:

• The GROUP BY operation within the CTE has a time complexity of O(N log N).
• The RANK() window function usually has a time complexity of O(N log N) although the exact complexity can depend on the database system's optimization techniques. It might be linear in some implementations, but O(N log N) is a safe upper bound.

Therefore, the overall time complexity is O(N log N). However, in practice, Solution 2 tends to be faster than Solution 1 because window functions are often optimized efficiently by database systems.

Conclusion:

Both solutions correctly solve the problem. Solution 2 using the RANK() window function is generally preferred due to its potential for better performance, especially on larger datasets. The asymptotic time complexity is the same for both but the practical performance can differ.