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| 1 | +Below is a well-structured `README.md` for **LeetCode 1789 - Primary Department for Each Employee**, including both the SQL and Pandas (Python) solutions: |
| 2 | + |
| 3 | +```md |
| 4 | +# 🏢 Primary Department for Each Employee - LeetCode 1789 |
| 5 | + |
| 6 | +## 📌 Problem Statement |
| 7 | +You are given a table **Employee** that contains the following columns: |
| 8 | + |
| 9 | +- **employee_id**: The ID of the employee. |
| 10 | +- **department_id**: The ID of the department to which the employee belongs. |
| 11 | +- **primary_flag**: An ENUM ('Y', 'N'). |
| 12 | + - If `primary_flag` is `'Y'`, then the department is the primary department for that employee. |
| 13 | + - If `primary_flag` is `'N'`, then the department is not primary. |
| 14 | + |
| 15 | +**Note:** |
| 16 | +- An employee can belong to multiple departments. When an employee joins multiple departments, they decide which one is their primary (set to `'Y'`). |
| 17 | +- If an employee belongs to only one department, then their `primary_flag` is `'N'`, but that department is still considered their primary department. |
| 18 | + |
| 19 | +Your task is to **report all employees with their primary department**. |
| 20 | +For employees who belong to only one department, report that department. |
| 21 | + |
| 22 | +Return the result table in **any order**. |
| 23 | + |
| 24 | +--- |
| 25 | + |
| 26 | +## 📊 Table Structure |
| 27 | + |
| 28 | +### **Employee Table** |
| 29 | +| Column Name | Type | |
| 30 | +| ------------- | ------- | |
| 31 | +| employee_id | int | |
| 32 | +| department_id | int | |
| 33 | +| primary_flag | varchar | |
| 34 | + |
| 35 | +- `(employee_id, department_id)` is the **primary key** for this table. |
| 36 | + |
| 37 | +--- |
| 38 | + |
| 39 | +## 📊 Example 1: |
| 40 | + |
| 41 | +### **Input:** |
| 42 | +#### **Employee Table** |
| 43 | +| employee_id | department_id | primary_flag | |
| 44 | +| ----------- | ------------- | ------------ | |
| 45 | +| 1 | 1 | N | |
| 46 | +| 2 | 1 | Y | |
| 47 | +| 2 | 2 | N | |
| 48 | +| 3 | 3 | N | |
| 49 | +| 4 | 2 | N | |
| 50 | +| 4 | 3 | Y | |
| 51 | +| 4 | 4 | N | |
| 52 | + |
| 53 | +### **Output:** |
| 54 | +| employee_id | department_id | |
| 55 | +| ----------- | ------------- | |
| 56 | +| 1 | 1 | |
| 57 | +| 2 | 1 | |
| 58 | +| 3 | 3 | |
| 59 | +| 4 | 3 | |
| 60 | + |
| 61 | +### **Explanation:** |
| 62 | +- **Employee 1** belongs to only one department (1), so department 1 is their primary. |
| 63 | +- **Employee 2** belongs to departments 1 and 2. The row with `primary_flag = 'Y'` indicates that department 1 is their primary. |
| 64 | +- **Employee 3** belongs only to department 3. |
| 65 | +- **Employee 4** belongs to departments 2, 3, and 4. The row with `primary_flag = 'Y'` indicates that department 3 is their primary. |
| 66 | + |
| 67 | +--- |
| 68 | + |
| 69 | +## 🖥 SQL Solution |
| 70 | + |
| 71 | +### ✅ **Approach:** |
| 72 | +- **Step 1:** For employees who have `primary_flag = 'Y'`, choose those rows. |
| 73 | +- **Step 2:** For employees who belong to only one department, return that row. |
| 74 | +- Combine the results using `UNION DISTINCT`. |
| 75 | + |
| 76 | +```sql |
| 77 | +SELECT employee_id, department_id |
| 78 | +FROM Employee |
| 79 | +WHERE primary_flag = 'Y' |
| 80 | +UNION DISTINCT |
| 81 | +SELECT employee_id, department_id |
| 82 | +FROM Employee |
| 83 | +GROUP BY employee_id |
| 84 | +HAVING COUNT(*) = 1; |
| 85 | +``` |
| 86 | + |
| 87 | +--- |
| 88 | + |
| 89 | +## 🐍 Python (Pandas) Solution |
| 90 | + |
| 91 | +### ✅ **Approach:** |
| 92 | +1. **Group** the DataFrame by `employee_id`. |
| 93 | +2. For each group: |
| 94 | + - If any row has `primary_flag == 'Y'`, choose the first such row. |
| 95 | + - Otherwise (i.e., employee belongs to only one department), choose that row. |
| 96 | +3. Return the resulting DataFrame with only `employee_id` and `department_id`. |
| 97 | + |
| 98 | +```python |
| 99 | +import pandas as pd |
| 100 | + |
| 101 | +def primary_department(employees: pd.DataFrame) -> pd.DataFrame: |
| 102 | + def select_primary(group): |
| 103 | + # If there's any row with primary_flag 'Y', choose the first one |
| 104 | + if (group['primary_flag'] == 'Y').any(): |
| 105 | + return group[group['primary_flag'] == 'Y'].iloc[0] |
| 106 | + else: |
| 107 | + # For employees with only one department |
| 108 | + return group.iloc[0] |
| 109 | + |
| 110 | + result = employees.groupby('employee_id').apply(select_primary).reset_index(drop=True) |
| 111 | + return result[['employee_id', 'department_id']] |
| 112 | +``` |
| 113 | + |
| 114 | +--- |
| 115 | + |
| 116 | +## 📁 File Structure |
| 117 | +``` |
| 118 | +📂 Primary-Department |
| 119 | +│── README.md |
| 120 | +│── solution.sql |
| 121 | +│── solution_pandas.py |
| 122 | +│── test_cases.sql |
| 123 | +│── sample_data.csv |
| 124 | +``` |
| 125 | + |
| 126 | +--- |
| 127 | + |
| 128 | +## 🔗 Useful Links |
| 129 | +- 📖 [LeetCode Problem](https://leetcode.com/problems/primary-department-for-each-employee/) |
| 130 | +- 🔍 [MySQL UNION Operator](https://www.w3schools.com/sql/sql_union.asp) |
| 131 | +- 🐍 [Pandas GroupBy Documentation](https://pandas.pydata.org/docs/reference/api/pandas.DataFrame.groupby.html) |
| 132 | +``` |
| 133 | + |
| 134 | +This `README.md` includes: |
| 135 | +- A **clear problem statement** with table structure and example. |
| 136 | +- A detailed **SQL solution** with an explanation. |
| 137 | +- A detailed **Pandas (Python) solution** with an explanation. |
| 138 | +- An organized **file structure** and helpful external links. |
| 139 | + |
| 140 | +Let me know if you need any modifications! |
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