This is a common SQL (Joins) interview and practice question. Below is the full answer with a clear, step-by-step explanation.

Question

Problem You are given an employee table where each employee reports to a manager. Find all pairs of employees who report to the same manager . Each pair should appear only once. Employees employee_id employee_name manager_id 101 Alice 1 102 Bob 1 103 Charlie 1 104 David 2 105 Emma 2 106 Frank 3 Task Return every unique pair of employees who report to the same manager. Do not return duplicate pairs such as (Bob, Alice) if (Alice, Bob) already exists. Expected Output manager_id employee_1 employee_2 1 Alice Bob 1 Alice Charlie 1 Bob Charlie 2 David Emma

Answer & Explanation

**What the query has to do**

You need a result that lists every pair of employees who share the same manager, and you must list each pair only once.
The output columns are

* manager_id – the manager they both report to
* employee_1 – the name of the first employee in the pair
* employee_2 – the name of the second employee in the pair

**Why a self‑join is the right tool**

All the data lives in a single table (Employees). To compare two rows from the same table you join the table to itself – this is called a *self‑join*.
When you join on `manager_id` you guarantee that the two rows you are looking at belong to employees with the same manager.

**How to avoid duplicate and self‑pairs**

If you simply join on `manager_id` you will get three kinds of rows:

1. (Alice, Alice) – the same employee paired with herself
2. (Alice, Bob) and (Bob, Alice) – the same logical pair appearing twice, just in opposite order
3. The real pairs you want

You can eliminate the first two cases with a single additional condition in the `ON` clause:

```
e1.employee_id < e2.employee_id
```

Because employee_id is unique, this condition does three things:

* It never allows the same row to match itself (`e1.employee_id` can never be less than itself).
* For any two different employees it forces the smaller id to appear in `e1` and the larger id in `e2`.
* Consequently only one ordering of the pair survives, so (Alice, Bob) appears and (Bob, Alice) is filtered out.

**Putting it together – the complete query**

```sql
SELECT
e1.manager_id,
e1.employee_name AS employee_1,
e2.employee_name AS employee_2
FROM
Employees e1
JOIN
Employees e2
ON e1.manager_id = e2.manager_id -- same manager
AND e1.employee_id < e2.employee_id -- keep one direction only
ORDER BY
e1.manager_id,
employee_1,
employee_2;
```

**Why this query returns exactly the expected rows**

* The join condition `e1.manager_id = e2.manager_id` restricts the result set to rows where both employees report to the same manager.
* The inequality `e1.employee_id < e2.employee_id` removes self‑pairs and the reverse duplicate of each pair.
* The `SELECT` list projects the manager id and the two employee names, matching the required column names.
* The `ORDER BY` clause is optional for correctness but makes the output deterministic and easier to read.

Running the query on the sample data produces:

which is exactly the expected output.

**Alternative approaches (optional)**

If you prefer not to rely on the numeric ordering of employee_id, you could use the employee name for the same purpose:

```sql
... AND e1.employee_name < e2.employee_name
```

Or you could generate pairs with a window function (`ROW_NUMBER`) and then join on the row numbers, but the simple self‑join with the `<` filter is the most straightforward and works on any SQL dialect that supports basic joins.

**Key take‑aways for the interview**

1. Recognize that comparing rows within the same table calls for a self‑join.
2. Use the natural relationship (`manager_id`) as the join key.
3. Add a “less‑than” condition on a unique column to eliminate self‑pairs and duplicate orderings.
4. Project the columns you need and optionally order the result for readability.

Mastering this pattern will let you solve many “pairwise” problems in SQL quickly and correctly.