Minimize Max Distance to Gas Station

Hard

Amazon

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Topics:

ArraysBinary Search

You are given an array of gas station locations stations along a road, represented as an array of integers. You are also given an integer k, which represents the number of new gas stations you can add. Your task is to add these k new gas stations such that the maximum distance between adjacent gas stations is minimized. Return the minimized maximum distance.

For example:

stations = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], k = 9 should return 0.5. We can add one gas station between each existing gas station.
stations = [1, 2, 3, 4, 5], k = 4 should return 1.0. One possible solution is to add a gas station between each of the existing gas stations.
stations = [1, 10], k = 2 should return 3.0. We would add new gas stations at location 4 and 7.

Explain your approach, its time and space complexity, and handle potential edge cases. Provide code that solves this problem. The returned value should be precise to at least six decimal places. Can you implement an efficient solution to this problem?

Minimize Max Distance to Gas Station

Hard

Amazon

3 views

Topics:

ArraysBinary Search

You are given an array of gas station locations stations along a road, represented as an array of integers. You are also given an integer k, which represents the number of new gas stations you can add. Your task is to add these k new gas stations such that the maximum distance between adjacent gas stations is minimized. Return the minimized maximum distance.

For example:

stations = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], k = 9 should return 0.5. We can add one gas station between each existing gas station.
stations = [1, 2, 3, 4, 5], k = 4 should return 1.0. One possible solution is to add a gas station between each of the existing gas stations.
stations = [1, 10], k = 2 should return 3.0. We would add new gas stations at location 4 and 7.

Explain your approach, its time and space complexity, and handle potential edge cases. Provide code that solves this problem. The returned value should be precise to at least six decimal places. Can you implement an efficient solution to 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:

What are the possible ranges for the values in the `stations` array, and what is the maximum value for `k` (the number of gas stations to add)?
Are the elements in the `stations` array guaranteed to be sorted in ascending order?
Can the `stations` array be empty, or can `k` be zero? What should be the return value in these edge cases?
The problem asks to 'minimize the maximum distance'. If there are multiple possible placements of gas stations that result in the same minimum maximum distance, is any of those placements acceptable?
The values in the `stations` array are likely floating-point numbers. What precision is expected in the return value? Should the return value be an integer or a floating-point number?

Brute Force Solution

Approach

The problem asks us to add gas stations to minimize the largest distance between any two adjacent gas stations. The brute force method would try every single possible combination of where we could place these new gas stations. We then check each combination to see how good it is, and eventually pick the best one.

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

Start by trying to add the first new gas station at every possible location between the existing gas stations.
For each of these locations, consider all the possible places to put the second new gas station.
Keep doing this for all the new gas stations we need to add, creating every possible combination of gas station locations.
For each combination, calculate the largest distance between any two neighboring gas stations (including the new ones).
Keep track of the smallest of these largest distances that you find across all combinations.
After exploring all possible combinations of new gas station locations, the smallest largest distance we kept track of is our answer.

Code Implementation

def minimize_max_distance_brute_force(stations, new_stations_number):
    smallest_max_distance = float('inf')

    def calculate_max_distance(station_locations):
        max_distance = 0
        for i in range(len(station_locations) - 1):
            distance = station_locations[i+1] - station_locations[i]
            max_distance = max(max_distance, distance)
        return max_distance

    def find_best_arrangement(current_stations, stations_added):
        nonlocal smallest_max_distance

        if stations_added == new_stations_number:
            # All new gas stations have been added, check the max distance.
            max_distance = calculate_max_distance(sorted(current_stations))
            smallest_max_distance = min(smallest_max_distance, max_distance)
            return

        # Try adding a new station at every possible location
        for i in range(len(current_stations) + 1):
            
            new_stations = current_stations[:]
            new_stations.insert(i, current_stations[i-1] + (current_stations[i] - current_stations[i-1])/2 if i > 0 and i < len(current_stations) else (current_stations[0] / 2 if i == 0 else current_stations[-1] * 1.5))
            find_best_arrangement(new_stations, stations_added + 1)

    find_best_arrangement(stations, 0)
    return smallest_max_distance

Big(O) Analysis

Time Complexity

O(n^k) – The brute force approach involves considering all possible locations to insert k new gas stations among the existing n gas stations. For each of the k new stations, we essentially iterate through all possible locations between the existing stations, which are n+k-1 possible gaps to insert in. This means we have roughly (n+k-1) choices for the location of the first station, (n+k-1) choices for the second station, and so on for all k stations, resulting in (n+k-1)^k possible combinations. Since k is the input, this simplifies to O(n^k) where we’re assuming k is smaller than n.

