LeetCode #607 — EASY

Sales Person

Build confidence with an intuition-first walkthrough focused on core interview patterns fundamentals.

The Problem

Problem Statement

Table: SalesPerson

+-----------------+---------+
| Column Name     | Type    |
+-----------------+---------+
| sales_id        | int     |
| name            | varchar |
| salary          | int     |
| commission_rate | int     |
| hire_date       | date    |
+-----------------+---------+
sales_id is the primary key (column with unique values) for this table.
Each row of this table indicates the name and the ID of a salesperson alongside their salary, commission rate, and hire date.

Table: Company

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| com_id      | int     |
| name        | varchar |
| city        | varchar |
+-------------+---------+
com_id is the primary key (column with unique values) for this table.
Each row of this table indicates the name and the ID of a company and the city in which the company is located.

Table: Orders

+-------------+------+
| Column Name | Type |
+-------------+------+
| order_id    | int  |
| order_date  | date |
| com_id      | int  |
| sales_id    | int  |
| amount      | int  |
+-------------+------+
order_id is the primary key (column with unique values) for this table.
com_id is a foreign key (reference column) to com_id from the Company table.
sales_id is a foreign key (reference column) to sales_id from the SalesPerson table.
Each row of this table contains information about one order. This includes the ID of the company, the ID of the salesperson, the date of the order, and the amount paid.

Write a solution to find the names of all the salespersons who did not have any orders related to the company with the name "RED".

Return the result table in any order.

The result format is in the following example.

Example 1:

Input: 
SalesPerson table:
+----------+------+--------+-----------------+------------+
| sales_id | name | salary | commission_rate | hire_date  |
+----------+------+--------+-----------------+------------+
| 1        | John | 100000 | 6               | 4/1/2006   |
| 2        | Amy  | 12000  | 5               | 5/1/2010   |
| 3        | Mark | 65000  | 12              | 12/25/2008 |
| 4        | Pam  | 25000  | 25              | 1/1/2005   |
| 5        | Alex | 5000   | 10              | 2/3/2007   |
+----------+------+--------+-----------------+------------+
Company table:
+--------+--------+----------+
| com_id | name   | city     |
+--------+--------+----------+
| 1      | RED    | Boston   |
| 2      | ORANGE | New York |
| 3      | YELLOW | Boston   |
| 4      | GREEN  | Austin   |
+--------+--------+----------+
Orders table:
+----------+------------+--------+----------+--------+
| order_id | order_date | com_id | sales_id | amount |
+----------+------------+--------+----------+--------+
| 1        | 1/1/2014   | 3      | 4        | 10000  |
| 2        | 2/1/2014   | 4      | 5        | 5000   |
| 3        | 3/1/2014   | 1      | 1        | 50000  |
| 4        | 4/1/2014   | 1      | 4        | 25000  |
+----------+------------+--------+----------+--------+
Output: 
+------+
| name |
+------+
| Amy  |
| Mark |
| Alex |
+------+
Explanation: 
According to orders 3 and 4 in the Orders table, it is easy to tell that only salesperson John and Pam have sales to company RED, so we report all the other names in the table salesperson.

Step 01

Brute Force Baseline

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

{"headers": {"SalesPerson": ["sales_id", "name", "salary", "commission_rate","hire_date"], "Company": ["com_id", "name","city"],"Orders":["order_id","order_date","com_id","sales_id","amount"]}, "rows": {"SalesPerson": [[1, "John", 100000, 6, "4/1/2006"], [2, "Amy", 12000, 5,"5/1/2010"], [3, "Mark", 65000, 12, "12/25/2008"], [4, "Pam", 25000, 25,"1/1/2005"],[5,"Alex",5000,10,"2/3/2007"]], "Company": [[1, "RED","Boston"], [2, "ORANGE", "New York"],[3, "YELLOW", "Boston"],[4, "GREEN", "Austin"]],"Orders":[[1,"1/1/2014",3,4,10000],[2, "2/1/2014", 4, 5, 5000],[3, "3/1/2014", 1, 1, 50000],[4, "4/1/2014", 1, 4, 25000]]}}

Step 02

Core Insight

What unlocks the optimal approach

You need to query who sold to company 'RED' first, then output the sales person who is not in the first query result.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #607: Sales Person
// Auto-generated Java example from rust.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (rust):
// // Accepted solution for LeetCode #607: Sales Person
// pub fn sql_example() -> &'static str {
//     r#"
// -- Accepted solution for LeetCode #607: Sales Person
// # Write your MySQL query statement below
// SELECT s.name
// FROM
//     SalesPerson AS s
//     LEFT JOIN Orders USING (sales_id)
//     LEFT JOIN Company AS c USING (com_id)
// GROUP BY sales_id
// HAVING IFNULL(SUM(c.name = 'RED'), 0) = 0;
// "#
// }

// Accepted solution for LeetCode #607: Sales Person
// Auto-generated Go example from rust.
func exampleSolution() {
}
// Reference (rust):
// // Accepted solution for LeetCode #607: Sales Person
// pub fn sql_example() -> &'static str {
//     r#"
// -- Accepted solution for LeetCode #607: Sales Person
// # Write your MySQL query statement below
// SELECT s.name
// FROM
//     SalesPerson AS s
//     LEFT JOIN Orders USING (sales_id)
//     LEFT JOIN Company AS c USING (com_id)
// GROUP BY sales_id
// HAVING IFNULL(SUM(c.name = 'RED'), 0) = 0;
// "#
// }

# Accepted solution for LeetCode #607: Sales Person
# Auto-generated Python example from rust.
def example_solution() -> None:
    return
# Reference (rust):
# // Accepted solution for LeetCode #607: Sales Person
# pub fn sql_example() -> &'static str {
#     r#"
# -- Accepted solution for LeetCode #607: Sales Person
# # Write your MySQL query statement below
# SELECT s.name
# FROM
#     SalesPerson AS s
#     LEFT JOIN Orders USING (sales_id)
#     LEFT JOIN Company AS c USING (com_id)
# GROUP BY sales_id
# HAVING IFNULL(SUM(c.name = 'RED'), 0) = 0;
# "#
# }

// Accepted solution for LeetCode #607: Sales Person
pub fn sql_example() -> &'static str {
    r#"
-- Accepted solution for LeetCode #607: Sales Person
# Write your MySQL query statement below
SELECT s.name
FROM
    SalesPerson AS s
    LEFT JOIN Orders USING (sales_id)
    LEFT JOIN Company AS c USING (com_id)
GROUP BY sales_id
HAVING IFNULL(SUM(c.name = 'RED'), 0) = 0;
"#
}

// Accepted solution for LeetCode #607: Sales Person
// Auto-generated TypeScript example from rust.
function exampleSolution(): void {
}
// Reference (rust):
// // Accepted solution for LeetCode #607: Sales Person
// pub fn sql_example() -> &'static str {
//     r#"
// -- Accepted solution for LeetCode #607: Sales Person
// # Write your MySQL query statement below
// SELECT s.name
// FROM
//     SalesPerson AS s
//     LEFT JOIN Orders USING (sales_id)
//     LEFT JOIN Company AS c USING (com_id)
// GROUP BY sales_id
// HAVING IFNULL(SUM(c.name = 'RED'), 0) = 0;
// "#
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.