How do you solve Sales Person on LeetCode?

Sales Person (LeetCode #607) 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 sets can significantly reduce the time complexity for membership checks.

What is the brute force solution for Sales Person?

The brute force approach for Sales Person is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we will check each salesperson to see if they have any orders. If they don't, we will include them in our result. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Sales Person?

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

What are the interview tips for Sales Person?

Always clarify the requirements and edge cases before jumping into coding. Think about the efficiency of your solution — can it be improved? Practice writing SQL queries in a structured way, breaking them down into smaller parts.

What are common mistakes when solving Sales Person?

Forgetting to use DISTINCT when retrieving sales_ids, leading to incorrect results. Not considering edge cases, such as when there are no salespersons or no orders.

#607

Sales Person

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)

We can optimize by using a set to track salespersons who have orders, allowing us to quickly check which salespersons did not make any sales.

⚙️

Algorithm

3 steps

1Step 1: Retrieve all sales_ids from the Orders table and store them in a set.
2Step 2: Retrieve all salespersons from the SalesPerson table.
3Step 3: For each salesperson, check if their sales_id is in the set. If not, add their name to the result.

solution.py3 lines

1# Full working Python code
2WITH SalesWithOrders AS (SELECT DISTINCT sales_id FROM Orders)
3SELECT name FROM SalesPerson WHERE sales_id NOT IN (SELECT sales_id FROM SalesWithOrders);

ℹ

Complexity note: This complexity is due to the single pass to create the set of sales_ids and another pass to check against it, making it linear.

1Using sets can significantly reduce the time complexity for membership checks.
2Understanding how to leverage SQL's capabilities, like WITH clauses, can simplify complex queries.

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