Data Analytics

𝗜𝗻𝘁𝗲𝗿𝘃𝗶𝗲𝘄𝗲𝗿: You have 2 minutes to solve this SQL query. Find the department with the highest average salary. 𝗠𝗲: Challenge accepted! 💪 SELECT department, AVG(salary) AS average_salary FROM employees GROUP BY department ORDER BY average_salary DESC LIMIT 1; 💡 Explanation: The query calculates the average salary for each department and returns the department with the highest average salary. • GROUP BY department groups employees by department • AVG(salary) calculates the average salary for each department • ORDER BY average_salary DESC sorts departments from highest to lowest average salary • LIMIT 1 returns only the top department This question tests your understanding of: ✅ GROUP BY ✅ Aggregate Functions AVG ✅ ORDER BY ✅ LIMIT 🎯 Expected Output Department Average_Salary IT 88,500 🚀 Bonus Handles Ties If multiple departments share the highest average salary, use DENSE_RANK(): SELECT department, average_salary FROM ( SELECT department, AVG(salary) AS average_salary, DENSE_RANK() OVER ( ORDER BY AVG(salary) DESC ) AS rnk FROM employees GROUP BY department ) ranked WHERE rnk = 1; This version returns all departments tied for the highest average salary. 🚀 Whenever you see questions like highest, lowest, top N, or rank, think beyond LIMIT. Ask yourself: What if there's a tie? Window functions like DENSE_RANK() often provide a more complete solution. ❤️ React with ❤️ for more SQL interview challenges!

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𝗜𝗻𝘁𝗲𝗿𝘃𝗶𝗲𝘄𝗲𝗿: You have 2 minutes to solve this SQL query. Find customers who have never placed an order. Tables: customers(customer_id, customer_name) orders(order_id, customer_id, order_date) 𝗠𝗲: Challenge accepted! 💪 SELECT c.customer_id, c.customer_name FROM customers c LEFT JOIN orders o ON c.customer_id = o.customer_id WHERE o.customer_id IS NULL; 💡 Explanation: This query uses a LEFT JOIN to return all customers, whether or not they have placed an order. • LEFT JOIN keeps every customer in the result. • Customers without matching records in the orders table will have NULL values. • The WHERE o.customer_id IS NULL condition filters only customers who have never placed an order. This question tests your understanding of: ✅ LEFT JOIN ✅ Finding missing records ✅ NULL handling 🎯 Expected Output Example Customer ID Customer Name 105 Alice 112 David 118 Sarah 🚀 Questions about finding unmatched records are very common in interviews. Practice using: - LEFT JOIN ... IS NULL - NOT EXISTS - NOT IN (carefully, because NULL values can affect results) Among these, NOT EXISTS is often preferred for correctness and performance in many databases. ❤️ React with ❤️ for more SQL interview challenges!

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𝗜𝗻𝘁𝗲𝗿𝘃𝗶𝗲𝘄𝗲𝗿: You have 2 minutes to solve this SQL query. Find the top 3 highest-paid employees in each department. 𝗠𝗲: Challenge accepted! 💪 SELECT     employee_id,     employee_name,     department,     salary FROM (     SELECT         employee_id,         employee_name,         department,         salary,         DENSE_RANK() OVER (             PARTITION BY department             ORDER BY salary DESC         ) AS salary_rank     FROM employees ) ranked WHERE salary_rank <= 3 ORDER BY department, salary DESC; 💡 Explanation: The query uses the DENSE_RANK() window function to rank employees based on salary within each department. • PARTITION BY department creates a separate ranking for every department. • ORDER BY salary DESC ranks the highest salary first. • DENSE_RANK() assigns the same rank to employees with identical salaries. • The outer query returns only employees with a rank of 3 or less. This question tests your understanding of: ✅ Window Functions ✅ DENSE_RANK() ✅ Top N per Group ✅ Partitioning Data 🎯 Expected Output Example Employee  Department Salary Rank John  IT  95,000 1 Alice  IT  95,000 1 Bob  IT  90,000 2 Mike  IT  85,000 3 Sarah  HR  80,000 1 🚀 Know when to use each ranking function: • ROW_NUMBER() → No ties (unique ranking) • RANK() → Leaves gaps after ties • DENSE_RANK() → No gaps after ties (ideal for Top N with ties) ❤️ React with ❤️ for more SQL interview challenges!

