LeetCode #585 — MEDIUM

Investments in 2016

Move from brute-force thinking to an efficient approach using core interview patterns strategy.

The Problem

Problem Statement

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.

Step 01

Brute Force Baseline

Problem summary: 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

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

{"headers":{"Insurance":["pid","tiv_2015","tiv_2016","lat","lon"]},"rows":{"Insurance":[[1,10,5,10,10],[2,20,20,20,20],[3,10,30,20,20],[4,10,40,40,40]]}}

Step 02

Core Insight

What unlocks the optimal approach

Make the (LAT, LON) a pair to represent the location information

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #585: Investments in 2016
// Auto-generated Java example from rust.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (rust):
// // Accepted solution for LeetCode #585: Investments in 2016
// pub fn sql_example() -> &'static str {
//     r#"
// -- Accepted solution for LeetCode #585: Investments in 2016
// # Write your MySQL query statement below
// 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;
// "#
// }

// Accepted solution for LeetCode #585: Investments in 2016
// Auto-generated Go example from rust.
func exampleSolution() {
}
// Reference (rust):
// // Accepted solution for LeetCode #585: Investments in 2016
// pub fn sql_example() -> &'static str {
//     r#"
// -- Accepted solution for LeetCode #585: Investments in 2016
// # Write your MySQL query statement below
// 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;
// "#
// }

# Accepted solution for LeetCode #585: Investments in 2016
# Auto-generated Python example from rust.
def example_solution() -> None:
    return
# Reference (rust):
# // Accepted solution for LeetCode #585: Investments in 2016
# pub fn sql_example() -> &'static str {
#     r#"
# -- Accepted solution for LeetCode #585: Investments in 2016
# # Write your MySQL query statement below
# 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;
# "#
# }

// Accepted solution for LeetCode #585: Investments in 2016
pub fn sql_example() -> &'static str {
    r#"
-- Accepted solution for LeetCode #585: Investments in 2016
# Write your MySQL query statement below
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;
"#
}

// Accepted solution for LeetCode #585: Investments in 2016
// Auto-generated TypeScript example from rust.
function exampleSolution(): void {
}
// Reference (rust):
// // Accepted solution for LeetCode #585: Investments in 2016
// pub fn sql_example() -> &'static str {
//     r#"
// -- Accepted solution for LeetCode #585: Investments in 2016
// # Write your MySQL query statement below
// 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;
// "#
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.