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Merge pull request #93 from iamAntimPal/Branch-1
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  • LeetCode SQL 50 Solution

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# 🏢 Primary Department for Each Employee - LeetCode 1789
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## 📌 Problem Statement
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You are given a table **Employee** that contains the following columns:
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- **employee_id**: The ID of the employee.
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- **department_id**: The ID of the department to which the employee belongs.
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- **primary_flag**: An ENUM ('Y', 'N').
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- If `primary_flag` is `'Y'`, then the department is the primary department for that employee.
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- If `primary_flag` is `'N'`, then the department is not primary.
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**Note:**
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- An employee can belong to multiple departments. When an employee joins multiple departments, they decide which one is their primary (set to `'Y'`).
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- If an employee belongs to only one department, then their `primary_flag` is `'N'`, but that department is still considered their primary department.
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Your task is to **report all employees with their primary department**.
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For employees who belong to only one department, report that department.
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Return the result table in **any order**.
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---
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## 📊 Table Structure
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### **Employee Table**
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| Column Name | Type |
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| ------------- | ------- |
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| employee_id | int |
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| department_id | int |
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| primary_flag | varchar |
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- `(employee_id, department_id)` is the **primary key** for this table.
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---
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## 📊 Example 1:
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### **Input:**
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#### **Employee Table**
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| employee_id | department_id | primary_flag |
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| ----------- | ------------- | ------------ |
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| 1 | 1 | N |
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| 2 | 1 | Y |
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| 2 | 2 | N |
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| 3 | 3 | N |
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| 4 | 2 | N |
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| 4 | 3 | Y |
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| 4 | 4 | N |
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### **Output:**
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| employee_id | department_id |
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| ----------- | ------------- |
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| 1 | 1 |
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| 2 | 1 |
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| 3 | 3 |
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| 4 | 3 |
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### **Explanation:**
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- **Employee 1** belongs to only one department (1), so department 1 is their primary.
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- **Employee 2** belongs to departments 1 and 2. The row with `primary_flag = 'Y'` indicates that department 1 is their primary.
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- **Employee 3** belongs only to department 3.
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- **Employee 4** belongs to departments 2, 3, and 4. The row with `primary_flag = 'Y'` indicates that department 3 is their primary.
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---
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## 🖥 SQL Solution
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### ✅ **Approach:**
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- **Step 1:** For employees who have `primary_flag = 'Y'`, choose those rows.
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- **Step 2:** For employees who belong to only one department, return that row.
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- Combine the results using `UNION DISTINCT`.
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```sql
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SELECT employee_id, department_id
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FROM Employee
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WHERE primary_flag = 'Y'
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UNION DISTINCT
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SELECT employee_id, department_id
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FROM Employee
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GROUP BY employee_id
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HAVING COUNT(*) = 1;
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```
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---
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## 🐍 Python (Pandas) Solution
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### ✅ **Approach:**
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1. **Group** the DataFrame by `employee_id`.
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2. For each group:
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- If any row has `primary_flag == 'Y'`, choose the first such row.
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- Otherwise (i.e., employee belongs to only one department), choose that row.
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3. Return the resulting DataFrame with only `employee_id` and `department_id`.
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```python
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import pandas as pd
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def primary_department(employees: pd.DataFrame) -> pd.DataFrame:
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def select_primary(group):
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# If there's any row with primary_flag 'Y', choose the first one
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if (group['primary_flag'] == 'Y').any():
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return group[group['primary_flag'] == 'Y'].iloc[0]
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else:
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# For employees with only one department
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return group.iloc[0]
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result = employees.groupby('employee_id').apply(select_primary).reset_index(drop=True)
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return result[['employee_id', 'department_id']]
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```
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---
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## 📁 File Structure
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```
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📂 Primary-Department
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│── README.md
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│── solution.sql
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│── solution_pandas.py
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│── test_cases.sql
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│── sample_data.csv
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```
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---
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## 🔗 Useful Links
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- 📖 [LeetCode Problem](https://leetcode.com/problems/primary-department-for-each-employee/)
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- 🔍 [MySQL UNION Operator](https://www.w3schools.com/sql/sql_union.asp)
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- 🐍 [Pandas GroupBy Documentation](https://pandas.pydata.org/docs/reference/api/pandas.DataFrame.groupby.html)
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# 💰 Count Salary Categories - LeetCode 907
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## 📌 Problem Statement
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You are given a table **Accounts** that contains information about bank accounts, including their monthly income.
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Your task is to calculate the number of bank accounts in each salary category.
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The salary categories are defined as follows:
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- **"Low Salary"**: Salaries strictly less than \$20,000.
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- **"Average Salary"**: Salaries in the inclusive range [\$20,000, \$50,000].
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- **"High Salary"**: Salaries strictly greater than \$50,000.
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The result table must contain **all three categories**. If there are no accounts in a category, return 0.
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Return the result in **any order**.
