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Medium/Active Users.sql

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+--Question 94
+-- Table Accounts:
+
+-- +---------------+---------+
+-- | Column Name   | Type    |
+-- +---------------+---------+
+-- | id            | int     |
+-- | name          | varchar |
+-- +---------------+---------+
+-- the id is the primary key for this table.
+-- This table contains the account id and the user name of each account.
+ 
+
+-- Table Logins:
+
+-- +---------------+---------+
+-- | Column Name   | Type    |
+-- +---------------+---------+
+-- | id            | int     |
+-- | login_date    | date    |
+-- +---------------+---------+
+-- There is no primary key for this table, it may contain duplicates.
+-- This table contains the account id of the user who logged in and the login date. A user may log in multiple times in the day.
+ 
+
+-- Write an SQL query to find the id and the name of active users.
+
+-- Active users are those who logged in to their accounts for 5 or more consecutive days.
+
+-- Return the result table ordered by the id.
+
+-- The query result format is in the following example:
+
+-- Accounts table:
+-- +----+----------+
+-- | id | name     |
+-- +----+----------+
+-- | 1  | Winston  |
+-- | 7  | Jonathan |
+-- +----+----------+
+
+-- Logins table:
+-- +----+------------+
+-- | id | login_date |
+-- +----+------------+
+-- | 7  | 2020-05-30 |
+-- | 1  | 2020-05-30 |
+-- | 7  | 2020-05-31 |
+-- | 7  | 2020-06-01 |
+-- | 7  | 2020-06-02 |
+-- | 7  | 2020-06-02 |
+-- | 7  | 2020-06-03 |
+-- | 1  | 2020-06-07 |
+-- | 7  | 2020-06-10 |
+-- +----+------------+
+
+-- Result table:
+-- +----+----------+
+-- | id | name     |
+-- +----+----------+
+-- | 7  | Jonathan |
+-- +----+----------+
+-- User Winston with id = 1 logged in 2 times only in 2 different days, so, Winston is not an active user.
+-- User Jonathan with id = 7 logged in 7 times in 6 different days, five of them were consecutive days, so, Jonathan is an active user.
+
+-- Solution
+with t1 as (
+select id,login_date,
+lead(login_date,4) over(partition by id order by login_date) date_5
+from (select distinct * from Logins) b
+)
+
+select distinct a.id, a.name from t1
+inner join accounts a 
+on t1.id = a.id
+where datediff(t1.date_5,login_date) = 4
+order by id

Medium/Customers who bought all products.sql

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+-- Question 93
+-- Table: Customer
+
+-- +-------------+---------+
+-- | Column Name | Type    |
+-- +-------------+---------+
+-- | customer_id | int     |
+-- | product_key | int     |
+-- +-------------+---------+
+-- product_key is a foreign key to Product table.
+-- Table: Product
+
+-- +-------------+---------+
+-- | Column Name | Type    |
+-- +-------------+---------+
+-- | product_key | int     |
+-- +-------------+---------+
+-- product_key is the primary key column for this table.
+ 
+
+-- Write an SQL query for a report that provides the customer ids from the Customer table that bought all the products in the Product table.
+
+-- For example:
+
+-- Customer table:
+-- +-------------+-------------+
+-- | customer_id | product_key |
+-- +-------------+-------------+
+-- | 1           | 5           |
+-- | 2           | 6           |
+-- | 3           | 5           |
+-- | 3           | 6           |
+-- | 1           | 6           |
+-- +-------------+-------------+
+
+-- Product table:
+-- +-------------+
+-- | product_key |
+-- +-------------+
+-- | 5           |
+-- | 6           |
+-- +-------------+
+
+-- Result table:
+-- +-------------+
+-- | customer_id |
+-- +-------------+
+-- | 1           |
+-- | 3           |
+-- +-------------+
+-- The customers who bought all the products (5 and 6) are customers with id 1 and 3.
+
+-- Solution
+select customer_id
+from customer
+group by customer_id
+having count(distinct product_key) = (select COUNT(distinct product_key) from product)

