Gas Station

class Solution {
public:
    int canTravel(vector<int>& fuel, vector<int>& expenses) {
        int currentSurplus = 0, totalSurplus = 0, startIndex = 0;

        for (int j = 0; j < fuel.size(); ++j) {
            // Calculating surplus fuel after expenses
            totalSurplus += fuel[j] - expenses[j];
            currentSurplus += fuel[j] - expenses[j];

            // Resetting to next station as start point when deficit occurs
            if (currentSurplus < 0) {
                startIndex = j + 1;
                currentSurplus = 0;
            }
        }

        return totalSurplus >= 0 ? startIndex : -1;
    }
};

The provided C++ code defines a solution for determining whether a full circle tour is possible among a circuit of gas stations, where each station provides fuel and incurs a certain travel expense to reach the next station. Here is a concise explanation:

• Function Definition: This solution contains a public method canTravel belonging to the class Solution. This method takes two vectors as arguments: fuel and expenses, representing the amount of fuel each gas station provides and the cost in fuel to travel to the next station, respectively.

• Variables Utilized:

  • currentSurplus: Tracks fuel surplus while traversing through the stations.
  • totalSurplus: Cumulative fuel surplus to check feasibility after completing one loop.
  • startIndex: Identifies a valid starting station index if a complete circuit is possible.

• Algorithm:

  1. Initialize currentSurplus and totalSurplus to zero, and set startIndex to zero.
  2. Iterate through each gas station:
     1. Update currentSurplus and totalSurplus by subtracting the expenses from fuel at each station.
     2. If currentSurplus becomes negative, it indicates an inability to travel from the current starting station:
        1. Set the next station as the new starting station.
        2. Reset currentSurplus to zero.
  3. Evaluate if the complete circuit is feasible:
     1. If totalSurplus is non-negative, the circuit can be completed starting from startIndex.
     2. Otherwise, return -1, indicating no valid starting point allows completing the circuit.

• Result Analysis:

  • -1 returned means it is not possible to start from any station and return to it without running out of fuel.
  • Any other returned index indicates the station number from which one can begin the circuit to successfully complete it without fuel shortage.

This function would be effective in scenarios like route planning based on fuel availability ensuring that journey continuity is maintained.

class Solution {
    public int canTravelCompleteCircuit(int[] fuel, int[] expense) {
        int currentSurplus = 0, totalSurplus = 0, startingStation = 0;

        for (int j = 0; j < fuel.length; ++j) {
            int netGain = fuel[j] - expense[j];
            totalSurplus += netGain;
            currentSurplus += netGain;

            if (currentSurplus < 0) {
                startingStation = j + 1;
                currentSurplus = 0;
            }
        }

        return totalSurplus >= 0 ? startingStation : -1;
    }
}

Understanding the given Java solution for finding whether there's a complete circuit that a vehicle can travel around a circle of gas stations without running out of gas involves a well-thought-out approach:

• Determine the feasibility of starting and completing the circuit at any given gas station.
• Let fuel and expense arrays represent the amount of gas at each station and the gas required to travel to the next station, respectively.

The solution uses a single pass approach:

1. Initialize currentSurplus to store the current surplus of gas, totalSurplus to accumulate the net total surplus, and startingStation to remember the current starting point as you iterate.

2. Loop through each station:

   • Compute netGain for each station which is the difference between fuel available and fuel needed (fuel[j] - expense[j]).
   • Add this netGain to totalSurplus and currentSurplus.
   • If currentSurplus becomes negative, it indicates that starting from the previous station and coming to current station j, there isn't enough fuel to proceed further. Thus, set startingStation to the next station (j + 1) and reset currentSurplus to zero.

3. Finally, return the startingStation if the totalSurplus across all stations is non-negative, which means a complete circuit is possible from that station. Otherwise, return -1 indicating no such starting station exists that allows a complete circuit.

This clever use of linear scan and net surplus calculation makes the code efficient and robust for scenarios involving any number of gas stations and corresponding costs.

int findStartingStation(int* fuel, int numStations, int* expense, int expenseSize) {
    int currentProfit = 0, totalProfit = 0, startPoint = 0;
    for (int index = 0; index < numStations; ++index) {
        int netGain = fuel[index] - expense[index];
        totalProfit += netGain;
        currentProfit += netGain;
        if (currentProfit < 0) {
            startPoint = index + 1;
            currentProfit = 0;
        }
    }
    return totalProfit >= 0 ? startPoint : -1;
}

The task involves finding a starting point from which you can travel through all the gas stations in a circular route without running out of fuel. The given C function findStartingStation effectively determines the best starting gas station index using two input arrays fuel and expense and their size.

