Data Scientists, Analysts, and Data Engineers at Bloomin' Brands write SQL queries to analyze customer dining patterns, helping them tailor menu offerings and promotions to fit customer preferences and improve the dining experience at their restaurants. They also use SQL to optimize supply chain efficiency, ensuring that each location is stocked with the right ingredients to meet customer demand, this is why Bloomin' Brands includes SQL problems in interviews.
Thus, to help you study, here’s 8 Bloomin' Brands SQL interview questions – able to answer them all?
Bloomin' Brands wishes to analyze their monthly sales. They want you to write a SQL query to calculate the total sales for each restaurant in their portfolio (Outback Steakhouse, Carrabba's Italian Grill, Bonefish Grill, and Fleming's Prime Steakhouse & Wine Bar) for each month, and rank them based on these sales for each month.
They are specifically interested in 2020. They also want the report to show each restaurant's rank in any given month, as compared to the other months in that year.
Provide the restaurant's id, name, month, year, total sales for that month and its rank compared to the other months in the year 2020.
We have 2 table:
sales Table:| sale_id | restaurant_id | sale_date | sale_amount |
|---|---|---|---|
| 1 | 1 | 2020-01-20 | 2500 |
| 2 | 2 | 2020-02-15 | 3000 |
| 3 | 1 | 2020-01-25 | 3500 |
| 4 | 3 | 2020-03-10 | 4000 |
| 5 | 1 | 2020-02-10 | 2000 |
restaurants Table:| restaurant_id | name |
|---|---|
| 1 | Outback Steakhouse |
| 2 | Carrabba's Italian Grill |
| 3 | Bonefish Grill |
| 4 | Fleming's Prime Steakhouse & Wine Bar |
WITH monthly_sales AS ( SELECT restaurant_id, date_part('month', sale_date) AS month, date_part('year', sale_date) AS year, SUM(sale_amount) AS total_sales FROM sales WHERE date_part('year', sale_date) = 2020 GROUP BY restaurant_id, date_part('month', sale_date), date_part('year', sale_date) ), sales_ranking AS ( SELECT restaurant_id, month, year, total_sales, RANK() OVER (PARTITION BY restaurant_id ORDER BY total_sales DESC) as sales_rank FROM monthly_sales ) SELECT sr.restaurant_id, res.name, sr.month, sr.year, sr.total_sales, sr.sales_rank FROM sales_ranking sr JOIN restaurants res ON sr.restaurant_id = res.restaurant_id ORDER BY sr.restaurant_id, sr.sales_rank;
This query first calculates the total sales for each restaurant per month in the year 2020 using a common table expression (monthly_sales), then ranks these sales within each restaurant using the RANK() window function. Finally, it joins the results with the restaurants table to show the restaurant names and sorts the results by restaurant_id and sales rank.
To solve a related window function SQL problem on DataLemur's free interactive SQL code editor, solve this Amazon SQL Interview Question:
Check out Bloomin' Brands' newsroom to see how they are enhancing dining experiences across their restaurant brands! Keeping up with their news can provide valuable insights into the evolving trends in the food and beverage sector.
Imagine you had a table of Bloomin' Brands employee salaries. Write a SQL query to find the employees who earn more than their direct manager.
employees Example Input:| employee_id | name | salary | department_id | manager_id |
|---|---|---|---|---|
| 1 | Emma Thompson | 3800 | 1 | |
| 2 | Daniel Rodriguez | 2230 | 1 | 10 |
| 3 | Olivia Smith | 8000 | 1 | 8 |
| 4 | Noah Johnson | 6800 | 2 | 8 |
| 5 | Sophia Martinez | 1750 | 1 | 10 |
| 8 | William Davis | 7000 | 2 | NULL |
| 10 | James Anderson | 4000 | 1 | NULL |
| employee_id | employee_name |
|---|---|
| 3 | Olivia Smith |
This is the output because Olivia Smith earns $8,000, surpassing her manager, William Davis who earns 7,800.
You can solve this question interactively on DataLemur:
First, we perform a SELF-JOIN where we treat the first employee table (mgr) as the managers' table and the second employee table (emp) as the employees' table. Then we use a WHERE clause to filter the results, ensuring we only get employees whose salaries are higher than their manager's salary.
