refactor 585:

Author: fishercoder1534

Diff for: database/_585.sql

@@ -1,47 +1,3 @@
---585. Investments in 2016
---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.
-
 select sum(TIV_2016) as TIV_2016
 from insurance a where
 (