𝗜𝗻𝘁𝗲𝗿𝘃𝗶𝗲𝘄𝗲𝗿:
You have 2 minutes to solve this SQL query.

Find the employees who joined in the last 30 days.

Assume the table structure:
employees(employee_id, employee_name, department, joining_date)

𝗠𝗲: Challenge accepted! 💪

SELECT
employee_id,
employee_name,
department,
joining_date
FROM employees
WHERE joining_date >= CURRENT_DATE - INTERVAL '30 days';

💡 Explanation:
This query filters employees whose joining date falls within the last 30 days.

• CURRENT_DATE returns today's date.
• INTERVAL '30 days' subtracts 30 days from the current date.
• The WHERE clause returns employees who joined on or after that date.

This question tests your understanding of:
✅ Date Functions
✅ Date Arithmetic
✅ Filtering Records with Dates

🎯 Expected Output Example
Employee Department Joining Date
John IT 2026-06-10
Sarah HR 2026-06-22

🚀 Database-Specific Alternatives

MySQL
SELECT *
FROM employees
WHERE joining_date >= CURDATE() - INTERVAL 30 DAY;

SQL Server
SELECT *
FROM employees
WHERE joining_date >= DATEADD(DAY, -30, GETDATE());

Oracle
SELECT *
FROM employees
WHERE joining_date >= SYSDATE - 30;

🚀 Tip for SQL Job Seekers:
Date-related SQL questions are very common. Make sure you're comfortable with:
• Finding records from the last N days
• Current month/year filters
• Date differences
• Date formatting
• Database-specific date functions

❤️ React with ❤️ for more SQL interview challenges!

𝗜𝗻𝘁𝗲𝗿𝘃𝗶𝗲𝘄𝗲𝗿:
You have 2 minutes to solve this SQL query.

Find the customer(s) who placed the highest number of orders.

Tables:
customers(customer_id, customer_name)
orders(order_id, customer_id, order_date)

𝗠𝗲: Challenge accepted! 💪

SELECT
    customer_id,
    customer_name,
    total_orders
FROM (
    SELECT
        c.customer_id,
        c.customer_name,
        COUNT(o.order_id) AS total_orders,
        DENSE_RANK() OVER (
            ORDER BY COUNT(o.order_id) DESC
        ) AS rnk
    FROM customers c
    JOIN orders o
        ON c.customer_id = o.customer_id
    GROUP BY
        c.customer_id,
        c.customer_name
) ranked
WHERE rnk = 1;

💡 Explanation:
The query first counts the total number of orders placed by each customer and then ranks them based on the order count.

✅ COUNT(o.order_id) calculates the number of orders per customer.

✅ GROUP BY ensures one row per customer.

✅ DENSE_RANK() ranks customers from highest to lowest order count.

✅ The outer query returns all customers with rnk = 1, including ties.

This question tests your understanding of:

✅ JOIN

✅ GROUP BY

✅ Aggregate Functions COUNT

✅ Window Functions DENSE_RANK

🎯 Expected Output Example

+----------+--------------+
| Customer | Total Orders |
+----------+--------------+
| John     | 25           |
| Sarah    | 25           |
+----------+--------------+

Both customers are returned because they are tied for the highest number of orders.

🚀 Tip for SQL Job Seekers:
Whenever an interview asks for the highest, lowest, most, or least, think about whether multiple records could tie for first place. Using DENSE_RANK() instead of LIMIT 1 makes your solution more robust and interview-ready.

❤️ React with ❤️ for more SQL interview challenges!

𝗜𝗻𝘁𝗲𝗿𝘃𝗶𝗲𝘄𝗲𝗿:
You have 2 minutes to solve this SQL query.
Find all employees who have the same manager.

