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Investments in 2016

Table: Insurance

+-------------+-------+
| Column Name | Type  |
+-------------+-------+
| pid         | int   |
| tiv_2015    | float |
| tiv_2016    | float |
| lat         | float |
| lon         | float |
+-------------+-------+
pid is the primary key (column with unique values) 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 a solution 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 as any other policyholder (i.e., the (lat, lon) attribute pairs must be unique).

Round tiv_2016 to two decimal places.

The 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.

Solution Explanation:

This problem requires finding the sum of tiv_2016 for policyholders who meet two conditions: their tiv_2015 value is shared by at least one other policyholder, and their location (lat, lon) is unique. The solution uses a window function in MySQL to efficiently identify policyholders satisfying these conditions.

Approach:

  1. Identify Shared tiv_2015: A window function COUNT(*) OVER (PARTITION BY tiv_2015) is used. This counts the number of policyholders with the same tiv_2015 value for each row. If the count (cnt1) is greater than 1, the policyholder shares their tiv_2015 value.

  2. Identify Unique Locations: Another window function COUNT(*) OVER (PARTITION BY lat, lon) is used. This counts the number of policyholders with the same latitude and longitude for each row. If the count (cnt2) is equal to 1, the policyholder's location is unique.

  3. Filtering and Aggregation: A WHERE clause filters the results to include only rows where cnt1 > 1 (shared tiv_2015) and cnt2 = 1 (unique location). Finally, SUM(tiv_2016) calculates the sum of tiv_2016 for the filtered rows, and ROUND(..., 2) rounds the result to two decimal places.

MySQL Code:

WITH
    T AS (
        SELECT
            tiv_2016,
            COUNT(1) OVER (PARTITION BY tiv_2015) AS cnt1,
            COUNT(1) OVER (PARTITION BY lat, lon) AS cnt2
        FROM Insurance
    )
SELECT ROUND(SUM(tiv_2016), 2) AS tiv_2016
FROM T
WHERE cnt1 > 1 AND cnt2 = 1;

Time Complexity Analysis:

The time complexity is dominated by the window function operations. Window functions typically have a time complexity of O(N log N) or O(N) depending on the database system's optimization strategies, where N is the number of rows in the Insurance table. The filtering and aggregation steps have a linear time complexity, O(N). Therefore, the overall time complexity is approximately O(N log N) or O(N), depending on the database implementation.

Space Complexity Analysis:

The space complexity depends on the size of the intermediate result set T. In the worst case, T will have the same number of rows as the original table. Therefore, the space complexity is O(N), where N is the number of rows in the Insurance table.

Other Languages (Note): This problem is inherently SQL-based. Directly translating it to other languages like Python or Java would require handling the database interaction separately, which is beyond the scope of this problem as presented. The core logic, however, using window functions or equivalent approaches to count and filter, would be analogous.