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SQL
🔹 Basic SQL
Table 1
Question
Concepts
1. Get the second highest salary from the Employee table.
There are no rows in this table
sql
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SELECT MAX(salary)
FROM Employee
WHERE salary < (SELECT MAX(salary) FROM Employee);
| Subquery, MAX |
| 2. Find all duplicate emails in the Users table.
sql
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SELECT email, COUNT(*)
FROM Users
GROUP BY email
HAVING COUNT(*) > 1;
| GROUP BY, HAVING |
| 3. Retrieve employees with names starting with ‘A’.
sql
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SELECT * FROM Employee WHERE name LIKE 'A%';
| LIKE |
🔸 Joins & Relationships
Table 2
Question
Concepts
4. List all customers and the orders they made (including customers who didn’t make any orders).
There are no rows in this table
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SELECT c.name, o.order_id
FROM Customers c
LEFT JOIN Orders o ON c.id = o.customer_id;
| LEFT JOIN |
| 5. Find customers who didn’t place any orders.
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SELECT c.name
FROM Customers c
LEFT JOIN Orders o ON c.id = o.customer_id
WHERE o.order_id IS NULL;
| Anti-Join |
🔹 Aggregations & GROUP BY
Table 3
Question
Concepts
6. Get department-wise average salary.
There are no rows in this table
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SELECT department, AVG(salary)
FROM Employee
GROUP BY department;
| AVG, GROUP BY |
| 7. Count number of employees hired each year.
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SELECT YEAR(hire_date) AS year, COUNT(*)
FROM Employee
GROUP BY YEAR(hire_date);
| DATE functions |
🔸 Window Functions
Table 4
Question
Concepts
8. Find the top 3 highest salaries per department.
There are no rows in this table
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SELECT * FROM (
SELECT *,
DENSE_RANK() OVER (PARTITION BY department ORDER BY salary DESC) as rank
FROM Employee
) t WHERE rank <= 3;
| DENSE_RANK, PARTITION BY |
| 9. Running total of sales by date.
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SELECT order_date,
SUM(sale_amount) OVER (ORDER BY order_date) AS running_total
FROM Sales;
| SUM, OVER |
🔹 CTEs & Subqueries
Table 5
Question
Concepts
10. With clause to get departments with avg salary > 50K.
There are no rows in this table
sql
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WITH dept_avg AS (
SELECT department, AVG(salary) AS avg_sal
FROM Employee
GROUP BY department
)
SELECT * FROM dept_avg WHERE avg_sal > 50000;
| CTE, WITH |
| 11. Find employees who earn more than their manager.
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SELECT e.name
FROM Employee e
JOIN Employee m ON e.manager_id = m.id
WHERE e.salary > m.salary;
| Self-join |
🧠 Bonus: SQL Interview Scenarios
Table 6
Scenario
What You Practice
Design a leaderboard system
Window functions, ranks
Recommend friends (2nd degree connections)
Self-joins
Analyze churn rate over months
Joins, CASE, date manipulation
Product orders by category
GROUP BY, COUNT, JOINs
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