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SQL Interview Series — Part 8📌 Question 8: Find Customers Who Have Never Placed an OrderSuppose you have two tables:Customerscustomer_id | customer_name------------|--------------1 | Amit2 | Rahul3 | Priya4 | Neha5 | RajOrdersorder_id | customer_id---------|------------101 | 1102 | 3103 | 1104 | 5❓ Find all customers who have never placed an order.💡 ApproachWe need to find customers from the "Customers" table who do not have a matching record in the "Orders" table.A "LEFT JOIN" is a common and interview-friendly approach.📌 SQL SolutionSELECT c.customer_id, c.customer_nameFROM Customers cLEFT JOIN Orders o ON c.customer_id = o.customer_idWHERE o.customer_id IS NULL;🔎 How does it work?"LEFT JOIN" keeps every customer, even if they don't have an order.For customers without an order, the columns from "Orders" become "NULL".Then:WHERE o.customer_id IS NULLkeeps only those customers.📌 Expected Result:customer_id | customer_name------------|--------------2 | Rahul4 | Neha⚠️ Interview TrapYou can also solve this using "NOT EXISTS":SELECT c.customer_id, c.customer_nameFROM Customers cWHERE NOT EXISTS ( SELECT 1 FROM Orders o WHERE o.customer_id = c.customer_id);"NOT EXISTS" is often preferred when you specifically want to test whether a matching row exists.Be careful with "NOT IN" when NULL values can exist in the subquery, because NULL can lead to unexpected results.🎯 Double Tap ❤️ For More

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SQL Interview Series — Part 6📌 Question 6: Find the Top 2 Highest-Paid Employees in Each DepartmentSuppose you have:Employeeemployee_id employee_name department salary1 Amit IT 600002 Rahul IT 900003 Priya IT 800004 Neha HR 700005 Raj HR 500006 Karan HR 90000❓ Find the top 2 highest-paid employees from each department.💡 ApproachThis is a classic window-function interview question.We need to:1️⃣ Divide employees by department.2️⃣ Rank employees by salary within each department.3️⃣ Keep only employees with rank ≤ 2.SQL SolutionSELECT employee_id, employee_name, department, salaryFROM ( SELECT employee_id, employee_name, department, salary, ROW_NUMBER() OVER ( PARTITION BY department ORDER BY salary DESC ) AS rn FROM Employee) eWHERE rn <= 2;🔎 How does it work?PARTITION BY departmentCreates a separate ranking for each department.ORDER BY salary DESCPlaces the highest salary first.ROW_NUMBER()Assigns 1, 2, 3... within each department.Finally:WHERE rn <= 2keeps the top two employees from each department.Expected Resultemployee_name department salaryRahul IT 90000Priya IT 80000Karan HR 90000Neha HR 70000⚠️ Interview TrapWhat if two employees have the same salary?For example:Rahul 90000Priya 90000Amit 80000ROW_NUMBER() will still assign unique numbers:Rahul 1Priya 2Amit 3Double Tap ❤️ For More

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SQL Interview Series — Part 5📌 Question 5: Find Employees Who Earn More Than Their ManagerSuppose you have an Employee table:Employee• employee_id• employee_name• manager_id• salaryExample:employee_id | employee_name | manager_id | salary------------|---------------|------------|-------1 | Amit | NULL | 1000002 | Rahul | 1 | 1200003 | Priya | 1 | 900004 | Neha | 2 | 1300005 | Raj | 2 | 110000❓ Find employees whose salary is greater than their manager's salary.━━━━━━━━━━━━━━━━━━💡 ApproachThe Employee table contains both employees and managers.So, we need to join the table with itself.This is called a Self Join.📌 SQL SolutionSELECT e.employee_name AS employee, e.salary AS employee_salary, m.employee_name AS manager, m.salary AS manager_salaryFROM Employee eJOIN Employee m ON e.manager_id = m.employee_idWHERE e.salary > m.salary;━━━━━━━━━━━━━━━━━━🔎 How does it work?We use the Employee table twice:"e" → represents the employee"m" → represents the managerThe join condition:e.manager_id = m.employee_idconnects each employee with their manager.Then:WHERE e.salary > m.salarykeeps only employees whose salary is higher than their manager's.Expected result:employee | employee_salary | manager | manager_salary---------|-----------------|---------|---------------Rahul | 120000 | Amit | 100000Neha | 130000 | Rahul | 120000━━━━━━━━━━━━━━━━━━🎯 Interview TipWhenever the same table contains a relationship such as:• Employee → Manager• Employee → Mentor• Employee → Supervisor• Employee → Parentthink about using a Self Join.A very common interview pattern is:FROM Employee eJOIN Employee m ON e.manager_id = m.employee_idThe key is to understand what each table alias represents before writing the query.Double Tap ❤️ For More