SELECT emp.employee_id AS employee_id, emp.name AS employee_name FROM employee AS mgr INNER JOIN employee AS emp ON mgr.employee_id = emp.manager_id WHERE emp.salary > mgr.salary;
If the solution above is tough, you can find a detailed solution here: Employees Earning More Than Managers.
Normalization involves breaking up your tables into smaller, more specialized ones and using primary and foreign keys to define relationships between them. Not only does this make your database more flexible and scalable, it also makes it easier to maintain. Plus, normalization helps to keep your data accurate by reducing the chance of inconsistencies and errors.
The only downside is now is that your queries will involve more joins, which are slow AF and often a DB performance botteleneck.
Bloomin' Brands, a casual dining company, wants to improve their menu management systems across their restaurants. You are tasked with designing the database model that will support this functionality and write a query.
The requirements are as follows:
Assume we have two tables, food_items and food_categories. The relationship between these two tables is many-to-many, thus resulting in a third table restaurant_menu.
food_items Sample Input:| food_item_id | name | cost |
|---|---|---|
| 1 | 'Chicken Wings' | 5.5 |
| 2 | 'Steak' | 15 |
| 3 | 'Chocolate Cake' | 7 |
food_categories Sample Input:| category_id | name |
|---|---|
| 1 | 'Appetizer' |
| 2 | 'Main Course' |
| 3 | 'Dessert' |
restaurant_menu Sample Input:| restaurant_id | food_item_id | category_id | sales_July_2022 |
|---|---|---|---|
| 1 | 1 | 1 | 100 |
| 1 | 2 | 2 | 30 |
| 1 | 3 | 3 | 50 |
| 2 | 1 | 1 | 200 |
| 2 | 2 | 2 | 100 |
| 2 | 3 | 3 | 150 |
You have to come up with a PostgreSQL query to find the most profitable food category in July 2022 for each restaurant based on the number of food items sold in that month.
The SQL query would be:
WITH profit_cal AS ( SELECT rm.restaurant_id, fc.name as category_name, SUM(fi.cost * rm.sales_July_2022) as total_profit FROM restaurant_menu rm JOIN food_items fi ON rm.food_item_id = fi.food_item_id JOIN food_categories fc ON rm.category_id = fc.category_id GROUP BY rm.restaurant_id, fc.name ) SELECT restaurant_id, category_name, MAX(total_profit) as max_profit FROM profit_cal GROUP BY restaurant_id;
This query calculates the total profit for each category in each restaurant by multiplying the cost of food items by the number of sales in July 2022. It then finds the maximum profit for each restaurant.
UNION is used to combine the output of multiple SELECT statements into one big result!
For a concrete example, say you were doing an HR Analytics project for Bloomin' Brands and needed to analyze both Bloomin' Brands's employees and contractors who were hired after 2022 started. You could use UNION in the following way:
SELECT name, job_title, hire_date, salary FROM bloomin_brands_employees WHERE hire_date > '2022-01-01'; UNION SELECT name, job_title, hire_date, salary FROM bloomin_brands_contractors WHERE hire_date > '2022-01-01'
Bloomin' Brands is a company that owns several American casual dining restaurant chains like Outback Steakhouse, Carrabba's Italian Grill, Bonefish Grill, and Fleming's Prime Steakhouse & Wine Bar. Let's say, in their database, they have a table called meals that has details of all the meals served in their different restaurants. The table records details like meal_id, restaurant_name, price, etc. Now, the company wants to find out the average price of meals served per restaurant.
meals Example Input:| meal_id | restaurant_name | price |
|---|---|---|
| 1 | Outback Steakhouse | 10.99 |
| 2 | Carrabba's Italian Grill | 15.5 |
| 3 | Fleming's Prime Steakhouse & Wine Bar | 25.5 |
| 4 | Bonefish Grill | 14.0 |
| 5 | Outback Steakhouse | 13.99 |
| restaurant_name | avg_price |
|---|---|
| Outback Steakhouse | 12.49 |
| Carrabba's Italian Grill | 15.5 |
| Fleming's Prime Steakhouse & Wine Bar | 25.5 |
| Bonefish Grill | 14.0 |
SELECT restaurant_name, AVG(price) as avg_price FROM meals GROUP BY restaurant_name;
This SQL query uses the AVG() function to find out the average price of meals per restaurant.
