How do you solve Reformat Department Table on LeetCode?

Reformat Department Table (LeetCode #1179) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using conditional aggregation can simplify complex queries.

What is the brute force solution for Reformat Department Table?

The brute force approach for Reformat Department Table is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves creating a new table for each department and manually filling in the revenue for each month. This is straightforward but inefficient as it requires multiple scans of the data. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Reformat Department Table?

Reformat Department Table uses the following concepts: Database. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Reformat Department Table?

Practice writing SQL queries without an IDE to improve fluency. Understand the underlying data structure and relationships.

What are common mistakes when solving Reformat Department Table?

Not using GROUP BY correctly. Overlooking the need for NULL handling in results.

#1179

Reformat Department Table

Easy

DatabaseHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses aggregation with conditional statements to transform the data in a single pass. This reduces the need for multiple iterations over the data, making it much more efficient.

⚙️

Algorithm

3 steps

1Step 1: Use a SQL query to group the data by department id.
2Step 2: Use conditional aggregation to create revenue columns for each month.
3Step 3: Return the result in the desired format.

solution.py15 lines

1SELECT id, 
2       MAX(CASE WHEN month = 'Jan' THEN revenue END) AS Jan_Revenue, 
3       MAX(CASE WHEN month = 'Feb' THEN revenue END) AS Feb_Revenue, 
4       MAX(CASE WHEN month = 'Mar' THEN revenue END) AS Mar_Revenue, 
5       MAX(CASE WHEN month = 'Apr' THEN revenue END) AS Apr_Revenue, 
6       MAX(CASE WHEN month = 'May' THEN revenue END) AS May_Revenue, 
7       MAX(CASE WHEN month = 'Jun' THEN revenue END) AS Jun_Revenue, 
8       MAX(CASE WHEN month = 'Jul' THEN revenue END) AS Jul_Revenue, 
9       MAX(CASE WHEN month = 'Aug' THEN revenue END) AS Aug_Revenue, 
10       MAX(CASE WHEN month = 'Sep' THEN revenue END) AS Sep_Revenue, 
11       MAX(CASE WHEN month = 'Oct' THEN revenue END) AS Oct_Revenue, 
12       MAX(CASE WHEN month = 'Nov' THEN revenue END) AS Nov_Revenue, 
13       MAX(CASE WHEN month = 'Dec' THEN revenue END) AS Dec_Revenue 
14FROM Department 
15GROUP BY id;

ℹ

Complexity note: The time complexity is linear because we only need to scan through the data once to aggregate it, while the space complexity is linear due to the storage of results.

1Using conditional aggregation can simplify complex queries.
2Understanding how to group data efficiently is crucial.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.