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𝗜𝗻𝘁𝗲𝗿𝘃𝗶𝗲𝘄𝗲𝗿: You have 2 minutes to solve this SQL query. Find employees whose salary is higher than their manager's salary. Assume the table structure is: employees(employee_id, employee_name, manager_id, salary) 𝗠𝗲: Challenge accepted! 💪 SELECT e.employee_id, e.employee_name, e.salary AS employee_salary, m.employee_name AS manager_name, m.salary AS manager_salary FROM employees e JOIN employees m ON e.manager_id = m.employee_id WHERE e.salary > m.salary; 💡 Explanation: This query uses a self join because both employees and managers are stored in the same table. • e represents the employee. • m represents the manager. • The join matches each employee with their manager using manager_id. • The WHERE clause filters employees whose salary is greater than their manager's salary. This question tests your understanding of: ✅ Self Joins ✅ Aliases (e and m) ✅ Comparing values across related rows 🎯 Expected Output Example Employee Employee Salary Manager Manager Salary John 90,000 David 80,000 Sarah 85,000 Michael 75,000 🚀 Self joins are one of the most frequently asked SQL interview topics. Practice scenarios involving employees, managers, organizational hierarchies, categories, and parent-child relationships. ❤️ React with ❤️ for more SQL interview challenges!

𝗜𝗻𝘁𝗲𝗿𝘃𝗶𝗲𝘄𝗲𝗿: You have 2 minutes to solve this SQL query. Find employees whose salary is higher than their manager's salary. Assume the table structure is: employees(employee_id, employee_name, manager_id, salary) 𝗠𝗲: Challenge accepted! 💪 SELECT e.employee_id, e.employee_name, e.salary AS employee_salary, m.employee_name AS manager_name, m.salary AS manager_salary FROM employees e JOIN employees m ON e.manager_id = m.employee_id WHERE e.salary > m.salary; 💡 Explanation: This query uses a self join because both employees and managers are stored in the same table. • e represents the employee. • m represents the manager. • The join matches each employee with their manager using manager_id. • The WHERE clause filters employees whose salary is greater than their manager's salary. This question tests your understanding of: ✅ Self Joins ✅ Aliases (e and m) ✅ Comparing values across related rows 🎯 Expected Output Example Employee Employee Salary Manager Manager Salary John 90,000 David 80,000 Sarah 85,000 Michael 75,000 🚀 Tip for SQL Job Seekers: Self joins are one of the most frequently asked SQL interview topics. Practice scenarios involving employees, managers, organizational hierarchies, categories, and parent-child relationships. ❤️ React with ❤️ for more SQL interview challenges!

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𝗜𝗻𝘁𝗲𝗿𝘃𝗶𝗲𝘄𝗲𝗿: You have 2 minutes to solve this SQL query. Find the employees who have the highest salary in each department. 𝗠𝗲: Challenge accepted! 💪

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

💡 Explanation: This query uses the DENSE_RANK() window function to rank employees by salary within each department. • PARTITION BY department creates separate rankings for each department. • ORDER BY salary DESC ranks the highest salary as 1. • DENSE_RANK() ensures that if multiple employees have the same highest salary, they all receive Rank 1. • The outer query filters only the employees with rnk = 1. This question tests your knowledge of: ✅ Window Functions ✅ DENSE_RANK() vs RANK() vs ROW_NUMBER() ✅ Partitioning Data 🎯 Output Example Employee | Department | Salary John | IT | 95,000 Sarah | HR | 80,000 David | Finance | 90,000 Alice | IT | 95,000 (John and Alice both appear because they share the highest salary in the IT department.) 🚀 Whenever an interview question asks for the top N records per group, think of window functions. DENSE_RANK(), RANK(), and ROW_NUMBER() are among the most commonly tested SQL concepts. ❤️ React with ❤️ for more SQL interview challenges!