It groups the meals by restaurant_name, and for each group of unique restaurant_name, it calculates the average price. The result is a list of each restaurant and the average price of the meals they serve.
To practice a very similar question try this interactive Amazon Highest-Grossing Items Question which is similar for requiring calculations of aggregate figures per category or this Amazon Average Review Ratings Question which is similar for involving average calculations grouped by different items.
A non-relational (NoSQL) database is any database that does not use the typical tabular format of rows and columns like in relational databases.
While knowing the four different types of NoSQL databases is beyond the scope of most Data Analyst & Data Science interviews, Data Engineers at Bloomin' Brands should vaguely refresh these concepts:
Assume that Bloomin' Brands owns several restaurants and each restaurant has a variety of dishes. Each meal has received a number of reviews. To improve the quality of their dishes, Bloomin' Brands wants to see the average rating of each dish per restaurant for the past year.
For this question, consider two tables, one for restaurants and dishes, and another for reviews:
restaurant_dishes Example Input:| restaurant_id | restaurant_name | dish_id | dish_name |
|---|---|---|---|
| 1 | Outback Steakhouse | 101 | Baby Back Ribs |
| 1 | Outback Steakhouse | 102 | Steak & Lobster |
| 2 | Carrabba's Italian Grill | 201 | Chicken Bryan |
| 2 | Carrabba's Italian Grill | 202 | Shrimp & Scallop Linguini |
reviews Example Input:| review_id | user_id | submit_date | dish_id | stars |
|---|---|---|---|---|
| 6171 | 123 | 06/10/2021 00:00:00 | 101 | 4 |
| 7802 | 265 | 06/14/2021 00:00:00 | 102 | 5 |
| 5293 | 362 | 07/18/2021 00:00:00 | 101 | 3 |
| 6352 | 192 | 11/26/2021 00:00:00 | 201 | 4 |
| 4517 | 981 | 12/05/2021 00:00:00 | 202 | 5 |
We are to find the average rating per dish for each restaurant:
SELECT rd.restaurant_name, rd.dish_name, AVG(r.stars) FROM restaurant_dishes AS rd JOIN reviews AS r ON rd.dish_id = r.dish_id WHERE r.submit_date >= '01/01/2021' AND r.submit_date < '01/01/2022' GROUP BY rd.restaurant_name, rd.dish_name ORDER BY rd.restaurant_name ASC, AVG(r.stars) DESC;
This PostgreSQL query performs a JOIN between restaurant_dishes and reviews based on dish_id. It filters the reviews for the past year using submit_date. It then groups the data by restaurant_name and dish_name, and calculates the average of stars.
The results are sorted by restaurant name and by descending order of the average stars so for each restaurant, the dish with the best average rating is presented first.
The best way to prepare for a Bloomin' Brands SQL interview is to practice, practice, practice. Besides solving the earlier Bloomin' Brands SQL interview questions, you should also solve the 200+ SQL questions on DataLemur which come from companies like Google, Facebook, Microsoft and Amazon.
Each problem on DataLemur has multiple hints, fully explained answers along with a discussion board to see how others solved it and crucially, there is an interactive coding environment so you can right online code up your query and have it checked.
To prep for the Bloomin' Brands SQL interview you can also be wise to solve SQL questions from other hospitality and restaurant companies like:
However, if your SQL skills are weak, forget about going right into solving questions – go learn SQL with this DataLemur SQL Tutorial.
This tutorial covers SQL topics like RANK vs. DENSE RANK and CASE/WHEN statements – both of these come up often in SQL interviews at Bloomin' Brands.
Besides SQL interview questions, the other question categories covered in the Bloomin' Brands Data Science Interview are:
To prepare for Bloomin' Brands Data Science interviews read the book Ace the Data Science Interview because it's got:
Also focus on the behavioral interview – prep for that with this list of behavioral interview questions for Data Scientists.