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SQL Interview Series — Part 4📌 Question 4: Find the Highest Salary in Each DepartmentSuppose you have an Employee table:Employee• employee_id• employee_name• department• salaryExample:employee_id | employee_name | department | salary------------|---------------|------------|-------1 | Amit | IT | 600002 | Rahul | IT | 900003 | Priya | HR | 500004 | Neha | HR | 700005 | Raj | Finance | 850006 | Karan | Finance | 95000❓ Find the employee(s) with the highest salary in each department.💡 ApproachWe need to find the maximum salary separately for every department.A common approach is to use a window function."RANK()" assigns a ranking within each department.SELECT employee_id, employee_name, department, salaryFROM ( SELECT employee_id, employee_name, department, salary, RANK() OVER ( PARTITION BY department ORDER BY salary DESC ) AS rnk FROM Employee) eWHERE rnk = 1;🔎 How does it work?"PARTITION BY department"→ Creates a separate ranking for each department."ORDER BY salary DESC"→ Places the highest salary first."RANK()"→ Gives the highest salary a rank of 1.Finally:WHERE rnk = 1→ Keeps only the highest-paid employee(s).Expected result:employee_name | department | salary--------------|------------|-------Rahul | IT | 90000Neha | HR | 70000Karan | Finance | 95000⚠️ Why use "RANK()"?If two employees have the same highest salary, both will receive rank 1 and both will be returned.For example:ITRahul 90000Amit 90000Raj 80000Both Rahul and Amit will be returned.Double Tap ❤️ For More

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SQL Interview Series — Part 2📌 Question 2: Find Duplicate RecordsSuppose you have an Employee table:Employee• employee_id• employee_name• department• email❓ Find all email addresses that appear more than once in the Employee table.Example:employee_id | employee_name | email------------|---------------|-------------------1 | Amit | amit@gmail.com2 | Rahul | rahul@gmail.com3 | Priya | amit@gmail.com4 | Neha | neha@gmail.com5 | Raj | rahul@gmail.comExpected result:emailamit@gmail.comrahul@gmail.com💡 ApproachWe need to:1️⃣ Group records by email.2️⃣ Count how many times each email appears.3️⃣ Keep only the emails whose count is greater than 1. 📌 SQL SolutionSELECT email, COUNT(*) AS occurrence_countFROM EmployeeGROUP BY emailHAVING COUNT(*) > 1;🔎 Why use HAVING instead of WHERE?"WHERE" filters individual rows before grouping."HAVING" filters groups after "GROUP BY".Since we want to filter based on "COUNT(*)", we use "HAVING".GROUP BY emailHAVING COUNT(*) > 1This means:"Group employees by email and return only those groups containing more than one record."🎯 Double Tap ❤️ For Part-3

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SQL Interview Series — Part 1Hi guys, let's start a SQL interview series covering frequently asked and important SQL interview questions.Each part will cover 1 practical interview question with:✅ Problem statement✅ SQL solution✅ Approach✅ Interview tip📌 Question 1: Find the Second Highest SalarySuppose you have an Employee table:Employee• employee_id• employee_name• salaryExample data:employee_id| employee_name| salary1| Amit| 500002| Rahul| 800003| Priya| 700004| Neha| 900005| Raj| 80000❓ Find the second-highest salary from the Employee table.💡 ApproachFirst, we need to identify the highest salary.Then, we need the highest salary that is less than the maximum salary.One simple approach is to use a subquery:SELECT MAX(salary) AS second_highest_salaryFROM EmployeeWHERE salary < ( SELECT MAX(salary) FROM Employee);Output:second_highest_salary80000🔎 Why does this work?The inner query:SELECT MAX(salary)FROM Employee;returns:90000Then the outer query considers only salaries below 90000:50000700008000080000Finally, MAX() returns:80000⚠️ Important Interview PointIf the question asks for the second-highest DISTINCT salary, this approach works because duplicate salaries are naturally treated as one value.For example:90000800008000070000The second-highest distinct salary is still 80000.💯 Double Tap ❤️ For Part-2

