Maximize the Profit as the Salesman

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ArraysDynamic Programming

You are given an integer n representing the number of houses on a number line, numbered from 0 to n - 1.

Additionally, you are given a 2D integer array offers where offers[i] = [start_i, end_i, gold_i], indicating that i^th buyer wants to buy all the houses from start_i to end_i for gold_i amount of gold.

As a salesman, your goal is to maximize your earnings by strategically selecting and selling houses to buyers.

Return the maximum amount of gold you can earn.

Note that different buyers can't buy the same house, and some houses may remain unsold.

Example 1:

Input: n = 5, offers = [[0,0,1],[0,2,2],[1,3,2]]
Output: 3
Explanation: There are 5 houses numbered from 0 to 4 and there are 3 purchase offers.
We sell houses in the range [0,0] to 1^st buyer for 1 gold and houses in the range [1,3] to 3^rd buyer for 2 golds.
It can be proven that 3 is the maximum amount of gold we can achieve.

Example 2:

Input: n = 5, offers = [[0,0,1],[0,2,10],[1,3,2]]
Output: 10
Explanation: There are 5 houses numbered from 0 to 4 and there are 3 purchase offers.
We sell houses in the range [0,2] to 2^nd buyer for 10 golds.
It can be proven that 10 is the maximum amount of gold we can achieve.

Constraints:

1 <= n <= 10⁵
1 <= offers.length <= 10⁵
offers[i].length == 3
0 <= start_i <= end_i <= n - 1
1 <= gold_i <= 10³

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the prices and costs? Can they be negative, zero, or very large?
Is it guaranteed that there will always be at least one sales opportunity (i.e., a price and a cost)? What should I return if there are no sales opportunities?
Are the prices and costs provided in the same units (e.g., dollars, euros)?
Are there any constraints on the relationship between the prices and costs? (e.g., can the cost ever exceed the price for a given item?)
Should I return the maximum total profit, or also the corresponding prices and costs that generate that profit?

Brute Force Solution

Approach

The brute force approach to maximizing profit as a salesman is like trying every possible route to see which one makes the most money. We explore every combination of cities visited in different orders.

Here's how the algorithm would work step-by-step:

Start by considering all possible sequences of cities you could visit.
For each possible sequence, calculate the total profit you would make by visiting the cities in that order.
To calculate profit, look at the profit from each city you visit in that sequence and add them all up.
After calculating the profit for every possible sequence of cities, compare the profits from all the sequences.
Finally, choose the sequence that gives you the highest profit. This is the best route for the salesman.

Code Implementation

import itertools

def maximize_profit_salesman_brute_force(city_profits):
    number_of_cities = len(city_profits)
    city_indices = list(range(number_of_cities))

    # Generate all possible permutations of city visits.
    all_city_permutations = itertools.permutations(city_indices)

    maximum_profit = float('-inf')

    for city_permutation in all_city_permutations:
        current_profit = 0

        # Calculate the total profit for the current permutation
        for city_index in city_permutation:
            current_profit += city_profits[city_index]

        # Compare current profit with the max profit
        if current_profit > maximum_profit:
            maximum_profit = current_profit

            # Track the sequence yielding the maximum profit.
            best_route = city_permutation

    return maximum_profit

Big(O) Analysis

Time Complexity

O(n!) – The algorithm explores all possible sequences of cities to find the route with the maximum profit. For 'n' cities, there are n! (n factorial) possible permutations or sequences. Calculating the profit for each sequence involves iterating through all 'n' cities in that specific order. Thus, the overall time complexity is dominated by generating and evaluating all n! permutations, making the time complexity O(n!).

Space Complexity

O(N!) – The brute force approach explores all possible sequences (permutations) of cities. To achieve this, we implicitly use a recursive call stack. The maximum depth of the recursion is N, where N is the number of cities. At each level, we are creating temporary lists or copies of the city sequence to explore different permutations which accumulates to O(N!). Therefore, the space complexity is proportional to the number of permutations, which is N!.

Optimal Solution

Approach

The problem involves maximizing profit when visiting cities with varying profits and travel costs. The core idea is to dynamically track the best possible profit at each city, considering the cost to travel there from previous cities. We avoid redundant calculations by remembering the best profit we've achieved so far at each location.

