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Hard/Find median given frequency of numbers.sql

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+-- Question 107
+-- The Numbers table keeps the value of number and its frequency.
+
+-- +----------+-------------+
+-- |  Number  |  Frequency  |
+-- +----------+-------------|
+-- |  0       |  7          |
+-- |  1       |  1          |
+-- |  2       |  3          |
+-- |  3       |  1          |
+-- +----------+-------------+
+-- In this table, the numbers are 0, 0, 0, 0, 0, 0, 0, 1, 2, 2, 2, 3, so the median is (0 + 0) / 2 = 0.
+
+-- +--------+
+-- | median |
+-- +--------|
+-- | 0.0000 |
+-- +--------+
+-- Write a query to find the median of all numbers and name the result as median.
+
+-- Solution
+with t1 as(
+select *,
+sum(frequency) over(order by number) as cum_sum, (sum(frequency) over())/2 as middle
+from numbers)
+
+select avg(number) as median
+from t1
+where middle between (cum_sum - frequency) and cum_sum

Hard/Find the quiet students in the exam.sql

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+-- Question 106
+-- Table: Student
+
+-- +---------------------+---------+
+-- | Column Name         | Type    |
+-- +---------------------+---------+
+-- | student_id          | int     |
+-- | student_name        | varchar |
+-- +---------------------+---------+
+-- student_id is the primary key for this table.
+-- student_name is the name of the student.
+ 
+
+-- Table: Exam
+
+-- +---------------+---------+
+-- | Column Name   | Type    |
+-- +---------------+---------+
+-- | exam_id       | int     |
+-- | student_id    | int     |
+-- | score         | int     |
+-- +---------------+---------+
+-- (exam_id, student_id) is the primary key for this table.
+-- Student with student_id got score points in exam with id exam_id.
+ 
+
+-- A "quite" student is the one who took at least one exam and didn't score neither the high score nor the low score.
+
+-- Write an SQL query to report the students (student_id, student_name) being "quiet" in ALL exams.
+
+-- Don't return the student who has never taken any exam. Return the result table ordered by student_id.
+
+-- The query result format is in the following example.
+
+ 
+
+-- Student table:
+-- +-------------+---------------+
+-- | student_id  | student_name  |
+-- +-------------+---------------+
+-- | 1           | Daniel        |
+-- | 2           | Jade          |
+-- | 3           | Stella        |
+-- | 4           | Jonathan      |
+-- | 5           | Will          |
+-- +-------------+---------------+
+
+-- Exam table:
+-- +------------+--------------+-----------+
+-- | exam_id    | student_id   | score     |
+-- +------------+--------------+-----------+
+-- | 10         |     1        |    70     |
+-- | 10         |     2        |    80     |
+-- | 10         |     3        |    90     |
+-- | 20         |     1        |    80     |
+-- | 30         |     1        |    70     |
+-- | 30         |     3        |    80     |
+-- | 30         |     4        |    90     |
+-- | 40         |     1        |    60     |
+-- | 40         |     2        |    70     |
+-- | 40         |     4        |    80     |
+-- +------------+--------------+-----------+
+
+-- Result table:
+-- +-------------+---------------+
+-- | student_id  | student_name  |
+-- +-------------+---------------+
+-- | 2           | Jade          |
+-- +-------------+---------------+
+
+-- For exam 1: Student 1 and 3 hold the lowest and high score respectively.
+-- For exam 2: Student 1 hold both highest and lowest score.
+-- For exam 3 and 4: Studnet 1 and 4 hold the lowest and high score respectively.
+-- Student 2 and 5 have never got the highest or lowest in any of the exam.
+-- Since student 5 is not taking any exam, he is excluded from the result.
+-- So, we only return the information of Student 2.
+
+-- Solution
+with t1 as(
+select student_id
+from
+(select *,
+min(score) over(partition by exam_id) as least,
+max(score) over(partition by exam_id) as most
+from exam) a
+where least = score or most = score)
+
+
+select distinct student_id, student_name
+from exam join student
+using (student_id)
+where student_id != all(select student_id from t1)
+order by 1

