Minimum Number of Refueling Stops

Problem Statement

In this scenario, a car aims to travel a specified distance eastward to a target destination. The car starts at a given position with an initial amount of fuel. Along its route to the destination, there are several gas stations, defined in an array where each gas station's position relative to the starting point and the amount of fuel it offers are provided. The car's fuel tank does not have a capacity limit but starts with a designated amount of fuel. It consumes one liter per mile. At any gas station along its route, the car can stop to replenish its fuel entirely from the station's reserves without limitation.

The challenge is to determine the minimum number of refueling stops required for the car to reach its destination. If it's impossible for the car to reach the destination with the available fuel and gas stations, the output should be -1. The nuance of this problem lies in deciding at which gas stations to refuel and ensuring that the car does not run out of fuel at any point before reaching a station or the destination.

Examples

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 <= positioni < positioni+1 < target
• 1 <= fueli < 109

Approach and Intuition

Given the characteristics of the provided examples, the primary aim is to minimize the number of stops for refueling while ensuring the destination is reachable:

1. Immediate Destination Infeasibility:

   • If the car cannot reach the first station or the target with its startFuel, it clearly won't make the journey. An example is when the car starts with less fuel than the distance to the first gas station or the target, resulting in an immediate output of -1.

2. Utilizing Stations Wisely:

   • Deciding when to refuel involves considering the fuel needed to travel to the next station or to the target from the current position. At each station, one should decide whether stopping there allows for a longer stretch of travel later or immediate refueling is necessary to proceed.

3. Maximizing Forward Movement:

   • Each decision to refuel should ideally maximize the car's ability to move forward as far as possible, considering both the current fuel reserves and potential refueling options ahead. This means sometimes bypassing stations with smaller amounts of fuel if reaching a further station provides a greater refueling benefit.

4. Edge Cases:

   • If no stations are available (stations.length == 0), then the trip feasibility lies solely in whether startFuel is greater than or equal to target.
   • A clear solution loop emerges in cases where the car reaches stations or the destination exactly as the fuel runs out, which are guided by the problem's rules permitting refueling in such scenarios.

Based on these insights, algorithms like greedy approaches or dynamic programming can be employed to evaluate the minimal stops by optimizing refueling strategies at every step of the journey.

Solutions

class Solution {
    public int minimumRefuels(int finalDistance, int fuel, int[][] refuelStations) {
        // Create a max-heap to store fuel capacities available at stations
        PriorityQueue<Integer> maxHeap = new PriorityQueue<>(Collections.reverseOrder());
        int refuels = 0, lastPosition = 0;
        for (int[] station : refuelStations) {
            int stationPosition = station[0];
            int fuelAvailable = station[1];
            fuel -= stationPosition - lastPosition;
            while (!maxHeap.isEmpty() && fuel < 0) {
                // Refuel the vehicle with the max available fuel from past stations
                fuel += maxHeap.poll();
                refuels++;
            }

            if (fuel < 0) return -1;
            maxHeap.offer(fuelAvailable);
            lastPosition = stationPosition;
        }

        // Final stretch to the target
        fuel -= finalDistance - lastPosition;
        while (!maxHeap.isEmpty() && fuel < 0) {
            fuel += maxHeap.poll();
            refuels++;
        }
        if (fuel < 0) return -1;

        return refuels;
    }
}