Minimum Amount of Time to Collect Garbage

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ArraysStrings

You are given a 0-indexed array of strings garbage where garbage[i] represents the assortment of garbage at the i<sup>th</sup> house. garbage[i] consists only of the characters 'M', 'P' and 'G' representing one unit of metal, paper and glass garbage respectively. Picking up one unit of any type of garbage takes 1 minute.

You are also given a 0-indexed integer array travel where travel[i] is the number of minutes needed to go from house i to house i + 1.

There are three garbage trucks in the city, each responsible for picking up one type of garbage. Each garbage truck starts at house 0 and must visit each house in order; however, they do not need to visit every house.

Only one garbage truck may be used at any given moment. While one truck is driving or picking up garbage, the other two trucks cannot do anything.

Return the minimum number of minutes needed to pick up all the garbage.

For example:

garbage = ["G","P","GP","GG"], travel = [2,4,3]

In this example, the output would be 21. The paper garbage truck travels from house 0 to house 1 (2 minutes), collects the paper garbage at house 1 (1 minute), travels from house 1 to house 2 (4 minutes), and collects the paper garbage at house 2 (1 minute). Altogether, it takes 8 minutes to pick up all the paper garbage.

The glass garbage truck collects the glass garbage at house 0 (1 minute), travels from house 0 to house 1 (2 minutes), travels from house 1 to house 2 (4 minutes), collects the glass garbage at house 2 (1 minute), travels from house 2 to house 3 (3 minutes), and collects the glass garbage at house 3 (2 minutes). Altogether, it takes 13 minutes to pick up all the glass garbage.

Since there is no metal garbage, we do not need to consider the metal garbage truck. Therefore, it takes a total of 8 + 13 = 21 minutes to collect all the garbage.

Can you provide an efficient algorithm to solve this problem?

Minimum Amount of Time to Collect Garbage

Medium

Amazon

3 views

Topics:

ArraysStrings

You are given a 0-indexed array of strings garbage where garbage[i] represents the assortment of garbage at the i<sup>th</sup> house. garbage[i] consists only of the characters 'M', 'P' and 'G' representing one unit of metal, paper and glass garbage respectively. Picking up one unit of any type of garbage takes 1 minute.

You are also given a 0-indexed integer array travel where travel[i] is the number of minutes needed to go from house i to house i + 1.

There are three garbage trucks in the city, each responsible for picking up one type of garbage. Each garbage truck starts at house 0 and must visit each house in order; however, they do not need to visit every house.

Only one garbage truck may be used at any given moment. While one truck is driving or picking up garbage, the other two trucks cannot do anything.

Return the minimum number of minutes needed to pick up all the garbage.

For example:

garbage = ["G","P","GP","GG"], travel = [2,4,3]

In this example, the output would be 21. The paper garbage truck travels from house 0 to house 1 (2 minutes), collects the paper garbage at house 1 (1 minute), travels from house 1 to house 2 (4 minutes), and collects the paper garbage at house 2 (1 minute). Altogether, it takes 8 minutes to pick up all the paper garbage.

The glass garbage truck collects the glass garbage at house 0 (1 minute), travels from house 0 to house 1 (2 minutes), travels from house 1 to house 2 (4 minutes), collects the glass garbage at house 2 (1 minute), travels from house 2 to house 3 (3 minutes), and collects the glass garbage at house 3 (2 minutes). Altogether, it takes 13 minutes to pick up all the glass garbage.

Since there is no metal garbage, we do not need to consider the metal garbage truck. Therefore, it takes a total of 8 + 13 = 21 minutes to collect all the garbage.

Can you provide an efficient algorithm to solve this problem?

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:

Can you provide the constraints on the length of the `garbage` array and the length of each string within it? Are there practical upper bounds?
What characters are allowed in the strings within the `garbage` array? Specifically, are there only 'M', 'P', and 'G' characters, or could there be others?
Can the `travel` array contain zero values, and what does a zero value signify in this context?
If a garbage truck doesn't need to collect a particular type of garbage (e.g., no 'M' in any house), do we still need to travel to those houses with that truck, or can that truck skip those houses?
If a house has multiple types of garbage (e.g., 'MPG'), should the cost of visiting that house be accounted for by each corresponding truck or only once?

Brute Force Solution

Approach

The brute force approach means we'll try every single possible path the garbage trucks can take. We will calculate the total time for each path and then pick the one that takes the least amount of time.

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

Consider all possible orderings in which the garbage trucks can visit the houses.
For each ordering, figure out the exact route each truck will take to collect its specific type of garbage, minimizing travel within that ordering.
Calculate the total time taken by all trucks for that specific ordering of houses and their routes.
Repeat the above steps for every possible ordering.
Compare the total time calculated for each ordering and select the one with the shortest time. This is our answer.

