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585\. Investments in 2016
Medium
SQL Schema
Table: `Insurance`
+-------------+-------+
| Column Name | Type |
+-------------+-------+
| pid | int |
| tiv_2015 | float |
| tiv_2016 | float |
| lat | float |
| lon | float |
+-------------+-------+
pid is the primary key column for this table.
Each row of this table contains information about one policy where:
pid is the policyholder's policy ID.
tiv\_2015 is the total investment value in 2015 and tiv\_2016 is the total investment value in 2016.
lat is the latitude of the policy holder's city. It's guaranteed that lat is not NULL.
lon is the longitude of the policy holder's city. It's guaranteed that lon is not NULL.
Write an SQL query to report the sum of all total investment values in 2016 `tiv_2016`, for all policyholders who:
* have the same `tiv_2015` value as one or more other policyholders, and
* are not located in the same city like any other policyholder (i.e., the (`lat, lon`) attribute pairs must be unique).
Round `tiv_2016` to **two decimal places**.
The query result format is in the following example.
**Example 1:**
**Input:** Insurance table:
+-----+----------+----------+-----+-----+
| 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 |
+-----+----------+----------+-----+-----+
**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 the same as the third and fourth records, and its location is 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 as 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.
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