585. Investments in 2016 - Solution & Explanation

The problem

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.

Approach overview

Problem Overview: You are given an Insurance table with pid, tiv_2015, tiv_2016, lat, and lon. The task is to sum tiv_2016 for policies where the tiv_2015 value appears more than once, but the (lat, lon) location is unique across the dataset.

Approach 1: SQL Query with Aggregation and Grouping (O(n log n) time, O(n) space)

This approach relies on SQL aggregation. First, group records by tiv_2015 and keep only values that appear more than once using GROUP BY tiv_2015 HAVING COUNT(*) > 1. Next, identify unique locations by grouping (lat, lon) and filtering where COUNT(*) = 1. Finally, sum tiv_2016 for rows that satisfy both conditions. Most SQL engines implement grouping using sorting or hashing internally, leading to roughly O(n log n) time with O(n) intermediate storage. This is the most direct solution when working with relational databases.

Approach 2: Structural Data Handling with In-Memory Computation (O(n) time, O(n) space)

Load the table into memory and compute frequencies using dictionaries. One map counts occurrences of tiv_2015, while another tracks occurrences of each (lat, lon) pair. After building both frequency maps, iterate through the records again and add tiv_2016 to the result when tiv_2015 frequency is greater than 1 and location frequency equals 1. Hash lookups make each operation constant time, producing overall O(n) complexity. This technique mirrors solutions built with hash tables.

Approach 3: Using Sorting and Two-Pointer Technique (O(n log n) time, O(1)–O(n) space)

Sort records by tiv_2015 to quickly detect duplicates and by (lat, lon) to detect unique locations. After sorting, use pointer scans to mark which entries share the same tiv_2015 and which locations repeat. Sorting costs O(n log n), but the scanning phase is linear. This approach avoids hash tables and works well when the dataset must be processed in deterministic order. It connects closely with patterns from sorting algorithms and pointer-based scans.

Approach 4: Hash Table for O(n) Complexity (O(n) time, O(n) space)

A more explicit version of the in-memory method. Maintain two hash tables: one mapping tiv_2015 to frequency and another mapping the string or tuple (lat, lon) to frequency. After the counting pass, iterate once more and accumulate tiv_2016 when tiv_2015 count > 1 and location count = 1. Each insertion and lookup in the hash table is constant time, so the total runtime stays O(n). This is the optimal algorithmic solution outside of SQL environments and relies on concepts from database-style filtering and hashing.

Recommended for interviews: Interviewers typically expect the hash-based counting approach or the SQL aggregation solution depending on the role. Showing the brute logic—count duplicates of tiv_2015 and filter unique coordinates—demonstrates understanding of the data constraints. Implementing it with hash maps in O(n) time shows strong practical problem-solving skills.

Approach 1: Approach 1: SQL Query with Aggregation and Grouping

This approach involves writing a SQL query to filter and compute the sum of all total investment values in 2016 (tiv_2016) for policyholders, ensuring they satisfy the given criteria.

In essence, we utilize SQL's aggregation functions alongside grouping features to systematically filter the data for our specific needs. This is accomplished by creating temporary tables/subqueries to collect and filter specific required datasets.

MySQL

SELECT ROUND(SUM(tiv_2016), 2) AS tiv_2016
FROM (
    SELECT tiv_2015, tiv_2016, lat, lon
    FROM Insurance
    WHERE tiv_2015 IN (
        SELECT tiv_2015
        FROM Insurance
        GROUP BY tiv_2015
        HAVING COUNT(DISTINCT pid) > 1
    )
    AND CONCAT(lat, '-', lon) IN (
        SELECT CONCAT(lat, '-', lon)
        FROM Insurance
        GROUP BY lat, lon
        HAVING COUNT(DISTINCT pid) = 1
    )
) AS filtered_records;

This SQL solution utilizes subqueries to filter polynomial datasets based on constraints.

The first subquery filters out groups where tiv_2015 is shared by more than one policyholder using HAVING COUNT(DISTINCT pid) > 1.
The second subquery ensures that the (lat, lon) combination is unique using HAVING COUNT(DISTINCT pid) = 1.
Finally, the main query sums up the tiv_2016 values and rounds it to two decimal places using SQL's ROUND function.

Complexity

Time Complexity: O(n log n) due to sorting in aggregation and grouping.

Space Complexity: O(n) to store intermediate datasets in memory.

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Approach 2: Approach 2: Structural Data Handling with In-Memory Computation

This approach involves using in-memory computations through programming languages with efficient data manipulation capabilities. We load the data into memory and use data structures like dictionaries and sets to maintain unique conditions and compute the resultant sum.

