Find Customer Referee — LeetCode #584 (Easy)
Tags: Database
Related patterns: Hash Map, Array
Brute Force approach
Time complexity: O(n²). Space complexity: O(1).
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 complexity is O(n²) because for each customer, we may need to check all others to see if they refer them, leading to a nested loop scenario.
- Step 1: Initialize an empty result set to store names of qualifying customers.
- Step 2: For each customer, check if their referee_id is NULL or if it is not equal to 2.
- Step 3: If either condition is met, add the customer's name to the result set.
- Step 4: Return the result set.
1. Start with Customer table: [1: Will, 2: Jane, 3: Alex, 4: Bill, 5: Zack, 6: Mark]
2. Check customer 1: referee_id is NULL -> add 'Will'
3. Check customer 2: referee_id is NULL -> add 'Jane'
4. Check customer 3: referee_id is 2 -> skip
5. Check customer 4: referee_id is NULL -> add 'Bill'
6. Check customer 5: referee_id is 1 -> add 'Zack'
Optimal Solution approach
Time complexity: O(n). Space complexity: 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.
The complexity is O(n) because we only scan through the Customer table once to gather the required data.
- Step 1: Select names from the Customer table where referee_id is NULL.
- Step 2: Union this result with the names of customers whose referee_id is not 2.
- Step 3: Return the combined results.
1. Start with Customer table: [1: Will, 2: Jane, 3: Alex, 4: Bill, 5: Zack, 6: Mark]
2. Check for NULL referee_id: 'Will', 'Jane', 'Bill' -> add to result
3. Check for referee_id not equal to 2: 'Will', 'Jane', 'Bill', 'Zack' -> add to result
4. Combine results: 'Will', 'Jane', 'Bill', 'Zack'
Key Insights
- Understanding NULL values is crucial when working with databases.
- Using UNION can simplify queries by combining results from multiple conditions.
Common Mistakes
- Overlooking NULL values when filtering results.
- Not considering the implications of using UNION vs. JOIN.
Interview Tips
- 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.