Medium/Monthly Transaction 2.sql

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+-- Question 95
+-- Table: Transactions
+
+-- +----------------+---------+
+-- | Column Name    | Type    |
+-- +----------------+---------+
+-- | id             | int     |
+-- | country        | varchar |
+-- | state          | enum    |
+-- | amount         | int     |
+-- | trans_date     | date    |
+-- +----------------+---------+
+-- id is the primary key of this table.
+-- The table has information about incoming transactions.
+-- The state column is an enum of type ["approved", "declined"].
+-- Table: Chargebacks
+
+-- +----------------+---------+
+-- | Column Name    | Type    |
+-- +----------------+---------+
+-- | trans_id       | int     |
+-- | charge_date    | date    |
+-- +----------------+---------+
+-- Chargebacks contains basic information regarding incoming chargebacks from some transactions placed in Transactions table.
+-- trans_id is a foreign key to the id column of Transactions table.
+-- Each chargeback corresponds to a transaction made previously even if they were not approved.
+ 
+
+-- Write an SQL query to find for each month and country, the number of approved transactions and their total amount, the number of chargebacks and their total amount.
+
+-- Note: In your query, given the month and country, ignore rows with all zeros.
+
+-- The query result format is in the following example:
+
+-- Transactions table:
+-- +------+---------+----------+--------+------------+
+-- | id   | country | state    | amount | trans_date |
+-- +------+---------+----------+--------+------------+
+-- | 101  | US      | approved | 1000   | 2019-05-18 |
+-- | 102  | US      | declined | 2000   | 2019-05-19 |
+-- | 103  | US      | approved | 3000   | 2019-06-10 |
+-- | 104  | US      | approved | 4000   | 2019-06-13 |
+-- | 105  | US      | approved | 5000   | 2019-06-15 |
+-- +------+---------+----------+--------+------------+
+
+-- Chargebacks table:
+-- +------------+------------+
+-- | trans_id   | trans_date |
+-- +------------+------------+
+-- | 102        | 2019-05-29 |
+-- | 101        | 2019-06-30 |
+-- | 105        | 2019-09-18 |
+-- +------------+------------+
+
+-- Result table:
+-- +----------+---------+----------------+-----------------+-------------------+--------------------+
+-- | month    | country | approved_count | approved_amount | chargeback_count  | chargeback_amount  |
+-- +----------+---------+----------------+-----------------+-------------------+--------------------+
+-- | 2019-05  | US      | 1              | 1000            | 1                 | 2000               |
+-- | 2019-06  | US      | 3              | 12000           | 1                 | 1000               |
+-- | 2019-09  | US      | 0              | 0               | 1                 | 5000               |
+-- +----------+---------+----------------+-----------------+-------------------+--------------------+
+
+-- Solution
+with t1 as
+(select country, extract('month' from trans_date), state, count(*) as approved_count, sum(amount) as approved_amount
+from transactions
+where state = 'approved'
+group by 1, 2, 3),
+t2 as(
+select t.country, extract('month' from c.trans_date), sum(amount) as chargeback_amount, count(*) as chargeback_count
+from chargebacks c left join transactions t 
+on trans_id = id
+group by t.country, extract('month' from c.trans_date)),
+
+t3 as(
+select t2.date_part, t2.country, coalesce(approved_count,0) as approved_count, coalesce(approved_amount,0) as approved_amount, coalesce(chargeback_count,0) as chargeback_count, coalesce(chargeback_amount,0) as chargeback_amount
+from t2 left join t1 
+on t2.date_part = t1.date_part and t2.country = t1.country),
+
+t4 as(
+select t1.date_part, t1.country, coalesce(approved_count,0) as approved_count, coalesce(approved_amount,0) as approved_amount, coalesce(chargeback_count,0) as chargeback_count, coalesce(chargeback_amount,0) as chargeback_amount
+from t2 right join t1 
+on t2.date_part = t1.date_part and t2.country = t1.country)
+
+select *
+from t3
+union
+select *
+from t4

Medium/New users daily count.sql

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+-- Question 92
+-- Table: Traffic
+
+-- +---------------+---------+
+-- | Column Name   | Type    |
+-- +---------------+---------+
+-- | user_id       | int     |
+-- | activity      | enum    |
+-- | activity_date | date    |
+-- +---------------+---------+
+-- There is no primary key for this table, it may have duplicate rows.
+-- The activity column is an ENUM type of ('login', 'logout', 'jobs', 'groups', 'homepage').
+ 
+
+-- Write an SQL query that reports for every date within at most 90 days from today, 
+-- the number of users that logged in for the first time on that date. Assume today is 2019-06-30.
+
+-- The query result format is in the following example:
+
+-- Traffic table:
+-- +---------+----------+---------------+
+-- | user_id | activity | activity_date |
+-- +---------+----------+---------------+
+-- | 1       | login    | 2019-05-01    |
+-- | 1       | homepage | 2019-05-01    |
+-- | 1       | logout   | 2019-05-01    |
+-- | 2       | login    | 2019-06-21    |
+-- | 2       | logout   | 2019-06-21    |
+-- | 3       | login    | 2019-01-01    |
+-- | 3       | jobs     | 2019-01-01    |
+-- | 3       | logout   | 2019-01-01    |
+-- | 4       | login    | 2019-06-21    |
+-- | 4       | groups   | 2019-06-21    |
+-- | 4       | logout   | 2019-06-21    |
+-- | 5       | login    | 2019-03-01    |
+-- | 5       | logout   | 2019-03-01    |
+-- | 5       | login    | 2019-06-21    |
+-- | 5       | logout   | 2019-06-21    |
+-- +---------+----------+---------------+
+
+-- Result table:
+-- +------------+-------------+
+-- | login_date | user_count  |
+-- +------------+-------------+
+-- | 2019-05-01 | 1           |
+-- | 2019-06-21 | 2           |
+-- +------------+-------------+
+-- Note that we only care about dates with non zero user count.
+-- The user with id 5 first logged in on 2019-03-01 so he's not counted on 2019-06-21.
+
+-- Solution
+with t1 as
+(
+    select user_id, min(activity_date) as login_date
+    from Traffic
+    where activity = 'login'
+    group by user_id
+)
+
+select login_date, count(distinct user_id) as user_count
+from t1
+where login_date between '2019-04-01' and '2019-06-30'
+group by login_date