Formatted question description: https://leetcode.ca/all/585.html
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).
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.
| 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 |
| TIV_2016 | |----------| | 45.00 |
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.
To output the sum of all total investment values in 2016, use
select sum(insurance.TIV_2016) as TIV_2016 from insurance.
where statement to select the entries that meet the two criteria. For criterion 1, there should be more than one entry with the same value of field
TIV_2015. For criterion 2, there should be exactly one entry of the values
# Write your MySQL query statement below select sum(insurance.TIV_2016) as TIV_2016 from insurance where insurance.TIV_2015 in ( select TIV_2015 from insurance group by TIV_2015 having count(*) > 1 and (LAT, LON) in ( select LAT, LON from insurance group by LAT, LON having count(*) = 1 ) );