Hard/Game Play Analysis 5.sql

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+-- Question 111
+-- Table: Activity
+
+-- +--------------+---------+
+-- | Column Name  | Type    |
+-- +--------------+---------+
+-- | player_id    | int     |
+-- | device_id    | int     |
+-- | event_date   | date    |
+-- | games_played | int     |
+-- +--------------+---------+
+-- (player_id, event_date) is the primary key of this table.
+-- This table shows the activity of players of some game.
+-- Each row is a record of a player who logged in and played a number of games (possibly 0) before logging out on some day using some device.
+ 
+
+-- We define the install date of a player to be the first login day of that player.
+
+-- We also define day 1 retention of some date X to be the number of players whose install date is X and they logged back in on the day right after X, divided by the number of players whose install date is X, rounded to 2 decimal places.
+
+-- Write an SQL query that reports for each install date, the number of players that installed the game on that day and the day 1 retention.
+
+-- The query result format is in the following example:
+
+-- Activity table:
+-- +-----------+-----------+------------+--------------+
+-- | player_id | device_id | event_date | games_played |
+-- +-----------+-----------+------------+--------------+
+-- | 1         | 2         | 2016-03-01 | 5            |
+-- | 1         | 2         | 2016-03-02 | 6            |
+-- | 2         | 3         | 2017-06-25 | 1            |
+-- | 3         | 1         | 2016-03-01 | 0            |
+-- | 3         | 4         | 2016-07-03 | 5            |
+-- +-----------+-----------+------------+--------------+
+
+-- Result table:
+-- +------------+----------+----------------+
+-- | install_dt | installs | Day1_retention |
+-- +------------+----------+----------------+
+-- | 2016-03-01 | 2        | 0.50           |
+-- | 2017-06-25 | 1        | 0.00           |
+-- +------------+----------+----------------+
+-- Player 1 and 3 installed the game on 2016-03-01 but only player 1 logged back in on 2016-03-02 so the
+-- day 1 retention of 2016-03-01 is 1 / 2 = 0.50
+-- Player 2 installed the game on 2017-06-25 but didn't log back in on 2017-06-26 so the day 1 retention of 2017-06-25 is 0 / 1 = 0.00
+
+-- Solution
+with t1 as(
+select *,
+row_number() over(partition by player_id order by event_date) as rnk,
+min(event_date) over(partition by player_id) as install_dt,
+lead(event_date,1) over(partition by player_id order by event_date) as nxt
+from Activity)
+
+select distinct install_dt,
+count(distinct player_id) as installs,
+round(sum(case when nxt=event_date+1 then 1 else 0 end)/count(distinct player_id),2) as Day1_retention
+from t1
+where rnk = 1
+group by 1
+order by 1