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𝗜𝗻𝘁𝗲𝗿𝘃𝗶𝗲𝘄𝗲𝗿: You have 2 minutes to solve this SQL query. Find employees who earn more than the average salary of their own department. 𝗠𝗲: Challenge accepted! 💪

SELECT
    employee_id,
    employee_name,
    department,
    salary
FROM employees e
WHERE salary > (
    SELECT AVG(salary)
    FROM employees
    WHERE department = e.department
);

💡 Explanation: The query uses a correlated subquery to calculate the average salary for each employee's department. • The outer query iterates through each employee. • The inner query calculates the average salary of that employee's department. • If an employee's salary is greater than their department's average, they're included in the result. This is a classic SQL interview question that tests your understanding of: ✅ Correlated Subqueries ✅ Aggregate Functions (AVG) ✅ Filtering with WHERE 🎯 Expected Output Example

+----------+------------+--------+
| Employee | Department | Salary |
+----------+------------+--------+
| John     | IT         | 90,000 |
| Sarah    | HR         | 70,000 |
| David    | Finance    | 85,000 |
+----------+------------+--------+

(Only employees earning above their department's average salary.) 🚀 Correlated subqueries are asked frequently in interviews. Learn when to use them—and also know how to rewrite them using window functions for better performance on large datasets. ❤️ React with ❤️ for more SQL interview challenges!

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🚀 SQL Scenario Based Interview Questions with Answers Part 2 📌 Question 11: Find the Second Highest Salary in Each Department  Table: employees (employee_id, department_id, salary)

WITH ranked_salary AS (
    SELECT *,
           DENSE_RANK() OVER (
               PARTITION BY department_id
               ORDER BY salary DESC
           ) AS rnk
    FROM employees
)
SELECT department_id, employee_id, salary
FROM ranked_salary
WHERE rnk = 2;

📌 Question 12: Identify Users Who Purchased on Their First Visit  Tables: visits (user_id, visit_date) | orders (user_id, order_date)

WITH first_visit AS (
    SELECT user_id,
           MIN(visit_date) AS first_visit_date
    FROM visits
    GROUP BY user_id
)
SELECT DISTINCT f.user_id
FROM first_visit f
JOIN orders o
    ON f.user_id = o.user_id
   AND f.first_visit_date = o.order_date;

📌 Question 13: Find Products Never Sold  Tables: products (product_id, product_name) | sales (product_id)

SELECT p.product_id, p.product_name
FROM products p
LEFT JOIN sales s
       ON p.product_id = s.product_id
WHERE s.product_id IS NULL;

📌 Question 14: Calculate Month-over-Month Revenue Growth  Table: orders (order_date, revenue)

WITH monthly_revenue AS (
    SELECT DATE_TRUNC('month', order_date) AS month,
           SUM(revenue) AS total_revenue
    FROM orders
    GROUP BY 1
)
SELECT month,
       total_revenue,
       LAG(total_revenue) OVER (ORDER BY month) AS previous_month,
       ROUND(
           100.0 *
           (total_revenue - LAG(total_revenue) OVER (ORDER BY month))
           /
           LAG(total_revenue) OVER (ORDER BY month),
           2
       ) AS growth_pct
FROM monthly_revenue;

📌 Question 15: Find Employees Earning More Than Department Average  Table: employees (employee_id, department_id, salary)

SELECT employee_id, department_id, salary
FROM (
    SELECT *,
           AVG(salary) OVER (
               PARTITION BY department_id
           ) AS dept_avg
    FROM employees
) t
WHERE salary > dept_avg;

📌 Question 16: Find Longest Consecutive Login Streak  Table: logins (user_id, login_date)

WITH cte AS (
    SELECT user_id,
           login_date,
           login_date -
           ROW_NUMBER() OVER (
               PARTITION BY user_id
               ORDER BY login_date
           ) * INTERVAL '1 day' AS grp
    FROM logins
)
SELECT user_id, COUNT(*) AS streak_days
FROM cte
GROUP BY user_id, grp
ORDER BY streak_days DESC;

📌 Question 17: Find Peak Sales Day of Every Month  Table: sales (sale_date, amount)

WITH daily_sales AS (
    SELECT DATE(sale_date) AS sale_day,
           SUM(amount) AS revenue
    FROM sales
    GROUP BY DATE(sale_date)
)
SELECT *
FROM (
    SELECT *,
           ROW_NUMBER() OVER (
               PARTITION BY DATE_TRUNC('month', sale_day)
               ORDER BY revenue DESC
           ) rn
    FROM daily_sales
) t
WHERE rn = 1;

📌 Question 18: Find Customers Who Ordered Every Month  Table: orders (customer_id, order_date)

WITH customer_months AS (
    SELECT customer_id,
           COUNT(DISTINCT DATE_TRUNC('month', order_date)) AS months_active
    FROM orders
    GROUP BY customer_id
),
total_months AS (
    SELECT COUNT(DISTINCT DATE_TRUNC('month', order_date)) AS total_months
    FROM orders
)
SELECT customer_id
FROM customer_months c
CROSS JOIN total_months t
WHERE c.months_active = t.total_months;

