LeetCode – Gas Station (Java)

There are N gas stations along a circular route, where the amount of gas at station i is gas[i].

You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from station i to its next station (i+1). You begin the journey with an empty tank at one of the gas stations.

Return the starting gas station's index if you can travel around the circuit once, otherwise return -1.


To solve this problem, we need to understand and use the following 2 facts:
1) if the sum of gas >= the sum of cost, then the circle can be completed.
2) if A can not reach C in a the sequence of A-->B-->C, then B can not make it either.

Proof of fact 2:

If gas[A] < cost[A], then A can not even reach B. 
So to reach C from A, gas[A] must >= cost[A]. 
Given that A can not reach C, we have gas[A] + gas[B] < cost[A] + cost[B],
and gas[A] >= cost[A],
Therefore, gas[B] < cost[B], i.e., B can not reach C. 

In the following solution, sumRemaining tracks the sum of remaining to the current index. If sumRemaining < 0, then every index between old start and current index is bad, and we need to update start to be the current index. You can use the following example to visualize the solution. leetcode-gas-station-java

Java Solution

public int canCompleteCircuit(int[] gas, int[] cost) {
	int sumRemaining = 0; // track current remaining
	int total = 0; // track total remaining
	int start = 0; 
	for (int i = 0; i < gas.length; i++) {
		int remaining = gas[i] - cost[i];
		//if sum remaining of (i-1) >= 0, continue 
		if (sumRemaining >= 0) { 
			sumRemaining += remaining;
		//otherwise, reset start index to be current
		} else {
			sumRemaining = remaining;
			start = i;
		total += remaining;
	if (total >= 0){
		return start;
		return -1;
Category >> Algorithms >> Interview >> Java  
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  1. Dev Dev on 2015-11-29

    Check this simpler solution.

    public static int findStart(int[] gas, int[] cost) {

    if (gas.length != cost.length) {

    return -1;


    int remaining = 0, start = 0;

    for (int i = 0; i < gas.length; i++) {

    remaining += gas[i] – cost[i];

    if (remaining < 0) {

    start = i + 1;

    remaining = 0;



    return start;


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