Assume the table structure:
employees(employee_id, employee_name, manager_id)

𝗠𝗲: Challenge accepted! 💪

SELECT
e.employee_id,
e.employee_name,
e.manager_id,
m.employee_name AS manager_name
FROM employees e
JOIN employees m
ON e.manager_id = m.employee_id
WHERE e.manager_id IN (
SELECT manager_id
FROM employees
WHERE manager_id IS NOT NULL
GROUP BY manager_id
HAVING COUNT(*) > 1
)
ORDER BY e.manager_id, e.employee_name;

💡 Explanation:
This query identifies managers who supervise more than one employee and returns all employees reporting to those managers.

✅ The subquery groups records by manager_id.

✅ HAVING COUNT(*) > 1 finds managers with multiple direct reports.

✅ A self join retrieves the manager's name.

✅ The outer query returns every employee reporting to those managers.

This question tests your understanding of:
✅ Self Joins
✅ GROUP BY and HAVING
✅ Subqueries
✅ Organizational Hierarchies

🎯 Expected Output Example
Employee Manager
John David
Alice David
Sarah Michael
Bob Michael

🚀 Alternative Using Window Functions
SELECT
employee_id,
employee_name,
manager_id
FROM (
SELECT
*,
COUNT(*) OVER (
PARTITION BY manager_id
) AS team_size
FROM employees
WHERE manager_id IS NOT NULL
) t
WHERE team_size > 1;

This approach uses a window function to count the number of employees under each manager without using GROUP BY.

🚀 Tip for SQL Job Seekers:
Hierarchy-based questions are very common in interviews. Practice problems involving:
Employees and managers
Parent-child relationships
Organizational charts
Category trees
Recursive queries WITH RECURSIVE or recursive CTEs

❤️ React with ❤️ for more SQL interview challenges!

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𝗜𝗻𝘁𝗲𝗿𝘃𝗶𝗲𝘄𝗲𝗿:
You have 2 minutes to solve this SQL query.
Find the departments where the average salary is greater than the company's overall average salary.

𝗠𝗲: Challenge accepted! 💪
SELECT
department,
AVG(salary) AS average_salary
FROM employees
GROUP BY department
HAVING AVG(salary) > (
SELECT AVG(salary)
FROM employees
);

💡 Explanation:
The query compares each department's average salary with the company's overall average salary.

• GROUP BY department calculates the average salary for each department.
• The subquery computes the overall average salary across all employees.
• HAVING filters only those departments whose average salary exceeds the company average.

This question tests your understanding of:
✅ GROUP BY
✅ HAVING
✅ Aggregate Functions AVG
✅ Subqueries

🎯 Expected Output Example
Department: Average Salary
IT: 88,500
Finance: 84,000

HR is excluded because its average salary is below the company average.

🚀 Alternative Using a Common Table Expression CTE
WITH company_avg AS (
SELECT AVG(salary) AS avg_salary
FROM employees
)
SELECT
department,
AVG(salary) AS average_salary
FROM employees, company_avg
GROUP BY department, company_avg.avg_salary
HAVING AVG(salary) > company_avg.avg_salary;

Using a CTE can improve readability, especially when the same calculated value is reused in larger queries.

🚀 Tip for SQL Job Seekers:
Interviewers often ask questions that compare group-level aggregates with overall aggregates. Master the use of HAVING with subqueries—it’s a key SQL pattern.

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Hope this helps you 😊

𝗜𝗻𝘁𝗲𝗿𝘃𝗶𝗲𝘄𝗲𝗿:
You have 2 minutes to solve this SQL query.
Find the month with the highest total sales.

Assume the table structure: sales(sale_id, sale_date, amount)

𝗠𝗲: Challenge accepted! 💪
SELECT
EXTRACT(YEAR FROM sale_date) AS year,
EXTRACT(MONTH FROM sale_date) AS month,
SUM(amount) AS total_sales
FROM sales
GROUP BY
EXTRACT(YEAR FROM sale_date),
EXTRACT(MONTH FROM sale_date)
ORDER BY total_sales DESC
LIMIT 1;

💡 Explanation:
This query groups sales by year and month, calculates the total sales for each month, and returns the month with the highest sales.

Key parts:
• EXTRACT YEAR FROM sale_date gets the year.
• EXTRACT MONTH FROM sale_date gets the month.
• SUM amount calculates total monthly sales.
• ORDER BY total_sales DESC sorts from highest to lowest.
• LIMIT 1 returns the top-performing month.