Hard/Sales by day of the week.sql

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+-- Question 112
+-- Table: Orders
+
+-- +---------------+---------+
+-- | Column Name   | Type    |
+-- +---------------+---------+
+-- | order_id      | int     |
+-- | customer_id   | int     |
+-- | order_date    | date    | 
+-- | item_id       | varchar |
+-- | quantity      | int     |
+-- +---------------+---------+
+-- (ordered_id, item_id) is the primary key for this table.
+-- This table contains information of the orders placed.
+-- order_date is the date when item_id was ordered by the customer with id customer_id.
+ 
+
+-- Table: Items
+
+-- +---------------------+---------+
+-- | Column Name         | Type    |
+-- +---------------------+---------+
+-- | item_id             | varchar |
+-- | item_name           | varchar |
+-- | item_category       | varchar |
+-- +---------------------+---------+
+-- item_id is the primary key for this table.
+-- item_name is the name of the item.
+-- item_category is the category of the item.
+ 
+
+-- You are the business owner and would like to obtain a sales report for category items and day of the week.
+
+-- Write an SQL query to report how many units in each category have been ordered on each day of the week.
+
+-- Return the result table ordered by category.
+
+-- The query result format is in the following example:
+
+ 
+
+-- Orders table:
+-- +------------+--------------+-------------+--------------+-------------+
+-- | order_id   | customer_id  | order_date  | item_id      | quantity    |
+-- +------------+--------------+-------------+--------------+-------------+
+-- | 1          | 1            | 2020-06-01  | 1            | 10          |
+-- | 2          | 1            | 2020-06-08  | 2            | 10          |
+-- | 3          | 2            | 2020-06-02  | 1            | 5           |
+-- | 4          | 3            | 2020-06-03  | 3            | 5           |
+-- | 5          | 4            | 2020-06-04  | 4            | 1           |
+-- | 6          | 4            | 2020-06-05  | 5            | 5           |
+-- | 7          | 5            | 2020-06-05  | 1            | 10          |
+-- | 8          | 5            | 2020-06-14  | 4            | 5           |
+-- | 9          | 5            | 2020-06-21  | 3            | 5           |
+-- +------------+--------------+-------------+--------------+-------------+
+
+-- Items table:
+-- +------------+----------------+---------------+
+-- | item_id    | item_name      | item_category |
+-- +------------+----------------+---------------+
+-- | 1          | LC Alg. Book   | Book          |
+-- | 2          | LC DB. Book    | Book          |
+-- | 3          | LC SmarthPhone | Phone         |
+-- | 4          | LC Phone 2020  | Phone         |
+-- | 5          | LC SmartGlass  | Glasses       |
+-- | 6          | LC T-Shirt XL  | T-Shirt       |
+-- +------------+----------------+---------------+
+
+-- Result table:
+-- +------------+-----------+-----------+-----------+-----------+-----------+-----------+-----------+
+-- | Category   | Monday    | Tuesday   | Wednesday | Thursday  | Friday    | Saturday  | Sunday    |
+-- +------------+-----------+-----------+-----------+-----------+-----------+-----------+-----------+
+-- | Book       | 20        | 5         | 0         | 0         | 10        | 0         | 0         |
+-- | Glasses    | 0         | 0         | 0         | 0         | 5         | 0         | 0         |
+-- | Phone      | 0         | 0         | 5         | 1         | 0         | 0         | 10        |
+-- | T-Shirt    | 0         | 0         | 0         | 0         | 0         | 0         | 0         |
+-- +------------+-----------+-----------+-----------+-----------+-----------+-----------+-----------+
+-- On Monday (2020-06-01, 2020-06-08) were sold a total of 20 units (10 + 10) in the category Book (ids: 1, 2).
+-- On Tuesday (2020-06-02) were sold a total of 5 units  in the category Book (ids: 1, 2).
+-- On Wednesday (2020-06-03) were sold a total of 5 units in the category Phone (ids: 3, 4).
+-- On Thursday (2020-06-04) were sold a total of 1 unit in the category Phone (ids: 3, 4).
+-- On Friday (2020-06-05) were sold 10 units in the category Book (ids: 1, 2) and 5 units in Glasses (ids: 5).
+-- On Saturday there are no items sold.
+-- On Sunday (2020-06-14, 2020-06-21) were sold a total of 10 units (5 +5) in the category Phone (ids: 3, 4).
+-- There are no sales of T-Shirt.
+
+-- Solution
+with t1 as(
+select distinct item_category,
+case when dayname(order_date)='Monday' then sum(quantity) over(partition by item_category,dayname(order_date)) else 0 end as Monday,
+Case when dayname(order_date)='Tuesday' then sum(quantity) over(partition by item_category,dayname(order_date)) else 0 end as Tuesday,
+Case when dayname(order_date)='Wednesday' then sum(quantity) over(partition by item_category,dayname(order_date)) else 0 end as Wednesday,
+Case when dayname(order_date)='Thursday' then sum(quantity) over(partition by item_category,dayname(order_date)) else 0 end as Thursday,
+Case when dayname(order_date)='Friday' then sum(quantity) over(partition by item_category,dayname(order_date)) else 0 end as Friday,
+Case when dayname(order_date)='Saturday' then sum(quantity) over(partition by item_category,dayname(order_date)) else 0 end as Saturday,
+Case when dayname(order_date)='Sunday' then sum(quantity) over(partition by item_category,dayname(order_date)) else 0 end as Sunday
+from orders o
+right join items i
+using (item_id))
+
+select item_category as category, sum(Monday) as Monday, sum(Tuesday) as Tuesday, sum(Wednesday) Wednesday, sum(Thursday) Thursday,
+sum(Friday) Friday, sum(Saturday) Saturday, sum(Sunday) Sunday
+from t1
+group by item_category

