Gas Station

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.

Note: The solution is guaranteed to be unique.

To solve this question, you need to understand three rules:

1. If gas[0] + gas[1] + gas[2] + ... + gas[n]>= cost[0] + cost[1] + cost[2] + ... + cost[n], you will always be able to do one circle, otherwise you can not complete a circle.
2. If at any given gas station, leftGas + (gas[i]-cost[i])<0, you can not go to next gas station and all previous gas station can not be start gas station.

How to prove 2? A->B->C and A can't go to C

If gas[A] < cost[A], then A can not go to B. Therefore, gas[A] >=cost[A].

We already know A can not go to 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 go to C.

public int canCompleteCircuit(int[] gas, int[] cost) {
        int leftGas = 0;
        int sum = 0;
        int startIndex = 0;

        for (int i=0; i<gas.length; i++) {
            leftGas += gas[i] - cost[i];
            sum += gas[i] - cost[i];
            if (sum<0) {
                startIndex = i+1;
                sum = 0;
            }
        }
        return leftGas>=0?startIndex:-1;       
    }
}