This question tests your understanding of:
Date Functions, Aggregate Functions SUM, GROUP BY, ORDER BY

🎯 Expected Output Example
Year: 2026, Month: 5, Total Sales: 245,000

🚀 Alternative Handles Ties
SELECT
year,
month,
total_sales
FROM (
SELECT
EXTRACT(YEAR FROM sale_date) AS year,
EXTRACT(MONTH FROM sale_date) AS month,
SUM(amount) AS total_sales,
DENSE_RANK() OVER (
ORDER BY SUM(amount) DESC
) AS rnk
FROM sales
GROUP BY
EXTRACT(YEAR FROM sale_date),
EXTRACT(MONTH FROM sale_date)
) ranked
WHERE rnk = 1;

This version returns all months tied for the highest total sales.

🚀 Date-based aggregation questions are among the most common in SQL interviews. Practice grouping data by Day, Week, Month, Quarter, Year. You'll encounter these patterns frequently in analytics and reporting roles.

❤️ React with ❤️ for more SQL interview challenges!

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𝗜𝗻𝘁𝗲𝗿𝘃𝗶𝗲𝘄𝗲𝗿:
You have 2 minutes to solve this SQL query.

Find the customers who placed orders on three or more consecutive days.

Assume the table structure:
orders(order_id, customer_id, order_date)

𝗠𝗲: Challenge accepted! 💪

WITH consecutive_orders AS (
SELECT
customer_id,
order_date,
DATE_SUB(
order_date,
INTERVAL ROW_NUMBER() OVER (
PARTITION BY customer_id
ORDER BY order_date
) DAY
) AS grp
FROM (
SELECT DISTINCT
customer_id,
order_date
FROM orders
) t
)
SELECT
customer_id
FROM consecutive_orders
GROUP BY
customer_id,
grp
HAVING COUNT(*) >= 3;

💡 Explanation:
This query identifies sequences of consecutive order dates for each customer.

• ROW_NUMBER() assigns a sequence number to each order date per customer.
• Subtracting the row number from the order date creates the same grp value for consecutive dates.
• GROUP BY customer_id, grp groups each consecutive streak.
• **HAVING COUNT(*) >= 3** returns customers with a streak of at least three consecutive days.

🎯 Expected Output Example
Customer ID
101
205

(Customer 101 ordered on June 1, 2, and 3. Customer 205 ordered on July 10, 11, and 12.)

🚀 Tip for SQL Job Seekers:
The Gaps and Islands pattern is one of the most advanced and frequently discussed SQL interview topics. Master it for solving:
Consecutive login days, Consecutive purchases, Attendance streaks, Consecutive transactions, User activity analysis

Being comfortable with this pattern can set you apart in technical interviews.

❤️ React with ❤️ for more SQL interview challenges!

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𝗜𝗻𝘁𝗲𝗿𝘃𝗶𝗲𝘄𝗲𝗿:
You have 2 minutes to solve this SQL query.

Find the customers who placed orders on three or more consecutive days.

Assume the table structure:
orders(order_id, customer_id, order_date)

𝗠𝗲: Challenge accepted! 💪

WITH consecutive_orders AS (
SELECT
customer_id,
order_date,
DATE_SUB(
order_date,
INTERVAL ROW_NUMBER() OVER (
PARTITION BY customer_id
ORDER BY order_date
) DAY
) AS grp
FROM (
SELECT DISTINCT
customer_id,
order_date
FROM orders
) t
)
SELECT
customer_id
FROM consecutive_orders
GROUP BY
customer_id,
grp
HAVING COUNT(*) >= 3;

💡 Explanation:
This query identifies sequences of consecutive order dates for each customer.

• ROW_NUMBER() assigns a sequence number to each order date per customer.
• Subtracting the row number from the order date creates the same grp value for consecutive dates.
• GROUP BY customer_id, grp groups each consecutive streak.
• **HAVING COUNT(*) >= 3** returns customers with a streak of at least three consecutive days.

