How do you solve Find Customer Referee on LeetCode?

Find Customer Referee (LeetCode #584) 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 NULL values is crucial when working with databases.

What is the time complexity of Find Customer Referee?

The optimal solution for Find Customer Referee runs in O(n) time and uses O(n) space. The complexity is O(n) because we only scan through the Customer table once to gather the required data.

What is the brute force solution for Find Customer Referee?

The brute force approach for Find Customer Referee is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute-force approach, we will check each customer and see if they are either referred by someone who is not customer 2 or not referred by anyone at all. This method is straightforward but inefficient as it involves multiple scans of the data. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find Customer Referee?

Find Customer Referee uses the following concepts: Database. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Find Customer Referee?

Always clarify the requirements and edge cases, especially with NULLs. Think about how to optimize your solution from the start. Practice writing SQL queries to become familiar with different clauses.

What are common mistakes when solving Find Customer Referee?

Overlooking NULL values when filtering results. Not considering the implications of using UNION vs. JOIN.

#584

Find Customer Referee

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 solution uses a single scan of the Customer table to gather the necessary information. We can use a simple SQL query to filter customers based on their referee_id efficiently.

⚙️

Algorithm

3 steps

1Step 1: Select names from the Customer table where referee_id is NULL.
2Step 2: Union this result with the names of customers whose referee_id is not 2.
3Step 3: Return the combined results.

solution.py2 lines

1# Full working Python code
2SELECT name FROM Customer WHERE referee_id IS NULL UNION SELECT name FROM Customer WHERE referee_id != 2;

ℹ

Complexity note: The complexity is O(n) because we only scan through the Customer table once to gather the required data.

1Understanding NULL values is crucial when working with databases.
2Using UNION can simplify queries by combining results from multiple conditions.

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