Minimum Number of Refueling Stops

Hard

DE Shaw

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Topics:

ArraysDynamic ProgrammingGreedy Algorithms

A car travels from a starting position to a destination which is target miles east of the starting position.

There are gas stations along the way. The gas stations are represented as an array stations where stations[i] = [position_i, fuel_i] indicates that the i^th gas station is position_i miles east of the starting position and has fuel_i liters of gas.

The car starts with an infinite tank of gas, which initially has startFuel liters of fuel in it. It uses one liter of gas per one mile that it drives. When the car reaches a gas station, it may stop and refuel, transferring all the gas from the station into the car.

Return the minimum number of refueling stops the car must make in order to reach its destination. If it cannot reach the destination, return -1.

Note that if the car reaches a gas station with 0 fuel left, the car can still refuel there. If the car reaches the destination with 0 fuel left, it is still considered to have arrived.

Example 1:

Input: target = 1, startFuel = 1, stations = []
Output: 0
Explanation: We can reach the target without refueling.

Example 2:

Input: target = 100, startFuel = 1, stations = [[10,100]]
Output: -1
Explanation: We can not reach the target (or even the first gas station).

Example 3:

Input: target = 100, startFuel = 10, stations = [[10,60],[20,30],[30,30],[60,40]]
Output: 2
Explanation: We start with 10 liters of fuel.
We drive to position 10, expending 10 liters of fuel.  We refuel from 0 liters to 60 liters of gas.
Then, we drive from position 10 to position 60 (expending 50 liters of fuel),
and refuel from 10 liters to 50 liters of gas.  We then drive to and reach the target.
We made 2 refueling stops along the way, so we return 2.

Constraints:

1 <= target, startFuel <= 10⁹
0 <= stations.length <= 500
1 <= position_i < position_i+1 < target
1 <= fuel_i < 10⁹

Minimum Number of Refueling Stops

Hard

DE Shaw

0 views

Topics:

ArraysDynamic ProgrammingGreedy Algorithms

A car travels from a starting position to a destination which is target miles east of the starting position.

There are gas stations along the way. The gas stations are represented as an array stations where stations[i] = [position_i, fuel_i] indicates that the i^th gas station is position_i miles east of the starting position and has fuel_i liters of gas.

The car starts with an infinite tank of gas, which initially has startFuel liters of fuel in it. It uses one liter of gas per one mile that it drives. When the car reaches a gas station, it may stop and refuel, transferring all the gas from the station into the car.

Return the minimum number of refueling stops the car must make in order to reach its destination. If it cannot reach the destination, return -1.

Note that if the car reaches a gas station with 0 fuel left, the car can still refuel there. If the car reaches the destination with 0 fuel left, it is still considered to have arrived.

Example 1:

Input: target = 1, startFuel = 1, stations = []
Output: 0
Explanation: We can reach the target without refueling.

Example 2:

Input: target = 100, startFuel = 1, stations = [[10,100]]
Output: -1
Explanation: We can not reach the target (or even the first gas station).

Example 3:

Input: target = 100, startFuel = 10, stations = [[10,60],[20,30],[30,30],[60,40]]
Output: 2
Explanation: We start with 10 liters of fuel.
We drive to position 10, expending 10 liters of fuel.  We refuel from 0 liters to 60 liters of gas.
Then, we drive from position 10 to position 60 (expending 50 liters of fuel),
and refuel from 10 liters to 50 liters of gas.  We then drive to and reach the target.
We made 2 refueling stops along the way, so we return 2.

Constraints:

1 <= target, startFuel <= 10⁹
0 <= stations.length <= 500
1 <= position_i < position_i+1 < target
1 <= fuel_i < 10⁹

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 starting fuel, target distance, and each station's position and fuel capacity? Should I expect integers or floating-point numbers?
Can the starting fuel be zero? What if the target distance is zero, or if the initial starting location is already at or beyond the target?
If it's impossible to reach the target, what should I return? Should I return -1, or is there another specified value?
Are the refueling stations guaranteed to be sorted by their location along the route, or do I need to sort them first?
If I reach a station with zero fuel capacity, can I still refuel there (essentially meaning I can't get any fuel)? Or should I skip it?

Brute Force Solution

Approach

The brute force approach to finding the fewest refueling stops involves trying every possible combination of gas stations. We essentially explore every possible path, checking if it leads to our destination, and then pick the shortest path, i.e., the path with the fewest stops.

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

First, consider traveling to the destination without stopping at any gas stations. See if you can make it.
If you can't, try stopping at just one gas station. Consider each gas station one at a time and see if stopping there allows you to reach the destination.
If stopping at only one gas station doesn't work, try stopping at two gas stations. Consider every possible pair of gas stations and see if any combination allows you to reach the destination.
Continue this process, trying every possible combination of three gas stations, then four, and so on, until you've tried stopping at all of them.
For each combination of gas stations, check if you can reach the destination from the starting point by stopping at those stations in a valid order. A valid order means you can reach each station from the previous one (either the starting point or another gas station in your combination).
Keep track of the minimum number of refueling stops that allows you to reach the destination.
If you've tried every combination of gas stations and you still can't reach the destination, then it's impossible to reach the destination, so return -1.