Code Implementation

import itertools

def minimum_time_to_collect_garbage_brute_force(garbage, travel):
    shortest_time = float('inf')
    number_of_houses = len(garbage)

    # Generate all possible orderings of houses
    for house_ordering in itertools.permutations(range(number_of_houses)):
        total_time_for_ordering = 0
        
        # Iterate through each type of garbage
        for garbage_type in ['M', 'P', 'G']:
            time_for_this_garbage = 0
            current_house_index = 0
            last_house_with_garbage = -1

            # Find the last house with this type of garbage
            for i in range(number_of_houses):
                if garbage_type in garbage[house_ordering[i]]:
                    last_house_with_garbage = i

            if last_house_with_garbage == -1:
                continue
            
            # Calculate time to collect garbage from each house.
            for i in range(number_of_houses):
                house_index_in_ordering = i

                if garbage_type in garbage[house_ordering[i]]:
                    time_for_this_garbage += garbage[house_ordering[i]].count(garbage_type)

                # Calculate travel time to the next house.  Stop at last house
                if i < last_house_with_garbage:

                    travel_index = min(house_ordering[i],house_ordering[i+1])
                    travel_index_two = max(house_ordering[i],house_ordering[i+1])
                    houses_between = travel_index_two-travel_index

                    # Travel time is calculated
                    for j in range(travel_index,travel_index_two):
                        time_for_this_garbage += travel[j]

            # Add up travel and garbage collection time
            total_time_for_ordering += time_for_this_garbage

        # Keep track of the shortest time
        shortest_time = min(shortest_time, total_time_for_ordering)

    return shortest_time

Big(O) Analysis

Time Complexity

O(n! * n^2) – The core of the brute force approach involves exploring all possible orderings of houses, which contributes a factorial component of O(n!) where n is the number of houses. For each ordering, we then need to determine the optimal route for each garbage truck. This route calculation involves iterating through the houses for each garbage type (metal, paper, glass), which takes O(n) time. Since we may need to do this for up to three garbage types, the route calculation within each ordering is O(n). The overall time calculation will be approximately n! * n * n, thus the algorithm has a time complexity of O(n! * n^2).

Space Complexity

O(N!) – The algorithm considers all possible orderings of houses for garbage collection. Generating all permutations of houses, where N is the number of houses, requires storing these permutations. The number of permutations grows factorially with the number of houses, specifically N!. Therefore, the space needed to store these permutations is proportional to N!. Thus the space complexity is O(N!).

Optimal Solution

Approach

The key to solving this efficiently is to realize we only need to visit each house once for each garbage type. We can precompute the last house each garbage truck needs to visit and the total garbage of each type. Then, for each garbage type, we simply add up the travel time to the last house and the amount of garbage of that type.

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

First, figure out the very last house that each type of garbage truck (metal, paper, glass) needs to visit.
Next, calculate the total amount of each type of garbage there is in all the houses.
Now, for each type of garbage, add up the travel time from the starting house to the last house that garbage truck visits.
Finally, add the total amount of each type of garbage to the travel time for that type to get the total time for each garbage truck.
Add up the total time for each garbage truck to find the overall minimum time.

Code Implementation

def minimum_garbage_collection_time(garbage, travel):
    last_house_metal = 0
    last_house_paper = 0
    last_house_glass = 0
    total_metal = 0
    total_paper = 0
    total_glass = 0

    for i in range(len(garbage)):
        if 'M' in garbage[i]:
            last_house_metal = i
            total_metal += garbage[i].count('M')
        if 'P' in garbage[i]:
            last_house_paper = i
            total_paper += garbage[i].count('P')
        if 'G' in garbage[i]:
            last_house_glass = i
            total_glass += garbage[i].count('G')

    travel_time_metal = 0
    travel_time_paper = 0
    travel_time_glass = 0

    # Calculate travel time to the last house for each garbage type
    for i in range(last_house_metal):
        travel_time_metal += travel[i]

    for i in range(last_house_paper):
        travel_time_paper += travel[i]

    for i in range(last_house_glass):
        travel_time_glass += travel[i]

    # Sum garbage collection time and travel time for each type
    total_time_metal = total_metal + travel_time_metal
    total_time_paper = total_paper + travel_time_paper
    total_time_glass = total_glass + travel_time_glass

    # Add up all the times to get the final result.
    total_time = total_time_metal + total_time_paper + total_time_glass

    return total_time

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the `garbage` array of size `n` (number of houses) to find the last occurrence of each garbage type. This takes O(n) time. Then, it iterates again through the same array to calculate the total amount of each garbage type, which is also O(n). Finally, it iterates through the `travel` array (also of size proportional to n) to calculate the travel time to the last house for each garbage type, which again is O(n). Since these are sequential linear operations, the overall time complexity is O(n + n + n) which simplifies to O(n).

Space Complexity

O(1) – The algorithm stores a fixed number of variables to keep track of the last house visited by each garbage type and the total amount of each garbage type. These variables consume a constant amount of space regardless of the number of houses or the amount of garbage at each house. Therefore, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Null or empty garbage array	Return 0 immediately, as no garbage needs collection.
Null or empty travel array	If garbage array is not empty, treat travel as if it had zero travel time for all trucks.
Garbage array with only one string	Calculate the time to collect that single garbage string plus travel time if not the first house.
Travel array shorter than the number of houses with garbage	Assume zero travel time for houses beyond the defined travel array size, essentially staying at the last valid house.
Garbage array containing empty strings	Treat empty strings as houses with no garbage of that type, skipping collection for that specific truck type at that house.
Travel array containing very large travel times	Ensure integer overflow doesn't occur when summing travel times; use appropriate data types (e.g., long).
Garbage array with all garbage of only one type	Only the truck corresponding to that garbage type is needed, and the algorithm should correctly compute the travel time.
Houses with no garbage, represented by empty strings in the garbage array.	The algorithm should skip these houses when calculating the travel distance for each garbage truck, avoiding unnecessary travel.