Hard/Total sales amount by year.sql

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+-- Question 114
+-- Table: Product
+
+-- +---------------+---------+
+-- | Column Name   | Type    |
+-- +---------------+---------+
+-- | product_id    | int     |
+-- | product_name  | varchar |
+-- +---------------+---------+
+-- product_id is the primary key for this table.
+-- product_name is the name of the product.
+ 
+
+-- Table: Sales
+
+-- +---------------------+---------+
+-- | Column Name         | Type    |
+-- +---------------------+---------+
+-- | product_id          | int     |
+-- | period_start        | varchar |
+-- | period_end          | date    |
+-- | average_daily_sales | int     |
+-- +---------------------+---------+
+-- product_id is the primary key for this table. 
+-- period_start and period_end indicates the start and end date for sales period, both dates are inclusive.
+-- The average_daily_sales column holds the average daily sales amount of the items for the period.
+
+-- Write an SQL query to report the Total sales amount of each item for each year, with corresponding product name, product_id, product_name and report_year.
+
+-- Dates of the sales years are between 2018 to 2020. Return the result table ordered by product_id and report_year.
+
+-- The query result format is in the following example:
+
+
+-- Product table:
+-- +------------+--------------+
+-- | product_id | product_name |
+-- +------------+--------------+
+-- | 1          | LC Phone     |
+-- | 2          | LC T-Shirt   |
+-- | 3          | LC Keychain  |
+-- +------------+--------------+
+
+-- Sales table:
+-- +------------+--------------+-------------+---------------------+
+-- | product_id | period_start | period_end  | average_daily_sales |
+-- +------------+--------------+-------------+---------------------+
+-- | 1          | 2019-01-25   | 2019-02-28  | 100                 |
+-- | 2          | 2018-12-01   | 2020-01-01  | 10                  |
+-- | 3          | 2019-12-01   | 2020-01-31  | 1                   |
+-- +------------+--------------+-------------+---------------------+
+
+-- Result table:
+-- +------------+--------------+-------------+--------------+
+-- | product_id | product_name | report_year | total_amount |
+-- +------------+--------------+-------------+--------------+
+-- | 1          | LC Phone     |    2019     | 3500         |
+-- | 2          | LC T-Shirt   |    2018     | 310          |
+-- | 2          | LC T-Shirt   |    2019     | 3650         |
+-- | 2          | LC T-Shirt   |    2020     | 10           |
+-- | 3          | LC Keychain  |    2019     | 31           |
+-- | 3          | LC Keychain  |    2020     | 31           |
+-- +------------+--------------+-------------+--------------+
+-- LC Phone was sold for the period of 2019-01-25 to 2019-02-28, and there are 35 days for this period. Total amount 35*100 = 3500. 
+-- LC T-shirt was sold for the period of 2018-12-01 to 2020-01-01, and there are 31, 365, 1 days for years 2018, 2019 and 2020 respectively.
+-- LC Keychain was sold for the period of 2019-12-01 to 2020-01-31, and there are 31, 31 days for years 2019 and 2020 respectively.
+
+-- Solution
+SELECT
+    b.product_id,
+    a.product_name,
+    a.yr AS report_year,
+    CASE 
+        WHEN YEAR(b.period_start)=YEAR(b.period_end) AND a.yr=YEAR(b.period_start) THEN DATEDIFF(b.period_end,b.period_start)+1
+        WHEN a.yr=YEAR(b.period_start) THEN DATEDIFF(DATE_FORMAT(b.period_start,'%Y-12-31'),b.period_start)+1
+        WHEN a.yr=YEAR(b.period_end) THEN DAYOFYEAR(b.period_end) 
+        WHEN a.yr>YEAR(b.period_start) AND a.yr<YEAR(b.period_end) THEN 365
+        ELSE 0
+    END * average_daily_sales AS total_amount
+FROM
+    (SELECT product_id,product_name,'2018' AS yr FROM Product
+    UNION
+    SELECT product_id,product_name,'2019' AS yr FROM Product
+    UNION
+    SELECT product_id,product_name,'2020' AS yr FROM Product) a
+    JOIN 
+    Sales b
+    ON a.product_id=b.product_id  
+HAVING total_amount > 0
+ORDER BY b.product_id,a.yr