How do you solve Rearrange Products Table on LeetCode?

Rearrange Products Table (LeetCode #1795) 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 UNION ALL in SQL can simplify data transformation tasks.

What is the brute force solution for Rearrange Products Table?

The brute force approach for Rearrange Products Table is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking each store's price for every product and manually constructing the output. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Rearrange Products Table?

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

What are the interview tips for Rearrange Products Table?

Understand how to manipulate data structures effectively. Practice writing SQL queries and understand their performance implications. Be prepared to explain your thought process clearly.

What are common mistakes when solving Rearrange Products Table?

Not filtering out null values properly. Overcomplicating the logic instead of using built-in functions.

#1795

Rearrange Products Table

Easy

DatabaseHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

The optimal approach leverages SQL's ability to handle data transformation efficiently using the UNION ALL operator, which allows us to combine results from multiple SELECT statements into one result set.

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Algorithm

2 steps

1Step 1: Use a SELECT statement to get product_id, store1, and its price, filtering out nulls.
2Step 2: Repeat for store2 and store3, using UNION ALL to combine results.

solution.py2 lines

1def rearrange_products(products):
2    return products.melt(id_vars=['product_id'], value_vars=['store1', 'store2', 'store3'], var_name='store', value_name='price').dropna()

ℹ

Complexity note: The time complexity is O(n) because we are processing each product once. The space complexity is O(n) due to the storage of results in a new list.

1Using UNION ALL in SQL can simplify data transformation tasks.
2Data can be reshaped efficiently using built-in functions in programming languages.

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