Here's how the algorithm would work step-by-step:

Imagine you start at the first city. Your initial profit is simply the profit of that first city.
Now, consider the second city. You have two options: either start fresh at the second city and only take its profit, or travel from the first city to the second city. Choose the option that gives you the most profit, considering the travel cost.
Continue this process for each subsequent city. For each city, calculate the potential profit by traveling from every previous city. Also, consider starting fresh at the current city.
Choose the best profit among all the travel options and the 'starting fresh' option. This becomes the best profit you can have at the current city.
Repeat this process until you've calculated the best profit for every city.
The maximum profit among all the cities is your final answer. This represents the highest profit you can achieve by strategically visiting cities.

Code Implementation

def maximize_profit(profits, costs):
    number_of_cities = len(profits)
    max_profit_at_city = [0] * number_of_cities

    max_profit_at_city[0] = profits[0]

    for current_city in range(1, number_of_cities):
        # Consider starting fresh at the current city.
        max_profit_at_city[current_city] = profits[current_city]
        for previous_city in range(current_city):
            # Calculate profit from traveling from previous cities
            profit_from_travel = max_profit_at_city[previous_city] - costs[previous_city][current_city]

            # Choose between starting fresh and traveling.
            max_profit_at_city[current_city] = max(max_profit_at_city[current_city], profit_from_travel)

    # The overall maximum profit.
    overall_max_profit = 0
    for profit in max_profit_at_city:
        overall_max_profit = max(overall_max_profit, profit)

    return overall_max_profit

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through each of the n cities. For each city, it considers traveling from every previous city to the current city, resulting in a nested loop structure. Therefore, for each of the n cities, we perform a maximum of n calculations (traveling from each prior city or starting fresh), resulting in approximately n * n operations. This simplifies to a time complexity of O(n²).

Space Complexity

O(N) – The algorithm keeps track of the best possible profit at each city. This requires storing intermediate results in an auxiliary data structure, specifically an array or list, to remember the maximum profit achieved at each of the N cities. Thus, we use an array of size N to store the maximum profit up to each city. The space required grows linearly with the number of cities.

Edge Cases

Case	How to Handle
Null or empty price/cost arrays	Return 0 immediately as no sales can be made.
Price and cost arrays have different lengths	Return 0 or throw an IllegalArgumentException since profits cannot be calculated.
Prices are all lower than costs (always a loss)	Return 0 indicating no profitable sales can be made.
Prices or costs contain negative numbers	Throw IllegalArgumentException or treat as invalid input, depending on problem specification.
Integer overflow in profit calculation (price - cost)	Use long to store intermediate profit values to avoid overflow.
Very large arrays exceeding memory constraints	Consider using an iterative solution and streaming data processing if possible or throw an exception if the size is not supported.
Costs are all zero, and prices are positive	Return sum of prices since all sales are profitable.
Prices and costs contain extremely large values near the maximum integer limit	Ensure that the difference between price and cost does not cause integer overflow, potentially by using long data types or adjusting values to fit the integer bounds.

Null or empty price/cost arrays

How to Handle:

Return 0 immediately as no sales can be made.

Price and cost arrays have different lengths

How to Handle:

Return 0 or throw an IllegalArgumentException since profits cannot be calculated.

Prices are all lower than costs (always a loss)

How to Handle:

Return 0 indicating no profitable sales can be made.

Prices or costs contain negative numbers

How to Handle:

Throw IllegalArgumentException or treat as invalid input, depending on problem specification.

Integer overflow in profit calculation (price - cost)

How to Handle:

Use long to store intermediate profit values to avoid overflow.

Very large arrays exceeding memory constraints

How to Handle:

Consider using an iterative solution and streaming data processing if possible or throw an exception if the size is not supported.

Costs are all zero, and prices are positive

How to Handle:

Return sum of prices since all sales are profitable.

Prices and costs contain extremely large values near the maximum integer limit

How to Handle:

Ensure that the difference between price and cost does not cause integer overflow, potentially by using long data types or adjusting values to fit the integer bounds.