🧠 Real-World SQL Scenario-Based Questions & Answers1. Get the 2nd highest salary from the Employees tableSELECT MAX(salary) AS SecondHighest FROM Employees WHERE salary < (SELECT MAX(salary) FROM Employees);2. Find employees without assigned managersSELECT * FROM Employees WHERE manager_id IS NULL;3. Retrieve departments with more than 5 employeesSELECT department_id, COUNT(*) AS employee_count FROM Employees GROUP BY department_id HAVING COUNT(*) > 5;4. List customers who made no ordersSELECT c.name FROM Customers c LEFT JOIN Orders o ON c.id = o.customer_id WHERE o.id IS NULL;5. Find the top 3 highest-paid employeesSELECT * FROM Employees ORDER BY salary DESC LIMIT 3;6. Display total sales for each productSELECT product, SUM(amount) AS total_sales FROM Sales GROUP BY product;7. Get employee names starting with 'A' and ending with 'n'SELECT name FROM Employees WHERE name LIKE 'A%n';8. Show employees who joined in the last 30 daysSELECT * FROM Employees WHERE join_date >= CURRENT_DATE - INTERVAL 30 DAY;💬 Tap ❤️ for more!

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will behave identically across every database.Always check the behavior of your specific database.21️⃣ Transaction Isolation LevelsIsolation is one of the deeper transaction concepts.Common isolation levels include:• READ UNCOMMITTED• READ COMMITTED• REPEATABLE READ• SERIALIZABLESome databases also support additional modes or implement these differently.The general idea is:• More isolation ↓ Stronger guarantees between concurrent transactions ↓ Potentially more locking/contention or reduced concurrencyThe exact behavior is database-specific.22️⃣ READ UNCOMMITTEDThis is the weakest commonly described isolation level.A transaction may potentially see changes that another transaction has not committed.This can lead to phenomena such as:• Dirty readsIt is not appropriate for every workload.23️⃣ READ COMMITTEDA transaction generally sees committed data rather than another transaction's uncommitted changes.This is a common default isolation level in some database systems.However, behavior across databases can differ.24️⃣ REPEATABLE READThe goal is to ensure that repeated reads within a transaction provide a stable view of previously read data under the database's isolation model.It provides stronger guarantees than READ COMMITTED.The exact implementation differs between database systems.25️⃣ SERIALIZABLEThis provides the strongest standard isolation level among these four.The goal is to make concurrent transactions behave as if they were executed serially.Conceptually: Transaction A ↓ Transaction B rather than allowing certain conflicting operations to interact concurrently.The trade-off can be reduced concurrency or increased contention.26️⃣ Common Transaction ProblemsWhen transactions run concurrently, several phenomena can occur depending on the isolation level and database.Dirty Read• Transaction A reads data changed by Transaction B before B commits.• B → UPDATE ↓ A → reads uncommitted value ↓ B → ROLLBACK• A saw a value that never became permanent.Non-Repeatable Read• A transaction reads the same row twice and gets different values because another transaction committed a change between the reads.• A → Read = 100• B → Update = 200, B → COMMIT• A → Read again = 200Phantom Read• A transaction repeats a query and sees additional or missing rows because another transaction inserted or deleted matching rows.• A → SELECT ... WHERE amount > 1000 → 5 rows• B → INSERT another matching row, B → COMMIT• A → same query → 6 rowsThe exact handling depends on the database and isolation level.27️⃣ Transaction vs QueryDon't confuse these concepts.A query is a SQL statement such as:SELECT * FROM customers;A transaction is a logical unit containing one or more operations.For example:Transaction │ ├── INSERT ├── UPDATE ├── UPDATE └── COMMITSo:• Query → Individual SQL operation• Transaction → Unit of work containing one or more operations28️⃣ Real-World Analytics ExampleSuppose an operations team needs to correct payment statuses.There are 10,000 affected records.Instead of blindly running:UPDATE paymentsSET status = 'Completed'WHERE payment_date IS NOT NULL;first inspect:SELECT payment_id, status, payment_dateFROM paymentsWHERE payment_date IS NOT NULL;Then, if the correction is confirmed:BEGIN;UPDATE paymentsSET status = 'Completed'WHERE payment_date IS NOT NULL;SELECT COUNT(*) AS updated_rowsFROM paymentsWHERE payment_date IS NOT NULL AND status = 'Completed';COMMIT;If the validation reveals an unexpected result:ROLLBACK;

