assd

LeetCode SQL 50 Solution/176. Second heighest salary/176. Second Highest Salary.py

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+import pandas as pd
+
+def second_highest_salary(employee: pd.DataFrame) -> pd.DataFrame:
+    unique_salaries = employee['salary'].drop_duplicates().sort_values(ascending=False)
+    second_highest = unique_salaries.iloc[1] if len(unique_salaries) > 1 else None
+    return pd.DataFrame({'SecondHighestSalary': [second_highest]})

LeetCode SQL 50 Solution/176. Second heighest salary/176. Second heighest salary.sql

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-
+# Solution 1
 # Write your MSSQL query statement below
 select max(salary) as SecondHighestSalary from employee
 where salary not in (select max(salary) from employee)
+
+
+
+# Solution 2
+# Write your MSSQL query statement below
+# Write your MySQL query statement below
+SELECT
+    (
+        SELECT DISTINCT salary
+        FROM Employee
+        ORDER BY salary DESC
+        LIMIT 1, 1
+    ) AS SecondHighestSalary;

LeetCode SQL 50 Solution/176. Second heighest salary/readme.md

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+# 🏆 LeetCode 176: Second Highest Salary (Pandas) 🚀  
+
+## **Problem Statement**  
+You are given an **Employee** table with the following schema:  
+
+```
++-------------+------+
+| Column Name | Type |
++-------------+------+
+| id          | int  |
+| salary      | int  |
++-------------+------+
+```
+- `id` is the **primary key** (unique for each employee).  
+- Each row contains information about an employee's salary.  
+
+Your task is to **find the second highest distinct salary**.  
+- If there is no second highest salary, return `None`.  
+
+---
+
+## **Example 1**  
+
+### **Input:**  
+```plaintext
++----+--------+
+| id | salary |
++----+--------+
+| 1  | 100    |
+| 2  | 200    |
+| 3  | 300    |
++----+--------+
+```
+### **Output:**  
+```plaintext
++---------------------+
+| SecondHighestSalary |
++---------------------+
+| 200                 |
++---------------------+
+```
+
+---
+
+## **Example 2**  
+
+### **Input:**  
+```plaintext
++----+--------+
+| id | salary |
++----+--------+
+| 1  | 100    |
++----+--------+
+```
+### **Output:**  
+```plaintext
++---------------------+
+| SecondHighestSalary |
++---------------------+
+| null                |
++---------------------+
+```
+
+---
+
+## **Approach**  
+
+1. **Remove duplicate salaries** using `.drop_duplicates()`.  
+2. **Sort salaries** in descending order.  
+3. **Retrieve the second highest salary**, if it exists.  
+4. If there is no second highest salary, return `None`.  
+
+---
+
+## **Code (Pandas Solution)**  
+
+```python
+import pandas as pd
+
+def second_highest_salary(employee: pd.DataFrame) -> pd.DataFrame:
+    # Get distinct salaries, sorted in descending order
+    unique_salaries = employee['salary'].drop_duplicates().sort_values(ascending=False)
+
+    # Check if there is a second highest salary
+    second_highest = unique_salaries.iloc[1] if len(unique_salaries) > 1 else None
+
+    # Return as a DataFrame with column name 'SecondHighestSalary'
+    return pd.DataFrame({'SecondHighestSalary': [second_highest]})
+```
+
+---
+
+## **Complexity Analysis**  
+- **Time Complexity:** $$O(n \log n)$$ due to sorting.  
+- **Space Complexity:** $$O(n)$$ for storing unique salaries.  
+
+---
+
+## **Why This Works?**  
+✅ Uses Pandas' `drop_duplicates()` for distinct values.  
+✅ Sorts efficiently using `sort_values()`.  
+✅ Handles cases where no second highest salary exists.  
+
+---
+
+## **SQL Equivalent Solution**  
+If solving in SQL, we can use **`DENSE_RANK()`**:  
+```sql
+WITH RankedEmployees AS (
+    SELECT salary, DENSE_RANK() OVER(ORDER BY salary DESC) AS `rank`
+    FROM Employee
+)
+SELECT MAX(salary) AS SecondHighestSalary
+FROM RankedEmployees
+WHERE `rank` = 2;
+```
+
+---
+
+### **Happy Coding! 🚀💡**  
+Hope this helps! Feel free to ⭐ **star** this repo if you found it useful. 😊