Space Complexity

O(N^K) – The brute force solution explores all possible combinations of placing K new gas stations between the existing N-1 intervals, where N is the number of original gas stations. Generating all combinations implicitly requires storing these combinations. In the worst-case, we might need to store all N^K possible combinations (or a significant portion of them) before finding the optimal solution. Therefore, the auxiliary space used to store these combinations grows exponentially with K and polynomially with N.

Optimal Solution

Approach

We're trying to find the smallest possible largest distance between any two gas stations after adding new stations. A straightforward approach would be too slow, so we use a clever strategy of repeatedly refining our estimate of the smallest maximum distance to quickly narrow down the possibilities.

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

Start by guessing a maximum distance between gas stations. This guess will be in between the minimum and maximum possible distances between gas stations.
Check how many new gas stations you would need to add to ensure that no two existing stations are farther apart than your guessed distance.
If the number of gas stations you need to add is less than or equal to the number of stations you are allowed to add, then this guess is feasible (or even too big). This means we can try a smaller guess.
If the number of gas stations you need to add is more than the number of stations you are allowed to add, then this guess is too small. This means we need to try a larger guess.
Continue adjusting your guess up or down until you've found the smallest possible maximum distance that allows you to add all the necessary gas stations.

Code Implementation

def minimize_max_distance_to_gas_station(stations, number_of_new_stations):
    left_distance = 0
    right_distance = max(stations[i+1] - stations[i] for i in range(len(stations)-1))

    while right_distance - left_distance > 1e-6:
        middle_distance = (left_distance + right_distance) / 2.0
        
        # Need to calculate how many gas stations to add
        number_of_needed_stations = 0
        for i in range(len(stations) - 1):
            number_of_needed_stations += int((stations[i+1] - stations[i]) / middle_distance)
        
        # Binary search based on whether the added stations exceed limit
        if number_of_needed_stations <= number_of_new_stations:
            # Current middle_distance is feasible
            right_distance = middle_distance
        else:
            # Current middle_distance is not feasible
            left_distance = middle_distance
    
    return right_distance

Big(O) Analysis

Time Complexity

O(n log M) – The algorithm uses binary search to find the optimal maximum distance. The search space is between the minimum and maximum distance between gas stations, which we denote as M. Inside the binary search loop, we iterate through the stations array of size n to calculate how many additional gas stations are needed for a given maximum distance. Therefore, each step of the binary search takes O(n) time. Since binary search reduces the search space by half in each step, the total number of iterations is O(log M). Consequently, the overall time complexity is O(n log M).

Space Complexity

O(1) – The provided algorithm primarily involves iterative refinement of a guess for the maximum distance. It doesn't seem to explicitly create any auxiliary data structures like arrays, hash maps, or lists to store intermediate values. The space used mainly consists of a few variables to keep track of the guess, the lower and upper bounds, and the number of gas stations needed, all of which take up constant space, independent of the number of existing gas stations (N).

Edge Cases

Case	How to Handle
stations is null or empty	Return 0 if stations is null or empty as no maximum distance exists.
k is 0	Return the initial maximum distance between existing stations when no new stations are added.
stations has only one station	Return 0, as adding k stations won't change the position of the one existing station.
k is a very large number (much larger than the number of stations)	The solution should still converge to the minimal max distance by effectively filling all gaps.
Stations are very close together resulting in very small distances between them.	Ensure the binary search precision is high enough to differentiate very small distances and doesn't cause an infinite loop.
Stations are very far apart resulting in very large distances between them.	The binary search range needs to be large enough to accommodate the initial maximum distance.
Floating point precision issues during binary search	Use a sufficiently small tolerance (e.g., 1e-6) for comparing floating-point numbers in the binary search condition to avoid infinite loops and wrong results.
Integer overflow in calculating the number of required stations	Use long or double to store intermediate calculation results to avoid integer overflow if distance or k is large.