How do you solve Seasonal Sales Analysis on LeetCode?

Seasonal Sales Analysis (LeetCode #3564) 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: Understanding seasonality can help in inventory management.

What is the time complexity of Seasonal Sales Analysis?

The optimal solution for Seasonal Sales Analysis runs in O(n) time and uses O(n) space. Single pass through sales data allows for linear complexity.

What is the brute force solution for Seasonal Sales Analysis?

The brute force approach for Seasonal Sales Analysis is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check each sale for every season, counting the quantities sold per category. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Seasonal Sales Analysis?

Seasonal Sales Analysis uses the following concepts: Database. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Seasonal Sales Analysis?

Clarify the problem requirements. Discuss your thought process aloud. Optimize your solution after the initial approach.

What are common mistakes when solving Seasonal Sales Analysis?

Overlooking the season mapping. Not grouping correctly by category.

#3564

Seasonal Sales Analysis

Medium

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)

By categorizing sales into seasons and aggregating quantities in a single pass, we achieve efficiency.

⚙️

Algorithm

3 steps

1Step 1: Create a mapping of seasons to their respective months.
2Step 2: Iterate through sales, summing quantities for each category in the corresponding season.
3Step 3: For each season, determine the category with the maximum quantity sold.

solution.py9 lines

1# Full working Python code
2SELECT season, category
3FROM (SELECT CASE WHEN MONTH(sale_date) IN (12, 1, 2) THEN 'Winter' WHEN MONTH(sale_date) IN (3, 4, 5) THEN 'Spring' WHEN MONTH(sale_date) IN (6, 7, 8) THEN 'Summer' ELSE 'Fall' END AS season,
4            p.category,
5            SUM(s.quantity) AS total_quantity
6      FROM sales s JOIN products p ON s.product_id = p.product_id
7      GROUP BY season, p.category) AS seasonal_sales
8GROUP BY season, category
9HAVING total_quantity = (SELECT MAX(total_quantity) FROM seasonal_sales WHERE season = seasonal_sales.season);

ℹ

Complexity note: Single pass through sales data allows for linear complexity.

1Understanding seasonality can help in inventory management.
2Aggregating data efficiently is crucial for performance.

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