Solution Explanation for LeetCode Problem 1571: Warehouse Manager

This problem requires calculating the total cubic feet of volume occupied by inventory in each warehouse. We achieve this by joining data from two tables: Warehouse and Products, performing calculations, and grouping the results.

Approach:

1. Join the Tables: We begin by joining the Warehouse and Products tables using an INNER JOIN. The join condition is Warehouse.product_id = Products.product_id. This links each warehouse entry with the corresponding product dimensions. USING (product_id) is a concise syntax for this join condition when the column name is identical in both tables.

2. Calculate Volume per Product: For each product in a warehouse, we calculate its volume: width * length * height * units. This takes into account the product's dimensions and the number of units present in that warehouse.

3. Group and Aggregate: We group the results using GROUP BY warehouse_name (or GROUP BY 1 as a shorthand to refer to the first column in the SELECT statement which is name aliased as warehouse_name). This ensures that we calculate a separate total volume for each warehouse. The SUM() function aggregates the individual product volumes for each warehouse.

4. Rename Columns (Optional): Finally, the query renames the name column to warehouse_name for clarity in the output and selects SUM(...) as volume.

Time Complexity Analysis:

The time complexity is dominated by the join operation and the group by aggregation. In general, a join operation on two tables with m and n rows can have a time complexity of O(m*n) in the worst case (nested loop join). However, database systems use optimized algorithms (like hash joins or merge joins) that often perform significantly better in practice, often achieving closer to O(m + n) complexity. The group by operation typically involves sorting or hashing which is O(k log k) where k is the number of rows after the join. Therefore the overall complexity is dominated by the Join O(m+n) and Group By O(k log k) operations where m is the number of rows in Warehouse, n is the number of rows in Products, and k is the number of rows in the result of the join. The exact complexity depends heavily on the database system's query optimizer and the indexing of the tables.

Space Complexity Analysis:

The space complexity is primarily determined by the intermediate result sets generated during the join and group by operations. In the worst case, it can be proportional to the number of rows in the joined table (O(m+n)). Again, database systems often optimize space usage, but this is an upper bound.

MySQL Code:

SELECT
    name AS warehouse_name,
    SUM(width * length * height * units) AS volume
FROM
    Warehouse
    JOIN Products USING (product_id)
GROUP BY 1;

Other Database Systems (Adaptations):

The core SQL logic remains the same for other database systems (PostgreSQL, SQL Server, Oracle, etc.). Minor syntax variations might be necessary, particularly in how you specify aliases or use the GROUP BY clause (e.g., using column index instead of column name). The fundamental operations of joining, calculating, and grouping are standard SQL features.