This question tests your understanding of: Common Table Expressions (CTEs), Window Functions (ROW_NUMBER()), Gaps and Islands Problem, Date Arithmetic.

🎯 Expected Output Example
Customer ID
101
205

(Customer 101 ordered on June 1, 2, and 3. Customer 205 ordered on July 10, 11, and 12.)

🚀 Tip for SQL Job Seekers:
The Gaps and Islands pattern is one of the most advanced and frequently discussed SQL interview topics.

Master it for solving:
• Consecutive login days
• Consecutive purchases
• Attendance streaks
• Consecutive transactions
• User activity analysis

Being comfortable with this pattern can set you apart in technical interviews.

❤️ React with ❤️ for more SQL interview challenges!

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𝗜𝗻𝘁𝗲𝗿𝘃𝗶𝗲𝘄𝗲𝗿: 
You have 2 minutes to solve this SQL query.

Find the product(s) that have never been ordered.

Tables: 
products(product_id, product_name) 
order_details(order_id, product_id, quantity)

𝗠𝗲: Challenge accepted! 💪 
SELECT 
    p.product_id, 
    p.product_name 
FROM products p 
LEFT JOIN order_details od 
    ON p.product_id = od.product_id 
WHERE od.product_id IS NULL;

💡 Explanation: 
This query finds all products that do not have a matching record in the order_details table.

LEFT JOIN returns all products, regardless of whether they've been ordered. Products without matching orders will have NULL values for columns from order_details. WHERE od.product_id IS NULL filters only those products that have never been ordered.

This question tests your understanding of: LEFT JOIN, NULL handling, Finding unmatched records

🎯 Expected Output Example 
Product ID   Product Name 
104   Wireless Mouse 
118   USB Hub 
125   Laptop Stand 

🚀 Alternative Using NOT EXISTS 
SELECT 
    p.product_id, 
    p.product_name 
FROM products p 
WHERE NOT EXISTS ( 
    SELECT 1 
    FROM order_details od 
    WHERE od.product_id = p.product_id 
);

NOT EXISTS is often preferred because it handles NULL values correctly and can perform better than other approaches in many database systems.

🚀 Tip for SQL Job Seekers: 
Whenever you're asked to find records that don't exist in another table, consider these approaches: 

LEFT JOIN ... IS NULL, NOT EXISTS ✅ (often the best choice), NOT IN (be cautious with NULL values)

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Interviewer:

You have 2 minutes to solve this SQL query. 

Find the employee or employees with the highest salary in the company without using MAX().

Me: Challenge accepted!

SELECT
    employee_id,
    employee_name,
    salary
FROM (
    SELECT
        employee_id,
        employee_name,
        salary,
        DENSE_RANK() OVER (ORDER BY salary DESC) AS salary_rank
    FROM employees
) ranked
WHERE salary_rank = 1;

Explanation:

This query finds the highest-paid employee or employees without using the MAX() aggregate function.

• DENSE_RANK() ranks salaries in descending order.

• The highest salary receives a rank of 1.

• The outer query filters only employees with salary_rank = 1.

• If multiple employees share the highest salary, they are all returned.

This question tests your understanding of:

• Window Functions using DENSE_RANK

• Ranking Data

• Handling Ties

• Alternatives to Aggregate Functions

Expected Output Example

Employee: John, Salary: 120,000

Employee: Alice, Salary: 120,000 

Both employees are returned because they share the highest salary.

Alternative Solution using NOT EXISTS

SELECT
    employee_id,
    employee_name,
    salary
FROM employees e1
WHERE NOT EXISTS (
    SELECT 1
    FROM employees e2
    WHERE e2.salary > e1.salary
);

This solution works by returning employees for whom no other employee has a higher salary.

Tip for SQL Job Seekers:

Interviewers often ask you to solve problems without using aggregate functions like MAX() or MIN(). Learn multiple approaches using: 

• Window Functions 

• NOT EXISTS 

• Correlated Subqueries 

• Self Joins 

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