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---
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## 📊 Table Structure
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### **Accounts Table**
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| Column Name | Type |
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| ----------- | ---- |
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| account_id | int |
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| income | int |
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- `account_id` is the **primary key** for this table.
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- Each row contains the monthly income for one bank account.
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---
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## 📊 Example 1:
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### **Input:**
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#### **Accounts Table**
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| account_id | income |
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| ---------- | ------ |
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| 3 | 108939 |
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| 2 | 12747 |
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| 8 | 87709 |
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| 6 | 91796 |
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### **Output:**
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| category | accounts_count |
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| -------------- | -------------- |
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| Low Salary | 1 |
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| Average Salary | 0 |
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| High Salary | 3 |
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### **Explanation:**
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- **Low Salary**: Account with income 12747.
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- **Average Salary**: No accounts have an income in the range [20000, 50000].
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- **High Salary**: Accounts with incomes 108939, 87709, and 91796.
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---
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## 🖥 SQL Solution
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### ✅ **Approach:**
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1. **CTE "S"**: Create a static table with the three salary categories.
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```sql
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WITH S AS (
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SELECT 'Low Salary' AS category
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UNION
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SELECT 'Average Salary'
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UNION
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SELECT 'High Salary'
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),
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```
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- This defines the three salary categories to ensure every category appears in the final result.
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2. **CTE "T"**: Categorize each account from the **Accounts** table using a `CASE` statement and count the number of accounts in each category.
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```sql
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T AS (
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SELECT
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CASE
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WHEN income < 20000 THEN 'Low Salary'
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WHEN income > 50000 THEN 'High Salary'
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ELSE 'Average Salary'
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END AS category,
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COUNT(1) AS accounts_count
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FROM Accounts
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GROUP BY 1
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)
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```
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- The `CASE` statement assigns a salary category based on the income.
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- `COUNT(1)` counts the number of accounts in each category.
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3. **Final SELECT with LEFT JOIN**: Combine the static category table `S` with the computed counts from `T` to ensure every category is included, using `IFNULL` to convert any missing count to 0.
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```sql
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SELECT S.category, IFNULL(T.accounts_count, 0) AS accounts_count
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FROM S
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LEFT JOIN T USING (category);
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```
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### ✅ **Complete SQL Query:**
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```sql
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WITH S AS (
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SELECT 'Low Salary' AS category
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UNION
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SELECT 'Average Salary'
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UNION
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SELECT 'High Salary'
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),
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T AS (
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SELECT
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CASE
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WHEN income < 20000 THEN 'Low Salary'
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WHEN income > 50000 THEN 'High Salary'
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ELSE 'Average Salary'
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END AS category,
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COUNT(1) AS accounts_count
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FROM Accounts
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GROUP BY 1
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)
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SELECT S.category, IFNULL(T.accounts_count, 0) AS accounts_count
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FROM S
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LEFT JOIN T USING (category);
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```
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---
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## 🐍 Python (Pandas) Solution
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### ✅ **Approach:**
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1. **Categorize Accounts**: Create a new column `category` in the DataFrame by applying the salary conditions.
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2. **Group and Count**: Group by the `category` column and count the number of accounts.
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3. **Merge with Static Categories**: Ensure all three salary categories appear by merging with a predefined DataFrame that contains all categories, filling missing counts with 0.
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```python
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import pandas as pd
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def count_salary_categories(accounts: pd.DataFrame) -> pd.DataFrame:
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# Define the salary categorization function
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def categorize(income):
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if income < 20000:
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return 'Low Salary'
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elif income > 50000:
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return 'High Salary'
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else:
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return 'Average Salary'
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# Apply categorization
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accounts['category'] = accounts['income'].apply(categorize)
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# Count accounts in each category
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counts = accounts.groupby('category').size().reset_index(name='accounts_count')
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# Define static categories DataFrame
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categories = pd.DataFrame({
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'category': ['Low Salary', 'Average Salary', 'High Salary']
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})
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# Merge to ensure all categories are present, fill missing values with 0
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result = categories.merge(counts, on='category', how='left')
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result['accounts_count'] = result['accounts_count'].fillna(0).astype(int)
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return result
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# Example usage:
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# df = pd.read_csv("sample_accounts.csv")
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# print(count_salary_categories(df))
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```
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---
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## 📁 File Structure
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```
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📂 Count-Salary-Categories
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│── README.md
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│── solution.sql
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│── solution_pandas.py
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│── test_cases.sql
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│── sample_accounts.csv
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```
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---
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## 🔗 Useful Links
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- 📖 [LeetCode Problem](https://leetcode.com/problems/count-salary-categories/)
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- 📝 [MySQL WITH Clause (CTE)](https://www.w3schools.com/sql/sql_with.asp)
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- 🔍 [MySQL IFNULL Function](https://www.w3schools.com/sql/func_mysql_ifnull.asp)
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- 🐍 [Pandas GroupBy Documentation](https://pandas.pydata.org/docs/reference/api/pandas.DataFrame.groupby.html)
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```
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