Formatted question description: https://leetcode.ca/all/585.html

# 585. Investments in 2016

## Level

Medium

## Description

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

To output the sum of all total investment values in 2016, use `select sum(insurance.TIV_2016) as TIV_2016 from insurance`

.

Use a `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 `LAT`

and `LON`

.

```
# 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
)
);
```