Utilize hashmaps to track matching tiv_2015 values and cities uniquely defined by (lat, lon).

Python

def calculate_investment(records):
    from collections import defaultdict
    
    tiv_2015_count = defaultdict(int)
    city_set = set()
    unique_cities = set()
    
    for record in records:
        pid, tiv_2015, tiv_2016, lat, lon = record
        tiv_2015_count[tiv_2015] += 1
        city = (lat, lon)
        if city in city_set:
            unique_cities.discard(city)
        else:
            city_set.add(city)
            unique_cities.add(city)
    
    result = 0.0
    for record in records:
        pid, tiv_2015, tiv_2016, lat, lon = record
        city = (lat, lon)
        if tiv_2015_count[tiv_2015] > 1 and city in unique_cities:
            result += tiv_2016
    return round(result, 2)

# Example usage
insurance_records = [
    (1, 10, 5, 10, 10),
    (2, 20, 20, 20, 20),
    (3, 10, 30, 20, 20),
    (4, 10, 40, 40, 40)
]
print(calculate_investment(insurance_records))

This solution involves several parts:

tiv_2015_count keeps track of how many policyholders share the same tiv_2015 value.
city_set and unique_cities are used to ensure uniqueness of the city based on the (lat, lon) combination.
After processing the records, we loop through them again to sum up the tiv_2016 value of those eligible policyholders.

Complexity

Time Complexity: O(n) as it involves two traversals over the dataset.

Space Complexity: O(n) due to the additional space required for hashmaps/databases for storing intermediate results.

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Approach 3: Using Sorting and Two-Pointer Technique

This approach involves sorting the array and then using two pointers to find pairs that satisfy the condition. Sorting helps to bring potential pairs closer, making it easier to identify valid combinations without unnecessary checks.

Python

def find_pairs(arr, sum):
    arr.sort()
    left, right = 0, len(arr) - 1
    while left < right:
        current_sum = arr[left] + arr[right]
        if current_sum == sum:
            print(f"Pair found: ({arr[left]}, {arr[right]})")
            left += 1
            right -= 1
        elif current_sum < sum:
            left += 1
        else:
            right -= 1

arr = [2, 4, 3, 1, 6, 5]
sum = 7
find_pairs(arr, sum)

Java

import java.util.Arrays;

public class TwoSum {
    public static void findPairs(int[] arr, int sum) {
        Arrays.sort(arr);
        int left = 0, right = arr.length - 1;
        while (left < right) {
            int currentSum = arr[left] + arr[right];
            if (currentSum == sum) {
                System.out.println("Pair found: (" + arr[left] + ", " + arr[right] + ")");
                left++;
                right--;
            } else if (currentSum < sum) {
                left++;
            } else {
                right--;
            }
        }
    }

    public static void main(String[] args) {
        int[] arr = {2, 4, 3, 1, 6, 5};
        int sum = 7;
        findPairs(arr, sum);
    }
}

C++

#include <iostream>
#include <vector>
#include <algorithm>

void findPairs(std::vector<int> &arr, int sum) {
    std::sort(arr.begin(), arr.end());
    int left = 0, right = arr.size() - 1;
    while (left < right) {
        int currentSum = arr[left] + arr[right];
        if (currentSum == sum) {
            std::cout << "Pair found: (" << arr[left] << ", " << arr[right] << ")\n";
            left++;
            right--;
        } else if (currentSum < sum) {
            left++;
        } else {
            right--;
        }
    }
}

int main() {
    std::vector<int> arr = {2, 4, 3, 1, 6, 5};
    int sum = 7;
    findPairs(arr, sum);
    return 0;
}

JavaScript

function findPairs(arr, sum) {
    arr.sort((a, b) => a - b);
    let left = 0, right = arr.length - 1;
    while (left < right) {
        let currentSum = arr[left] + arr[right];
        if (currentSum === sum) {
            console.log(`Pair found: (${arr[left]}, ${arr[right]})`);
            left++;
            right--;
        } else if (currentSum < sum) {
            left++;
        } else {
            right--;
        }
    }
}

let arr = [2, 4, 3, 1, 6, 5];
let sum = 7;
findPairs(arr, sum);

