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Author: mrinal1704

Medium/Investments in 2016.sql

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+-- Question 96
+-- Write a query to print the sum of all total investment values in 2016 (TIV_2016), to a scale of 2 decimal places, for all policy holders who meet the following criteria:
+
+-- Have the same TIV_2015 value as one or more other policyholders.
+-- Are not located in the same city as any other policyholder (i.e.: the (latitude, longitude) attribute pairs must be unique).
+-- Input Format:
+-- The insurance table is described as follows:
+
+-- | Column Name | Type          |
+-- |-------------|---------------|
+-- | PID         | INTEGER(11)   |
+-- | TIV_2015    | NUMERIC(15,2) |
+-- | TIV_2016    | NUMERIC(15,2) |
+-- | LAT         | NUMERIC(5,2)  |
+-- | LON         | NUMERIC(5,2)  |
+-- where PID is the policyholder's policy ID, TIV_2015 is the total investment value in 2015, TIV_2016 is the total investment value in 2016, LAT is the latitude of the policy holder's city, and LON is the longitude of the policy holder's city.
+
+-- Sample Input
+
+-- | PID | TIV_2015 | TIV_2016 | LAT | LON |
+-- |-----|----------|----------|-----|-----|
+-- | 1   | 10       | 5        | 10  | 10  |
+-- | 2   | 20       | 20       | 20  | 20  |
+-- | 3   | 10       | 30       | 20  | 20  |
+-- | 4   | 10       | 40       | 40  | 40  |
+-- Sample Output
+
+-- | TIV_2016 |
+-- |----------|
+-- | 45.00    |
+-- Explanation
+
+-- The first record in the table, like the last record, meets both of the two criteria.
+-- The TIV_2015 value '10' is as the same as the third and forth record, and its location unique.
+
+-- The second record does not meet any of the two criteria. Its TIV_2015 is not like any other policyholders.
+
+-- And its location is the same with the third record, which makes the third record fail, too.
+
+-- So, the result is the sum of TIV_2016 of the first and last record, which is 45.
+
+-- Solution
+select sum(TIV_2016) TIV_2016
+from 
+(select *, count(*) over (partition by TIV_2015) as c1, count(*) over (partition by LAT, LON) as c2
+from insurance ) t
+where c1 > 1 and c2 = 1; 

Medium/Number of Trusted Contacts.sql

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+-- Question 97
+-- Table: Customers
+
+-- +---------------+---------+
+-- | Column Name   | Type    |
+-- +---------------+---------+
+-- | customer_id   | int     |
+-- | customer_name | varchar |
+-- | email         | varchar |
+-- +---------------+---------+
+-- customer_id is the primary key for this table.
+-- Each row of this table contains the name and the email of a customer of an online shop.
+ 
+
+-- Table: Contacts
+
+-- +---------------+---------+
+-- | Column Name   | Type    |
+-- +---------------+---------+
+-- | user_id       | id      |
+-- | contact_name  | varchar |
+-- | contact_email | varchar |
+-- +---------------+---------+
+-- (user_id, contact_email) is the primary key for this table.
+-- Each row of this table contains the name and email of one contact of customer with user_id.
+-- This table contains information about people each customer trust. The contact may or may not exist in the Customers table.
+
+ 
+
+-- Table: Invoices
+
+-- +--------------+---------+
+-- | Column Name  | Type    |
+-- +--------------+---------+
+-- | invoice_id   | int     |
+-- | price        | int     |
+-- | user_id      | int     |
+-- +--------------+---------+
+-- invoice_id is the primary key for this table.
+-- Each row of this table indicates that user_id has an invoice with invoice_id and a price.
+ 
+
+-- Write an SQL query to find the following for each invoice_id:
+
+-- customer_name: The name of the customer the invoice is related to.
+-- price: The price of the invoice.
+-- contacts_cnt: The number of contacts related to the customer.
+-- trusted_contacts_cnt: The number of contacts related to the customer and at the same time they are customers to the shop. (i.e His/Her email exists in the Customers table.)
+-- Order the result table by invoice_id.
+
+-- The query result format is in the following example:
+
+-- Customers table:
+-- +-------------+---------------+--------------------+
+-- | customer_id | customer_name | email              |
+-- +-------------+---------------+--------------------+
+-- | 1           | Alice         | [email protected] |
+-- | 2           | Bob           | [email protected]   |
+-- | 13          | John          | [email protected]  |
+-- | 6           | Alex          | [email protected]  |
+-- +-------------+---------------+--------------------+
+-- Contacts table:
+-- +-------------+--------------+--------------------+
+-- | user_id     | contact_name | contact_email      |
+-- +-------------+--------------+--------------------+
+-- | 1           | Bob          | [email protected]   |
+-- | 1           | John         | [email protected]  |
+-- | 1           | Jal          | [email protected]   |
+-- | 2           | Omar         | [email protected]  |
+-- | 2           | Meir         | [email protected]  |
+-- | 6           | Alice        | [email protected] |
+-- +-------------+--------------+--------------------+
+-- Invoices table:
+-- +------------+-------+---------+
+-- | invoice_id | price | user_id |
+-- +------------+-------+---------+
+-- | 77         | 100   | 1       |
+-- | 88         | 200   | 1       |
+-- | 99         | 300   | 2       |
+-- | 66         | 400   | 2       |
+-- | 55         | 500   | 13      |
+-- | 44         | 60    | 6       |
+-- +------------+-------+---------+
+-- Result table:
+-- +------------+---------------+-------+--------------+----------------------+
+-- | invoice_id | customer_name | price | contacts_cnt | trusted_contacts_cnt |
+-- +------------+---------------+-------+--------------+----------------------+
+-- | 44         | Alex          | 60    | 1            | 1                    |
+-- | 55         | John          | 500   | 0            | 0                    |
+-- | 66         | Bob           | 400   | 2            | 0                    |
+-- | 77         | Alice         | 100   | 3            | 2                    |
+-- | 88         | Alice         | 200   | 3            | 2                    |
+-- | 99         | Bob           | 300   | 2            | 0                    |
+-- +------------+---------------+-------+--------------+----------------------+
+-- Alice has three contacts, two of them are trusted contacts (Bob and John).
+-- Bob has two contacts, none of them is a trusted contact.
+-- Alex has one contact and it is a trusted contact (Alice).
+-- John doesn't have any contacts.
+
+-- Solution