The exact workflow should follow your organization's production-change and approval procedures.29️⃣ COMMIT vs SAVEPOINT vs ROLLBACKRemember:• COMMIT → Save the transaction• ROLLBACK → Undo uncommitted transaction changes• SAVEPOINT → Mark a point inside a transaction• ROLLBACK TO SAVEPOINT → Undo changes after that pointExample:BEGIN;UPDATE customersSET status = 'Active'WHERE customer_id = 101;SAVEPOINT s1;UPDATE customersSET status = 'Inactive'WHERE customer_id = 102;ROLLBACK TO SAVEPOINT s1;COMMIT;The first update can remain while the second update is rolled back, subject to the database's transaction semantics.30️⃣ Common Mistakes❌ Mistake 1: Forgetting COMMIT — You may make changes but not persist them as intended.❌ Mistake 2: Assuming ROLLBACK always works — If the changes have already been committed, a normal rollback cannot undo them.❌ Mistake 3: Running UPDATE without checking the WHERE condition — Dangerous:UPDATE customers SET status = 'Inactive';Safer workflow:SELECT * FROM customers WHERE ...;UPDATE customers SET status = 'Inactive' WHERE ...;❌ Mistake 4: Assuming transaction behavior is identical everywhere — Database systems differ in areas such as: - Autocommit - DDL transactions - Isolation - Locking - Savepoints - Error handling🎯 Interview Questions• Q1. What is a transaction? A transaction is a logical unit of one or more database operations that are handled together.• Q2. What does COMMIT do? It commits the transaction's changes.• Q3. What does ROLLBACK do? It reverses uncommitted changes in the transaction.• Q4. What is SAVEPOINT? A savepoint marks a point inside a transaction to which you can potentially roll back without undoing the entire transaction.• Q5. What does ACID stand for? A → Atomicity, C → Consistency, I → Isolation, D → Durability• Q6. What is atomicity? It treats the transaction as a logical unit so that its operations are committed together or rolled back as appropriate.• Q7. What is isolation? It controls how concurrent transactions interact and what changes they can see.• Q8. What is durability? Committed changes are intended to survive failures according to the database's durability mechanisms.• Q9. What is autocommit? A mode where individual statements may be committed automatically.• Q10. What is a dirty read? Reading data changed by another transaction before that transaction commits.🧠 Practice QuestionsPractice 1 Write a transaction that updates a customer's status and commits it.BEGIN;UPDATE customers SET status = 'Active' WHERE customer_id = 101;COMMIT;Practice 2 Update a customer's status but roll back the change.BEGIN;UPDATE customers SET status = 'Inactive' WHERE customer_id = 101;ROLLBACK;Practice 3 Create a savepoint after the first update.BEGIN;UPDATE customers SET status = 'Active' WHERE customer_id = 101;SAVEPOINT customer_update;UPDATE customers SET status = 'Inactive' WHERE customer_id = 102;ROLLBACK TO SAVEPOINT customer_update;COMMIT;Practice 4 Before executing this UPDATE:UPDATE orders SET status = 'Cancelled' WHERE order_date < DATE '2025-01-01';write a SELECT that lets you inspect the affected records.SELECT * FROM orders WHERE order_date < DATE '2025-01-01';Practice 5 Explain what happens here:BEGIN;UPDATE accounts SET balance = balance - 5000 WHERE account_id = 101;UPDATE accounts SET balance = balance + 5000 WHERE account_id = 202;ROLLBACK;

Answer: Both uncommitted updates are rolled back, assuming the statements executed within the same transaction and the database/session supports the shown transaction behavior.🔥 Mini ChallengeImagine an order-processing system. You need to:1. Create an order.2. Add an order item.3. Reduce inventory.4. Commit everything if successful.5. Roll back if a critical operation fails.Write a transaction structure for this workflow.Solution:BEGIN;INSERT INTO orders (order_id, customer_id, order_amount)VALUES (1001, 101, 5000);INSERT INTO order_items (order_id, product_id, quantity)VALUES (1001, 501, 2);UPDATE productsSET stock_quantity = stock_quantity - 2WHERE product_id = 501 AND stock_quantity >= 2;COMMIT;In a real application, you would also validate that each critical operation succeeded and handle errors according to the database/application's transaction mechanism.If a critical operation fails:ROLLBACK;🎯 Key TakeawayRemember:• TRANSACTION → Group related operations into one unit• COMMIT → Save changes• ROLLBACK → Undo uncommitted changes• SAVEPOINT → Create a rollback point• ACID → Atomicity → Consistency → Isolation → DurabilityThe most important practical lesson:«Before making large UPDATE or DELETE changes, first run the corresponding SELECT and verify exactly which rows will be affected.»Transactions help protect data, but safe SQL also depends on careful query design, validation, permissions, and understanding your database's transaction behavior.Double Tap ❤️ For More