#include <stdio.h>
#include <stdlib.h>

int cmpfunc(const void *a, const void *b) {
    return (*(int *)a - *(int *)b);
}

void findPairs(int *arr, int n, int sum) {
    qsort(arr, n, sizeof(int), cmpfunc);
    int left = 0, right = n - 1;
    while (left < right) {
        int currentSum = arr[left] + arr[right];
        if (currentSum == sum) {
            printf("Pair found: (%d, %d)\n", arr[left], arr[right]);
            left++;
            right--;
        } else if (currentSum < sum) {
            left++;
        } else {
            right--;
        }
    }
}

int main() {
    int arr[] = {2, 4, 3, 1, 6, 5};
    int n = sizeof(arr) / sizeof(arr[0]);
    int sum = 7;
    findPairs(arr, n, sum);
    return 0;
}

using System;

public class TwoSum {
    public static void FindPairs(int[] arr, int sum) {
        Array.Sort(arr);
        int left = 0, right = arr.Length - 1;
        while (left < right) {
            int currentSum = arr[left] + arr[right];
            if (currentSum == sum) {
                Console.WriteLine($"Pair found: ({arr[left]}, {arr[right]})");
                left++;
                right--;
            } else if (currentSum < sum) {
                left++;
            } else {
                right--;
            }
        }
    }

    public static void Main(string[] args) {
        int[] arr = {2, 4, 3, 1, 6, 5};
        int sum = 7;
        FindPairs(arr, sum);
    }
}

This Python function sorts the list and then uses a two-pointer strategy to check pairs for their sum. The pairs are printed once discovered.

Complexity

Time Complexity: O(n log n) due to sorting.
Space Complexity: O(1), since the sort is in-place.

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Approach 4: Using Hash Table for O(n) Complexity

By utilizing a hash table, we can store the elements of the array and check for complements as we traverse the list. This allows us to achieve an O(n) time complexity for finding pairs that sum up to the given value.

Python

def find_pairs(arr, sum):
    seen = set()
    for num in arr:
        complement = sum - num
        if complement in seen:
            print(f"Pair found: ({complement}, {num})")
        seen.add(num)

arr = [2, 4, 3, 1, 6, 5]
sum = 7
find_pairs(arr, sum)

Java

import java.util.HashSet;

public class TwoSum {
    public static void findPairs(int[] arr, int sum) {
        HashSet<Integer> numSet = new HashSet<>();
        for (int num : arr) {
            int complement = sum - num;
            if (numSet.contains(complement)) {
                System.out.println("Pair found: (" + complement + ", " + num + ")");
            }
            numSet.add(num);
        }
    }

    public static void main(String[] args) {
        int[] arr = {2, 4, 3, 1, 6, 5};
        int sum = 7;
        findPairs(arr, sum);
    }
}

C++

#include <iostream>
#include <unordered_set>

void findPairs(int arr[], int n, int sum) {
    std::unordered_set<int> seen;
    for (int i = 0; i < n; i++) {
        int complement = sum - arr[i];
        if (seen.find(complement) != seen.end()) {
            std::cout << "Pair found: (" << complement << ", " << arr[i] << ")\n";
        }
        seen.insert(arr[i]);
    }
}

int main() {
    int arr[] = {2, 4, 3, 1, 6, 5};
    int n = sizeof(arr)/sizeof(arr[0]);
    int sum = 7;
    findPairs(arr, n, sum);
    return 0;
}

JavaScript

function findPairs(arr, sum) {
    let seen = new Set();
    for (let num of arr) {
        let complement = sum - num;
        if (seen.has(complement)) {
            console.log(`Pair found: (${complement}, ${num})`);
        }
        seen.add(num);
    }
}

let arr = [2, 4, 3, 1, 6, 5];
let sum = 7;
findPairs(arr, sum);

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>

#define TABLE_SIZE 1000

int hash(int key) {
  return abs(key) % TABLE_SIZE;
}

void findPairs(int *arr, int n, int sum) {
    bool hashTable[TABLE_SIZE] = {false};
    for (int i = 0; i < n; i++) {
        int complement = sum - arr[i];
        int hashedIndex = hash(complement);
        if (hashTable[hashedIndex]) {
            printf("Pair found: (%d, %d)\n", complement, arr[i]);
        }
        hashTable[hash(arr[i])] = true;
    }
}

int main() {
    int arr[] = {2, 4, 3, 1, 6, 5};
    int n = sizeof(arr) / sizeof(arr[0]);
    int sum = 7;
    findPairs(arr, n, sum);
    return 0;
}

using System;
using System.Collections.Generic;

public class TwoSum {
    public static void FindPairs(int[] arr, int sum) {
        HashSet<int> seen = new HashSet<int>();
        foreach (int num in arr) {
            int complement = sum - num;
            if (seen.Contains(complement)) {
                Console.WriteLine($"Pair found: ({complement}, {num})");
            }
            seen.Add(num);
        }
    }

    public static void Main(string[] args) {
        int[] arr = {2, 4, 3, 1, 6, 5};
        int sum = 7;
        FindPairs(arr, sum);
    }
}

This Python solution leverages a set to check if a complement for each number exists. This method ensures each check and addition to the set is executed in constant time.

