Problem

Table: Employees

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| employee_id | int     |
| name        | varchar |
| salary      | int     |
+-------------+---------+
employee_id is the column with unique values for this table.
Each row of this table indicates the employee ID, employee name, and salary.

A company wants to divide the employees into teams such that all the members on each team have the same salary. The teams should follow these criteria:

  • Each team should consist of at least two employees.
  • All the employees on a team should have the same salary.
  • All the employees of the same salary should be assigned to the same team.
  • If the salary of an employee is unique, we do not assign this employee to any team.
  • A team’s ID is assigned based on the rank of the team ’s salary relative to the other teams’ salaries, where the team with the lowest salary has team_id = 1. Note that the salaries for employees not on a team are not included in this ranking.

Write a solution to get the team_id of each employee that is in a team.

Return the result table ordered by team_id in ascending order. In case of a tie, order it by employee_id in ascending order.

The result format is in the following example.

Examples

Example 1:

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Input: 
Employees table:
+-------------+---------+--------+
| employee_id | name    | salary |
+-------------+---------+--------+
| 2           | Meir    | 3000   |
| 3           | Michael | 3000   |
| 7           | Addilyn | 7400   |
| 8           | Juan    | 6100   |
| 9           | Kannon  | 7400   |
+-------------+---------+--------+
Output: 
+-------------+---------+--------+---------+
| employee_id | name    | salary | team_id |
+-------------+---------+--------+---------+
| 2           | Meir    | 3000   | 1       |
| 3           | Michael | 3000   | 1       |
| 7           | Addilyn | 7400   | 2       |
| 9           | Kannon  | 7400   | 2       |
+-------------+---------+--------+---------+
Explanation: 
Meir (employee_id=2) and Michael (employee_id=3) are in the same team because they have the same salary of 3000.
Addilyn (employee_id=7) and Kannon (employee_id=9) are in the same team because they have the same salary of 7400.
Juan (employee_id=8) is not included in any team because their salary of 6100 is unique (i.e. no other employee has the same salary).
The team IDs are assigned as follows (based on salary ranking, lowest first):
- team_id=1: Meir and Michael, a salary of 3000
- team_id=2: Addilyn and Kannon, a salary of 7400
Juan's salary of 6100 is not included in the ranking because they are not on a team.

Solution

Method 1 – Group By and String Aggregation

Intuition

We need to group employees by salary and return the list of employee IDs for each salary group, sorted in ascending order. SQL’s GROUP BY and string aggregation functions can be used for this.

Approach

  1. Group the Employees table by salary.
  2. For each group, collect all employee IDs, sort them in ascending order, and concatenate them into a comma-separated string.
  3. Return the salary and the corresponding employee_id list as a string.

Code

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SELECT salary, GROUP_CONCAT(employee_id ORDER BY employee_id ASC) AS employee_id
FROM Employees
GROUP BY salary;
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SELECT salary, STRING_AGG(employee_id::text, ',' ORDER BY employee_id) AS employee_id
FROM Employees
GROUP BY salary;
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class Solution:
    def group_employees_by_salary(self, employees):
        # employees: DataFrame with columns ['employee_id', 'name', 'salary']
        df = employees.groupby('salary')['employee_id'].apply(lambda x: ','.join(map(str, sorted(x)))).reset_index()
        return df[['salary', 'employee_id']]

Complexity

  • ⏰ Time complexity: O(n log n) — Sorting employee IDs within each salary group.
  • 🧺 Space complexity: O(n) — For storing grouped and sorted employee IDs.