LeetCode 585: Investments in 2016 Solution in Python – A Step-by-Step Guide

What Is LeetCode 585: Investments in 2016?

In LeetCode 585: Investments in 2016, you’re given a list of tuples insurance where each tuple (pid, tiv_2015, tiv_2016, lat, lon) represents a customer’s policy ID, their total insured value (TIV) in 2015 and 2016, and their latitude and longitude coordinates. Your task is to return the sum of tiv_2016 for all records from 2016 where: (1) the customer’s location (lat, lon) is unique (no other customer at the exact same coordinates), and (2) there exists at least one other customer with the same tiv_2015 but a different location. For example, with insurance = [(1, 1000, 2000, 10, 20), (2, 1000, 3000, 10, 20), (3, 1000, 4000, 30, 40)], the result is 4000 (only customer 3 qualifies). This problem builds on LeetCode 349: Intersection of Two Arrays for data filtering but focuses on aggregation with multiple conditions.

Problem Statement

• Input: insurance (List[Tuple[int, float, float, float, float]])—list of (pid, tiv_2015, tiv_2016, lat, lon).
• Output: Float—sum of tiv_2016 for qualifying 2016 investments.
• Rules: Include tiv_2016 if location unique and tiv_2015 matches another customer at a different location; round to 2 decimal places.

Constraints

• 1 <= insurance.length <= 10⁴
• 1 <= pid <= 10⁴ (unique)
• 0 <= tiv_2015, tiv_2016 <= 10⁴
• -1000 ≤ lat, lon ≤ 1000 (coordinates)

Examples

• Input: insurance = [(1,1000,2000,10,20),(2,1000,3000,10,20),(3,1000,4000,30,40)]
• Output: 4000.00
• P1, P2: Same location (10,20), P3: Unique (30,40), has tiv_2015=1000 match elsewhere.
• Input: insurance = [(1,500,1000,10,20),(2,500,1500,20,30)]
• Output: 2500.00
• Both unique and match tiv_2015=500, sum = 1000 + 1500.
• Input: insurance = [(1,1000,2000,10,20)]
• Output: 0.00
• No tiv_2015 match at different location.

Understanding the Problem: Summing Investments

To solve LeetCode 585: Investments in 2016 in Python, we need to sum tiv_2016 values for customers meeting two conditions: their location (lat, lon) is unique, and their tiv_2015 matches at least one other customer at a different location, handling up to 10⁴ records efficiently. A brute-force approach checking each record against all others (O(n²)) could work but would be slow. Instead, a dictionary-based approach preprocesses data in O(n) time, grouping by location and tiv_2015 to filter and aggregate effectively. We’ll explore:

• Best Solution (Dictionary-Based Aggregation with Filtering): O(n) time, O(n) space—fast and optimal.
• Alternative Solution (Brute-Force Traversal): O(n²) time, O(1) space—simple but slow.

Let’s dive into the dictionary solution with a friendly breakdown!

Best Solution: Dictionary-Based Aggregation with Filtering

Why Dictionary Wins

The dictionary-based aggregation with filtering solution is the best for LeetCode 585 because it computes the sum in O(n) time and O(n) space by using dictionaries to group customers by location and tiv_2015, then applying the conditions in a single pass, efficiently avoiding redundant comparisons. It’s like organizing customer files by address and investment history, quickly picking out the unique earners—all in a neat, streamlined process!

How It Works

Think of this as an investment-summing organizer:

• Step 1: Build Location Dictionary:
• Map (lat, lon) to list of (pid, tiv_2015, tiv_2016) tuples.
• Step 2: Build TIV_2015 Dictionary:
• Map tiv_2015 to list of (pid, lat, lon) tuples.
• Step 3: Filter and Sum:
• For each location:
• If only one customer (unique), check tiv_2015 matches another location.
• Add tiv_2016 to sum if conditions met.
• Step 4: Return Result:
• Rounded sum of qualifying tiv_2016 values.
• Why It Works:
• Dictionaries enable O(1) lookups for grouping.
• Single pass ensures efficiency.

It’s like an investment-tallying maestro!

Step-by-Step Example

Example: insurance = [(1,1000,2000,10,20),(2,1000,3000,10,20),(3,1000,4000,30,40)]

• Step 1: Location dictionary:
• loc_data = {(10,20): [(1,1000,2000), (2,1000,3000)], (30,40): [(3,1000,4000)]}.
• Step 2: TIV_2015 dictionary:
• tiv_2015_data = {1000: [(1,10,20), (2,10,20), (3,30,40)]}.
• Step 3: Filter and sum:
• (10,20): 2 customers, not unique, skip.
• (30,40): 1 customer (PID 3), tiv_2015=1000 matches (1,2) at (10,20), add 4000.
• Sum = 4000.
• Step 4: Result: 4000.00.
• Result: 4000.00.

Example: insurance = [(1,500,1000,10,20),(2,500,1500,20,30)]

• Step 1: loc_data = {(10,20): [(1,500,1000)], (20,30): [(2,500,1500)]}.
• Step 2: tiv_2015_data = {500: [(1,10,20), (2,20,30)]}.
• Step 3:
• (10,20): Unique, tiv_2015=500 matches (2) at (20,30), add 1000.
• (20,30): Unique, tiv_2015=500 matches (1) at (10,20), add 1500.
• Sum = 2500.
• Result: 2500.00.

