# LeetCode 1076: Project Employees II

## Problem Restatement

Using the `Project(project_id, employee_id)` table, report the project(s) with the most employees.

## Edge Cases

- Match the aggregate to the question: use raw `COUNT` or `SUM` for rows/amounts, and `COUNT(DISTINCT ...)` only for unique entities.
- Tie cases should return every qualifying row unless the statement asks for a single row.
- For outer joins, keep filters on the joined table in the `ON` clause when unmatched rows must remain visible.

## Common Pitfalls

- Do not put aggregate filters in `WHERE`; use `HAVING` after `GROUP BY`.
- When the answer needs all tied winners, avoid `LIMIT 1` unless the statement explicitly asks for one row.

## Implementation

```sql
SELECT project_id
FROM Project
GROUP BY project_id
HAVING COUNT(employee_id) = (
    SELECT MAX(cnt)
    FROM (
        SELECT COUNT(employee_id) AS cnt
        FROM Project
        GROUP BY project_id
    ) AS sub
);
```

## Code Explanation

The inner subquery counts employees per project.

The outer `MAX(cnt)` finds the maximum employee count.

The main query returns all projects whose employee count equals the maximum.

This handles ties (multiple projects with the same maximum count).