• The function initializes three variables: currentProfit to track the surplus or deficit at the current station, totalProfit to calculate the overall feasibility of completing the circuit, and startPoint to determine the starting index of the station.
• It iterates over each gas station:
  • It calculates the net gain or loss at that station by subtracting expense[index] from fuel[index].
  • Both totalProfit and currentProfit are updated with this net gain.
  • If currentProfit becomes negative, it indicates that you cannot proceed from the current starting station, and so, startPoint is moved to the next station, and currentProfit resets to zero.
• Finally, the function checks if the totalProfit is non-negative, which indicates that a complete circuit is possible starting from the computed startPoint. If it's possible, startPoint is returned; otherwise, -1 indicates no possible start exists to complete the circuit without running out of fuel.

This method ensures efficiency by calculating the feasible starting point in a single pass through the stations, which is crucial for scenarios involving a large number of stations.

var findCircularRoute = function (fuel, expense) {
    let currentTotal = 0,
        overallTotal = 0,
        startPoint = 0;
    for (let i = 0; i < fuel.length; ++i) {
        let netGain = fuel[i] - expense[i];
        overallTotal += netGain;
        currentTotal += netGain;
        if (currentTotal < 0) {
            startPoint = i + 1;
            currentTotal = 0;
        }
    }
    return overallTotal >= 0 ? startPoint : -1;
};

The problem describes a scenario involving a circular route with gas stations, where you need to determine a starting gas station to complete the route without running out of fuel. The JavaScript function findCircularRoute takes two input arrays: fuel and expense. fuel indicates the amount of fuel available at each station, while expense represents the cost in fuel to travel to the next station.

This code operates as follows:

• Initialize currentTotal to track the net gain or loss of fuel starting from a potential station, and overallTotal to evaluate if the circuit can be completed.
• Iterate through each gas station using a for loop. Calculate the net gain by subtracting the travel expense from the available fuel.
• Accumulate the net gains into overallTotal and currentTotal.
• If currentTotal becomes negative, it means that starting from the current startPoint is not feasible, so set startPoint to the next station and reset currentTotal to zero.
• If overallTotal is non-negative at the end of the loop, the circuit can be completed starting from startPoint. Otherwise, return -1 indicating such a route is not possible.

In this solution, the key is to reset the starting position whenever a deficit is encountered due to accumulated expenses surpassing the fuel supply. This ensures that we only consider viable start points. If the total fuel across all stations can at least cover the total expenses, it guarantees there is a valid route. The function returns the starting index of this route, or -1 if no valid starting point exists.

class Solution:
    def completeCircuit(self, gas_stations: List[int], fuel_cost: List[int]) -> int:
        net_gain = 0
        current_gain = 0
        starting_point = 0

        for index in range(len(gas_stations)):
            net_gain += gas_stations[index] - fuel_cost[index]
            current_gain += gas_stations[index] - fuel_cost[index]

            if current_gain < 0:
                current_gain = 0
                starting_point = index + 1

        return starting_point if net_gain >= 0 else -1

The provided Python solution defines a method to determine if there's a valid starting point in a circular route where a vehicle can complete the circuit given the amounts of gas at each station and the cost in gas to travel to the next station.

Here's a breakdown of the solution:

• The function completeCircuit receives two lists: gas_stations detailing the amount of gas available at each station, and fuel_cost specifying the fuel necessary to travel to the next station.
• Two variables, net_gain and current_gain, are initialized to keep track of the total and the current surplus of gasoline after each station, respectively.
• The starting_point variable is utilized to mark the potential starting gas station index.

The method iterates through each gas station and updates the net_gain and current_gain by subtracting the fuel cost from the gas available. If at any station, current_gain falls below zero, which indicates that the journey can't continue from the previous starting point:

• Reset current_gain to zero.
• Update starting_point to the next station index, proposing a new starting potential.

After evaluating all stations:

• If net_gain is non-negative, it indicates that there's enough total gas to cover the fuel costs for the entire circuit, and the function returns the starting_point.
• Otherwise, it returns -1, indicating that completing the circuit is not possible with the given gas distribution and costs.

This solution employs a single-pass algorithm, thereby ensuring efficient computation with time complexity of O(n), where n is the number of gas stations. This efficiency is crucial for scenarios with a large number of gas stations.