Last updated on March 10th, 2025 at 10:36 pm

Here, we see an Investments in 2016 LeetCode Solution. This Leetcode problem is solved using MySQL and Pandas.

List of all LeetCode Solution

Companies

Twitter

Level of Question

Medium

Investments in 2016 LeetCode Solution

1. Problem Statement

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:

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.

2. Code Implementation in Different Languages

3.1 Investments in 2016 MySQL

SELECT 
  ROUND(
    SUM(tiv_2016), 
    2
  ) AS tiv_2016 
FROM 
  Insurance 
WHERE 
  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
  )

3.2 Investments in 2016 Pandas

import pandas as pd
def find_investments(insurance: pd.DataFrame) -> pd.DataFrame:
  uniq_lat_lon = insurance.drop_duplicates(subset = ['lat','lon'], keep = False).pid
  not_uniq_tiv_2015 = insurance.loc[insurance.duplicated(subset = 'tiv_2015', keep=False)].pid
  df = insurance.loc[insurance.pid.isin(uniq_lat_lon) & insurance.pid.isin(not_uniq_tiv_2015)]
  return df[['tiv_2016']].sum().to_frame('tiv_2016').round(2)