How do you solve Investments in 2016 on LeetCode?

Investments in 2016 (LeetCode #585) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using hash maps can significantly reduce time complexity.

What is the brute force solution for Investments in 2016?

The brute force approach for Investments in 2016 is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking each policyholder against every other policyholder to find those with the same tiv_2015 value and unique locations. This method is straightforward but inefficient for larger datasets. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Investments in 2016?

Investments in 2016 uses the following concepts: Database. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Investments in 2016?

Clearly explain your thought process as you write code. Consider edge cases, such as all policies having the same tiv_2015. Practice writing SQL queries if this problem is presented in a database context.

What are common mistakes when solving Investments in 2016?

Not checking for unique locations properly. Forgetting to round the final result.

#585

Investments in 2016

Medium

DatabaseHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses hash maps to efficiently track the counts of tiv_2015 values and unique locations. This reduces the need for nested loops, allowing us to process the data in a single pass.

⚙️

Algorithm

5 steps

1Step 1: Create a hash map to count occurrences of each tiv_2015 value.
2Step 2: Create a set to track unique (lat, lon) pairs.
3Step 3: Iterate through the insurance records to populate the hash map and set.
4Step 4: Iterate through the records again to sum tiv_2016 values for those with matching tiv_2015 and unique locations.
5Step 5: Round the total sum to two decimal places and return it.

solution.py11 lines

1# Full working Python code
2import pandas as pd
3
4def investment_sum(insurances):
5    tiv_count = insurances['tiv_2015'].value_counts()
6    unique_locations = set((insurances['lat'][i], insurances['lon'][i]) for i in range(len(insurances)))
7    total = 0.0
8    for i in range(len(insurances)):
9        if tiv_count[insurances['tiv_2015'][i]] > 1 and (insurances['lat'][i], insurances['lon'][i]) in unique_locations:
10            total += insurances['tiv_2016'][i]
11    return round(total, 2)

ℹ

Complexity note: The time complexity is O(n) because we are iterating through the records a constant number of times. The space complexity is O(n) due to the storage of counts and unique locations in hash maps and sets.

1Using hash maps can significantly reduce time complexity.
2Understanding unique constraints is crucial for filtering results.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.