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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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Focus on these key topics: 👇

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2) Learn Excel for data cleaning & analysis
3) Master SQL for data querying
4) Study data visualization principles
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7) Learn data wrangling and preprocessing
8) Understand data storytelling and report writing
9) Practice hypothesis testing & A/B testing
10) Get familiar with Python/R for analytics (optional but helpful)
11) Work on real datasets and case studies (Kaggle is great)
12) Build end-to-end projects from data collection to visualization
13) Learn how to communicate insights effectively
14) Practice problem-solving with datasets regularly
15) Optimize your resume with analytics keywords
16) Follow analytics experts and tutorials on YouTube/LinkedIn

Pro tip: Search each topic on YouTube and watch short 10-15 min videos. Practice alongside to build strong fundamentals.

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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.

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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.

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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.

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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 

Demonstrating alternative solutions shows strong SQL problem-solving skills.

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Interviewer: 
You have 2 minutes to solve this SQL query. 
Find the employee or employees who earn more than the average salary of their department and have been with the company for more than 5 years.

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

Me: Challenge accepted! 

SELECT
    employee_id,
    employee_name,
    department,
    salary,
    joining_date
FROM employees e
WHERE salary > (
    SELECT AVG(salary)
    FROM employees
    WHERE department = e.department
)
AND joining_date <= CURRENT_DATE - INTERVAL '5 years';

Explanation: 
This query applies two conditions to identify experienced, high-performing employees.

• The correlated subquery calculates the average salary for each employee's department.
• The first condition returns employees earning above their department's average salary.
• The second condition filters employees who joined the company more than 5 years ago.
• Only employees satisfying both conditions are included in the final result.

This question tests your understanding of: 
• Correlated Subqueries
• Aggregate Functions using AVG
• Date Arithmetic
• Multiple Filtering Conditions

Expected Output Example 
Employee: John, Department: IT, Salary: 95,000, Joining Date: 2018-01-10 

Employee: Sarah, Department: HR, Salary: 82,000, Joining Date: 2017-06-15 

Alternative Using Window Functions 

SELECT
    employee_id,
    employee_name,
    department,
    salary,
    joining_date
FROM (
    SELECT
        *,
        AVG(salary) OVER (PARTITION BY department) AS dept_avg_salary
    FROM employees
) e
WHERE salary > dept_avg_salary
AND joining_date <= CURRENT_DATE - INTERVAL '5 years';

This approach avoids a correlated subquery by calculating the departmental average once using a window function, which can be more efficient on large datasets.

Tip for SQL Job Seekers: 
Real-world interview questions often combine multiple SQL concepts in a single problem. Practice writing queries that use: 
• Window Functions
• Correlated Subqueries
• Date Functions
• Aggregate Functions
• Complex WHERE conditions

These combined-concept questions are common in mid-level and senior SQL interviews.

❤️ React with ❤️ for more SQL interview challenges!

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

Find the employee(s) with the highest salary in each department without using window functions.

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

Me: Challenge accepted! 💪 
SELECT 
    employee_id, 
    employee_name, 
    department, 
    salary 
FROM employees e1 
WHERE salary = ( 
    SELECT MAX(salary) 
    FROM employees e2 
    WHERE e2.department = e1.department 
);

💡 Explanation: 
This query uses a correlated subquery instead of a window function.

• The outer query processes each employee
• The correlated subquery finds the maximum salary within that employee's department
• If the employee's salary matches the maximum salary, the employee is returned
• If multiple employees share the highest salary in a department, they are all included

This question tests your understanding of: 
• Correlated Subqueries
• Aggregate Functions (MAX)
• Filtering with Subqueries
• Handling Ties

🎯 Expected Output Example: 
John     | IT      | 95,000 
Alice    | IT      | 95,000 
Sarah    | HR      | 82,000 
David    | Finance | 91,000 

John and Alice are both returned because they share the highest salary in the IT department.

🚀 Alternative Using a Self Join 
SELECT 
    e1.employee_id, 
    e1.employee_name, 
    e1.department, 
    e1.salary 
FROM employees e1 
LEFT JOIN employees e2 
    ON e1.department = e2.department 
   AND e1.salary < e2.salary 
WHERE e2.employee_id IS NULL; 

This solution works by eliminating employees who have someone in the same department with a higher salary. The remaining employees are the highest-paid in their respective departments.

🚀 Tip for SQL Job Seekers: 
Interviewers often restrict certain SQL features like window functions or CTEs to evaluate your understanding of alternative approaches. Be prepared to solve the same problem using: 
• Correlated Subqueries
• Self Joins
• CTEs
• Window Functions

Knowing multiple solutions demonstrates strong SQL fundamentals.

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

Find the second most recent order placed by each customer.

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

𝗠𝗲: Challenge accepted! 💪

SELECT
order_id,
customer_id,
order_date
FROM (
SELECT
order_id,
customer_id,
order_date,
ROW_NUMBER() OVER (
PARTITION BY customer_id
ORDER BY order_date DESC
) AS rn
FROM orders
) ranked
WHERE rn = 2;

💡 Explanation:

The query assigns a rank to each order based on its order date for every customer.

• PARTITION BY customer_id creates a separate ranking for each customer
• ORDER BY order_date DESC ranks the most recent order as 1
• ROW_NUMBER() ensures each order gets a unique rank
• The outer query returns only the order with rn = 2, i.e., the second most recent order

This question tests your understanding of:
✅ Window Functions (ROW_NUMBER)
✅ Ranking Records
✅ Partitioning Data
✅ Top N per Group

🎯 Expected Output Example
Customer ID | Order ID | Order Date
101 | 2056 | 2026-06-15
102 | 2074 | 2026-06-18

Customers with fewer than two orders are automatically excluded.

🚀 Alternative Using a Correlated Subquery
SELECT
o1.order_id,
o1.customer_id,
o1.order_date
FROM orders o1
WHERE 1 = (
SELECT COUNT(*)
FROM orders o2
WHERE o2.customer_id = o1.customer_id
AND o2.order_date > o1.order_date
);

This approach counts how many orders are more recent than the current order. If exactly one order is more recent, the current order is the second most recent.

🚀 Tip for SQL Job Seekers:
Questions involving the Nth latest or Nth earliest record appear frequently in interviews. Practice solving them using:
• ROW_NUMBER()
• RANK()
• DENSE_RANK()
• Correlated Subqueries

Understanding when to use each approach is a valuable interview skill.

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