SQL Programming Resources

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SQL Interview Series — Part 13 📌 Question 13: Find Employees Earning More Than the Company Average Suppose you have an Employee table:

employee_id | employee_name | department | salary
------------|---------------|------------|-------
1           | Amit          | IT         | 50000
2           | Rahul         | IT         | 80000
3           | Priya         | HR         | 60000
4           | Neha          | HR         | 90000
5           | Raj           | Finance    | 70000

❓ Find all employees whose salary is greater than the average salary of the entire company. 💡 Approach First, calculate the average salary across all employees. Then compare every employee's salary with that average. Company average: (50000 + 80000 + 60000 + 90000 + 70000) / 5 = 70000 Employees earning more than 70000: Rahul → 80000 Neha → 90000 📌 SQL Solution

SELECT
    employee_id,
    employee_name,
    department,
    salary
FROM Employee
WHERE salary > (
    SELECT AVG(salary)
    FROM Employee
);

🔎 How does it work? The subquery:

SELECT AVG(salary)
FROM Employee;

calculates the company-wide average: 70000 The outer query then checks:

WHERE salary > 70000

and returns only employees earning above that amount. 📌 Expected Result:

employee_name | department | salary
--------------|------------|-------
Rahul         | IT         | 80000
Neha          | HR         | 90000

Double Tap ❤️ For More ----- 1.33 ₽ · /balance_help

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SQL Interview Series — Part 12 📌 Question 12: Find the Department with the Highest Average Salary Suppose you have an Employee table:

employee_id | employee_name | department | salary
------------|---------------|------------|-------
1           | Amit          | IT         | 60000
2           | Rahul         | IT         | 80000
3           | Priya         | HR         | 70000
4           | Neha          | HR         | 90000
5           | Raj           | Finance    | 85000
6           | Karan         | Finance    | 95000

❓ Find the department that has the highest average salary. 💡 Approach First, calculate the average salary for every department. Then, find the department with the highest average. Department averages: IT → (60000 + 80000) / 2 = 70000 HR → (70000 + 90000) / 2 = 80000 Finance → (85000 + 95000) / 2 = 90000 Therefore: Finance → 90000 📌 SQL Solution

SELECT
    department,
    AVG(salary) AS avg_salary
FROM Employee
GROUP BY department
ORDER BY avg_salary DESC
LIMIT 1;

Expected result:

department | avg_salary
-----------|-----------
Finance    | 90000

🔎 How does it work? 1️⃣ "GROUP BY department" Creates one group for each department. 2️⃣ "AVG(salary)" Calculates the average salary within each department. 3️⃣ "ORDER BY avg_salary DESC" Places the department with the highest average first. 4️⃣ "LIMIT 1" Returns only the top department. ⚠️ Interview Trap What if two departments have the same highest average salary? For example: IT → 80000 HR → 80000 Finance → 75000 "LIMIT 1" would return only one department. If the interviewer asks: 👉 "Find all departments having the highest average salary" then use a ranking approach:

SELECT department, avg_salary
FROM (
    SELECT
        department,
        AVG(salary) AS avg_salary,
        DENSE_RANK() OVER (
            ORDER BY AVG(salary) DESC
        ) AS rnk
    FROM Employee
    GROUP BY department
) d
WHERE rnk = 1;

Now both IT and HR would be returned. 🎯 Double Tap ❤️ For More ----- 1.3 ₽ · /balance_help

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