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained for beginners:

Copy codefrom typing import List, Tuple
from collections import defaultdict

class Solution:
    def findInvestments(self, insurance: List[Tuple[int, float, float, float, float]]) -> float:
        # Step 1: Build dictionaries
        loc_data = defaultdict(list)  # (lat, lon) -> [(pid, tiv_2015, tiv_2016)]
        tiv_2015_data = defaultdict(list)  # tiv_2015 -> [(pid, lat, lon)]

        for pid, tiv_2015, tiv_2016, lat, lon in insurance:
            loc_data[(lat, lon)].append((pid, tiv_2015, tiv_2016))
            tiv_2015_data[tiv_2015].append((pid, lat, lon))

        # Step 2: Filter and sum tiv_2016
        total = 0.0
        for loc, records in loc_data.items():
            if len(records) == 1:  # Unique location
                pid, tiv_2015, tiv_2016 = records[0]
                # Check if tiv_2015 matches another location
                for other_pid, other_lat, other_lon in tiv_2015_data[tiv_2015]:
                    if (other_lat, other_lon) != loc:
                        total += tiv_2016
                        break

        # Step 3: Return rounded result
        return round(total, 2)

• Lines 7-9: Initialize defaultdicts for location and TIV_2015 grouping.
• Lines 11-13: Populate dictionaries with records.
• Lines 16-23: Filter:
• Check if location has one record (unique).
• Verify tiv_2015 matches another location, add tiv_2016.
• Line 26: Return sum rounded to 2 decimals.
• Time Complexity: O(n)—single pass to build, linear checks per unique location.
• Space Complexity: O(n)—dictionary storage.

It’s like an investment-summing wizard!

Alternative Solution: Brute-Force Traversal

Why an Alternative Approach?

The brute-force traversal approach checks each customer against all others to determine uniqueness and TIV_2015 matches, running in O(n²) time and O(1) space (excluding output). It’s simple but inefficient for large lists, making it a good baseline for small datasets or understanding.

How It Works

Picture this as an investment-scanning seeker:

• Step 1: Initialize sum.
• Step 2: For each customer, check uniqueness and TIV_2015 match.
• Step 3: Add qualifying tiv_2016 to sum.
• Step 4: Return rounded sum.

It’s like a manual investment checker!

Step-by-Step Example

Example: insurance = [(1,1000,2000,10,20),(2,1000,3000,10,20)]

• Step 1: total = 0.
• Step 2: Check:
• P1: (10,20) has 2, not unique, skip.
• P2: (10,20) has 2, not unique, skip.
• Step 3: Total: 0.
• Result: 0.00.

Code for Brute-Force Approach

Copy codefrom typing import List, Tuple

class Solution:
    def findInvestments(self, insurance: List[Tuple[int, float, float, float, float]]) -> float:
        total = 0.0

        for i, (pid1, tiv_2015_1, tiv_2016_1, lat1, lon1) in enumerate(insurance):
            # Check if location is unique
            is_unique = True
            for j, (_, _, _, lat2, lon2) in enumerate(insurance):
                if i != j and lat1 == lat2 and lon1 == lon2:
                    is_unique = False
                    break

            # If unique, check tiv_2015 match at different location
            if is_unique:
                has_match = False
                for j, (_, tiv_2015_2, _, lat2, lon2) in enumerate(insurance):
                    if i != j and tiv_2015_1 == tiv_2015_2 and (lat1 != lat2 or lon1 != lon2):
                        has_match = True
                        break
                if has_match:
                    total += tiv_2016_1

        return round(total, 2)

• Line 7: Initialize sum.
• Lines 10-15: Check location uniqueness with nested loop.
• Lines 18-24: If unique, check tiv_2015 match at different location.
• Line 27: Return rounded sum.
• Time Complexity: O(n²)—nested loops.
• Space Complexity: O(1)—minimal space.

It’s a brute-force investment summer!

Comparing the Two Solutions

• Dictionary (Best):
• Pros: O(n), O(n), fast.
• Cons: Extra space for dictionaries.
• Brute-Force (Alternative):
• Pros: O(n²), O(1), space-efficient.
• Cons: Slower.

Dictionary wins for efficiency!

Additional Examples and Edge Cases

• All same location: 0 if > 1 customer.
• No matches: 0 if no tiv_2015 pairs.
• Empty list: 0.

Dictionary handles them all!

Complexity Recap

• Dictionary: Time O(n), Space O(n).
• Brute-Force: Time O(n²), Space O(1).

Dictionary’s the speed champ!

Key Takeaways

• Dictionary: Aggregation finesse—learn at Python Basics!
• Brute-Force: Scan simplicity.
• Investments: Summing is fun.
• Python: Dict or loops, your pick!

Final Thoughts: Sum Those Investments!

LeetCode 585: Investments in 2016 in Python is a practical data challenge. Dictionary-based aggregation is your fast track, while brute-force offers a thorough dive. Want more? Try LeetCode 349: Intersection of Two Arrays or LeetCode 560: Subarray Sum Equals K. Ready to tally? Head to Solve LeetCode 585 on LeetCode and sum those 2016 investments today!