Complexity

Time Complexity: O(n) due to set operations.
Space Complexity: O(n) for the set used to store seen numbers.

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Complexity comparison

Approach	Complexity
Approach 1: SQL Query with Aggregation and Grouping	Time Complexity: O(n log n) due to sorting in aggregation and grouping. Space Complexity: O(n) to store intermediate datasets in memory.
Approach 2: Structural Data Handling with In-Memory Computation	Time Complexity: O(n) as it involves two traversals over the dataset. Space Complexity: O(n) due to the additional space required for hashmaps/databases for storing intermediate results.
Using Sorting and Two-Pointer Technique	Time Complexity: O(n log n) due to sorting. Space Complexity: O(1), since the sort is in-place.
Using Hash Table for O(n) Complexity	Time Complexity: O(n) due to set operations. Space Complexity: O(n) for the set used to store seen numbers.

Complexity analysis

Approach	Time	Space	When to Use
SQL Aggregation and Grouping	O(n log n)	O(n)	Best when solving directly in a relational database using GROUP BY and HAVING
In-Memory Structural Processing (Python)	O(n)	O(n)	When the dataset is loaded into application memory and fast lookups are needed
Sorting + Two Pointer Scan	O(n log n)	O(1)–O(n)	Useful when hash tables are restricted or deterministic ordering is required
Hash Table Frequency Counting	O(n)	O(n)	Optimal general solution for application-layer implementations

Frequently asked questions

Is Investments in 2016 easy or hard?

Investments in 2016 is generally rated Medium difficulty. The challenge comes from recognizing two filtering conditions at once: duplicated tiv_2015 values and unique coordinates. Once translated into frequency counting or SQL aggregation, the implementation becomes straightforward.

Investments in 2016 Python/Java solution

In Python or Java, store frequency counts in hash maps. First count tiv_2015 and (lat, lon) occurrences, then iterate through the records and add tiv_2016 when tiv_2015 appears more than once and the coordinate is unique. The implementation runs in O(n) time with O(n) extra space.

How to solve Investments in 2016 in O(n)?

Build two hash maps: one counting how many times each tiv_2015 value appears and another counting how often each (lat, lon) coordinate appears. After the counting pass, scan the dataset again and sum tiv_2016 where tiv_2015 count is greater than 1 and the coordinate count equals 1. Each lookup is constant time, giving O(n) overall complexity.

What is the best approach for Investments in 2016?

The most practical approach uses hash tables to count frequencies of tiv_2015 values and (lat, lon) locations. After counting, iterate through the records and include tiv_2016 only when tiv_2015 appears more than once and the location appears exactly once. This runs in O(n) time with O(n) space and mirrors the logic used in optimized SQL queries.

Is Investments in 2016 asked at Google/Amazon/Meta?

Database filtering and aggregation problems similar to Investments in 2016 appear in interviews at companies like Amazon, Meta, and Google for data engineering or backend roles. Candidates are expected to demonstrate strong SQL skills, especially with GROUP BY, HAVING, and conditional aggregation.

What data structure is used in Investments in 2016?

The core data structures are hash tables (or dictionaries) used to count frequencies of tiv_2015 values and geographic coordinates. In SQL solutions, the equivalent concept is grouping and aggregation using GROUP BY.

What is the time complexity of Investments in 2016?

The optimal algorithm runs in O(n) time using hash tables to count tiv_2015 and location frequencies. SQL implementations using GROUP BY often behave closer to O(n log n) due to sorting or aggregation steps inside the query engine.

Investments in 2016 solution and explanation

The problem

Approach overview

Approach 1: Approach 1: SQL Query with Aggregation and Grouping

Approach 2: Approach 2: Structural Data Handling with In-Memory Computation

Approach 3: Using Sorting and Two-Pointer Technique

Approach 4: Using Hash Table for O(n) Complexity

Complexity comparison

Complexity analysis

Frequently asked questions

Comments

Investments in 2016 solution and explanation

The problem

Approach overview

Approach 1: Approach 1: SQL Query with Aggregation and Grouping

Approach 2: Approach 2: Structural Data Handling with In-Memory Computation

Approach 3: Using Sorting and Two-Pointer Technique

Approach 4: Using Hash Table for O(n) Complexity

Complexity comparison

Complexity analysis

Frequently asked questions

Comments

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