Code Implementation

def min_refuel_stops_brute_force(target_distance, start_fuel, stations):
    number_of_stations = len(stations)

    # Iterate through all possible numbers of refueling stops, starting from 0
    for number_of_stops in range(number_of_stations + 1):
        # Iterate through all possible combinations of gas stations for the current number of stops
        for combination in combinations(range(number_of_stations), number_of_stops):
            # Check if the combination of stations allows reaching the target
            if can_reach_destination(target_distance, start_fuel, stations, combination):
                return number_of_stops

    return -1

from itertools import combinations

def can_reach_destination(target_distance, start_fuel, stations, combination):
    current_fuel = start_fuel
    current_position = 0
    stations_visited = sorted([stations[i] for i in combination])
    
    # Iterate through the stations and check if we can reach each one
    for station_index in range(len(stations_visited)):
        distance_to_station = stations_visited[station_index][0] - current_position

        # We can't reach this station with our current fuel, this path is invalid
        if current_fuel < distance_to_station:
            return False

        current_fuel -= distance_to_station
        current_position = stations_visited[station_index][0]
        current_fuel += stations_visited[station_index][1]

    # Check if we can reach the destination from the last station
    distance_to_target = target_distance - current_position

    if current_fuel >= distance_to_target:
        return True
    else:
        return False

# Testing
# target_distance = 1000
# start_fuel = 100
# stations = [[100, 200], [200, 400], [300, 600], [400, 800], [600, 500]]
# result = min_refuel_stops_brute_force(target_distance, start_fuel, stations)
# print(result)

# target_distance = 100
# start_fuel = 10
# stations = [[10,60],[20,30],[30,30],[60,40]]
# result = min_refuel_stops_brute_force(target_distance, start_fuel, stations)
# print(result)

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(2^n) – The algorithm explores all possible subsets of gas stations to find the minimum number of refueling stops. Given 'n' gas stations, there are 2^n possible subsets (each gas station can either be included or excluded in a stop). For each subset, we need to check if it's a valid path to the destination, which takes O(n) time in the worst case to check distances between the stations in the subset and from the start. Therefore, the overall time complexity is O(n * 2^n). Since exponential growth dominates, we simplify it to O(2^n).

Space Complexity

O(2^N) – The algorithm explores every possible combination of gas stations. In the worst-case scenario, where N is the number of gas stations, it generates all possible subsets of gas stations. Each subset represents a potential refueling strategy, and there can be 2^N such subsets. Therefore, the space complexity arises from the implicit storage needed to manage and potentially store these combinations during the recursive exploration, although not explicitly stored, the function calls and state management for these combinations contributes to the space. This results in exponential space complexity.

Optimal Solution

Approach

The best way to solve this problem is to think about it step-by-step, always making the smartest decision at each gas station. The key is to prioritize refueling at stations that let us travel the furthest.

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

First, figure out how far you can travel initially with your starting fuel.
As you travel, keep track of all the gas stations you've passed so far and how much fuel they offer.
If you run out of fuel before reaching your destination, find the gas station you've passed that has the most fuel and refuel there.
Add the fuel from that station to your tank and continue your journey.
Repeat the process of finding the best gas station to refuel at whenever you run out of fuel until you either reach your destination or you can't reach the next gas station.
If you can't reach the next gas station, it means you can't reach your destination. Return -1
Count how many times you refueled. That's the minimum number of refueling stops you need.

Code Implementation

def min_refuel_stops(target, start_fuel, stations):
    number_of_refuels = 0
    current_fuel_reach = start_fuel
    stations_visited = []

    stations.append((target, 0)) 

    for station_distance, station_fuel in stations:
        # If we can't reach the next station, refuel.
        while current_fuel_reach < station_distance:
            if not stations_visited:
                return -1

            # Always refuel at the best station seen so far.
            best_station_fuel = max(stations_visited)
            current_fuel_reach += best_station_fuel
            stations_visited.remove(best_station_fuel)
            number_of_refuels += 1

        stations_visited.append(station_fuel)

    return number_of_refuels

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n log n) – The dominant operation within the algorithm is the selection of the gas station with the maximum fuel among the stations passed. To efficiently track the available fuel at stations, a max-heap (priority queue) is used. In the worst case, we might add all n gas stations to the heap. For each stop needed, we extract the maximum fuel from the heap (log n operation) and potentially add a new station to the heap (log n operation). Since in the worst-case we might need to refuel at almost every gas station, leading to up to n refuels, the overall complexity is O(n log n) due to the heap operations.

Space Complexity

O(N) – The algorithm keeps track of all gas stations passed so far and their fuel amounts. This implies storing these stations in a data structure that grows with the number of stations, potentially up to N, where N is the number of gas stations. Therefore, the auxiliary space is proportional to the number of gas stations. This yields a space complexity of O(N).

Edge Cases

Case	How to Handle
Empty stations array.	If the stations array is empty and startFuel is enough to reach target, return 0; otherwise, return -1.
Initial fuel is already sufficient to reach the target without refueling.	Return 0 immediately as no refueling is needed.
It's impossible to reach the target even with refueling at all stations.	Return -1 when, after exhausting all possible refueling stops, we cannot reach the target.
A station has zero fuel.	Treat a station with zero fuel as a valid station if it's reachable, but it doesn't contribute any fuel to the tank.
Stations are very close to the starting point but far from the destination, requiring many small refills initially.	The solution should correctly prioritize nearer stations with higher fuel capacity to minimize refills.
Stations are sorted in reverse order of distance from the starting point.	The algorithm should still correctly find the minimum number of refueling stops even with reverse sorted stations.
Integer overflow when calculating the current fuel level.	Use long data type to prevent potential overflow during fuel calculations.
Large number of stations, potentially exceeding memory limits.	An efficient implementation, possibly using a priority queue, should handle a large number of stations without running out of memory.