In this article, we’ll walk through a Java solution for a classic problem often encountered in coding interviews: determining the starting gas station from which a vehicle can complete a circular route without running out of gas. This problem can be solved using an efficient algorithm with a time complexity of O(n).
Problem Statement
You are given two integer arrays:
gas[]: represents the amount of gas available at each gas station.cost[]: represents the gas cost required to travel to the next station.
The goal is to find the starting gas station index from which you can complete a full circuit of gas stations without running out of gas. If it’s not possible to complete the circuit, return -1.
Java Solution Overview
We’ll use a simple approach where we maintain two variables to keep track of:
- The total gas balance as we traverse through the gas stations.
- The current gas balance during the traversal.
Here’s the step-by-step breakdown of the implementation:
Step 1: Setting Up the Main Method
We begin with the main method, which initializes the gas and cost arrays. We then call the getStartingStation method and print the result.
public static void main(String[] args) {
int[] gas = {1,2,3,4,5};
int[] cost = {3,4,5,1,2};
int startStation = getStartingStation(gas, cost);
System.out.println("Starting station will be: "+startStation);
}
Step 2: Creating the getStartingStation Method
Next, we define the getStartingStation method, which contains the main logic for solving the problem.
private static int getStartingStation(int[] gas, int[] cost) {
int total = 0, current =0, start =0 ;
Variables Explained:
total: Tracks the overall gas balance (total gas - total cost).current: Keeps track of the current gas balance as we iterate through the stations.start: Indicates the potential starting station.
Step 3: Iterating Through the Gas Stations
We loop through each gas station and calculate the difference between gas and cost for each station.
for (int i = 0; i < cost.length; i++) {
int diff = gas[i] - cost[i];
total += diff;
current += diff;
diffis calculated asgas[i] - cost[i]. Ifdiffis positive, it means that the gas station provides more gas than the cost required to get to the next station.
Step 4: Checking Current Balance
We check if the current balance goes negative:
if (current<0) {
current=0;
start = i+1;
}
- If
currentdrops below 0, it means we can’t reach the next station from the current starting point. Therefore, we resetcurrentto 0 and setstartto the next station (i + 1).
Step 5: Returning the Result
Finally, we determine if a valid starting station exists by checking the total gas balance.
return total>=0?start:-1;
- If the total balance is non-negative, we return the starting station index. Otherwise, we return -1, indicating that it’s impossible to complete the circuit.
Full Code Implementation
Here’s the complete code for reference:
public class GasStation {
public static void main(String[] args) {
int[] gas = {1,2,3,4,5};
int[] cost = {3,4,5,1,2};
int startStation = getStartingStation(gas, cost);
System.out.println("Starting station will be: "+startStation);
}
private static int getStartingStation(int[] gas, int[] cost) {
int total = 0, current =0, start =0 ;
for (int i = 0; i < cost.length; i++) {
int diff = gas[i] - cost[i];
total += diff;
current += diff;
if (current<0) {
current=0;
start = i+1;
}
}
return total>=0?start:-1;
}
}
Conclusion
This Java solution effectively finds the optimal starting gas station by leveraging a linear traversal approach that checks gas availability against costs. The algorithm is efficient and runs in O(n) time, making it suitable for large datasets. By understanding each component of the solution, you can easily adapt this approach to similar problems in coding interviews or practical applications. Happy coding!
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