📌 Question 19: Find Top Selling Product Category  Tables: products (product_id, category) | sales (product_id, quantity)

SELECT category, SUM(quantity) AS total_sold
FROM sales s
JOIN products p ON s.product_id = p.product_id
GROUP BY category
ORDER BY total_sold DESC
LIMIT 1;

📌 Question 20: Calculate Median Salary  Table: employees (employee_id, salary)

SELECT PERCENTILE_CONT(0.5)
WITHIN GROUP (ORDER BY salary)
AS median_salary
FROM employees;

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🚀 Real-World SQL Scenario Based Interview Questions with Answers 📌 Question 1: Find Customers Who Purchased in Consecutive Months Table: orders customer_id, order_date Requirement: Identify customers who placed orders in consecutive months. WITH monthly_orders AS ( SELECT DISTINCT customer_id, DATE_TRUNC('month', order_date) AS order_month FROM orders ), consecutive_orders AS ( SELECT customer_id, order_month, LAG(order_month) OVER ( PARTITION BY customer_id ORDER BY order_month ) AS prev_month FROM monthly_orders ) SELECT customer_id FROM consecutive_orders WHERE order_month = prev_month + INTERVAL '1 month'; 📌 Question 2: Find the Top 3 Customers by Revenue Each Month Table: orders customer_id, amount, order_date WITH customer_revenue AS ( SELECT DATE_TRUNC('month', order_date) AS month, customer_id, SUM(amount) AS revenue FROM orders GROUP BY 1, 2 ) SELECT * FROM ( SELECT *, DENSE_RANK() OVER ( PARTITION BY month ORDER BY revenue DESC ) AS rnk FROM customer_revenue ) t WHERE rnk <= 3; 📌 Question 3: Calculate Running Total Revenue Table: sales sale_date, amount Requirement: Show cumulative revenue over time. SELECT sale_date, amount, SUM(amount) OVER ( ORDER BY sale_date ) AS running_revenue FROM sales; 📌 Question 4: Find Users Who Have Not Logged In During the Last 30 Days Tables: users user_id, logins user_id, login_date SELECT u.user_id FROM users u LEFT JOIN logins l ON u.user_id = l.user_id GROUP BY u.user_id HAVING MAX(login_date) < CURRENT_DATE - INTERVAL '30 days' OR MAX(login_date) IS NULL; 📌 Question 5: Detect Duplicate Transactions Table: transactions transaction_id, customer_id, amount, transaction_date Requirement: Find duplicate transactions based on customer, amount, and date. SELECT customer_id, amount, transaction_date, COUNT(*) AS duplicate_count FROM transactions GROUP BY customer_id, amount, transaction_date HAVING COUNT(*) > 1; 📌 Question 6: Calculate Average Order Value by Month Table: orders order_id, amount, order_date SELECT DATE_TRUNC('month', order_date) AS month, ROUND(AVG(amount), 2) AS avg_order_value FROM orders GROUP BY DATE_TRUNC('month', order_date) ORDER BY month; 📌 Question 7: Find the Most Recent Order for Each Customer Table: orders order_id, customer_id, order_date WITH ranked_orders AS ( SELECT *, ROW_NUMBER() OVER ( PARTITION BY customer_id ORDER BY order_date DESC ) AS rn FROM orders ) SELECT customer_id, order_id, order_date FROM ranked_orders WHERE rn = 1; 📌 Question 8: Calculate Product Contribution to Total Revenue Table: sales product_id, amount Requirement: Find percentage contribution of each product. SELECT product_id, SUM(amount) AS revenue, ROUND( 100.0 * SUM(amount) / SUM(SUM(amount)) OVER (), 2 ) AS contribution_pct FROM sales GROUP BY product_id; 📌 Question 9: Find Customers with No Orders Tables: customers customer_id, orders customer_id SELECT c.customer_id FROM customers c LEFT JOIN orders o ON c.customer_id = o.customer_id WHERE o.customer_id IS NULL; 📌 Question 10: Calculate 7-Day Moving Average Sales Table: sales sale_date, amount SELECT sale_date, amount, ROUND( AVG(amount) OVER ( ORDER BY sale_date ROWS BETWEEN 6 PRECEDING AND CURRENT ROW ), 2 ) AS moving_avg_7_days FROM sales; ❤️ Double Tap For More

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