134. 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.

https://discuss.leetcode.com/topic/1344/share-some-of-my-ideas

I have thought for a long time and got two ideas:

• If car starts at A and can not reach B. Any station between A and B
  can not reach B.(B is the first station that A can not reach.)
• If the total number of gas is bigger than the total number of cost. There must be a solution.
• (Should I prove them?)

public int canCompleteCircuit(int[] gas, int[] cost) {
    int sumGas = 0;
    int sumCost = 0;
    int start = 0;
    int tank = 0;
    for (int i = 0; i < gas.length; i++) {
        sumGas += gas[i];
        sumCost += cost[i];
        tank += gas[i] - cost[i];
        if (tank < 0) {
            start = i + 1;
            tank = 0;
        }
    }
    if (sumGas < sumCost) {
        return -1;
    } else {
        return start;
    }
}

The reason why I think this works:
1, if sum of gas is more than sum of cost, then there must be a solution. And the question guaranteed that the solution is unique(The first one I found is the right one).
2, The tank should never be negative, so restart whenever there is a negative number.

Proof to the first point: say there is a point C between A and B -- that is A can reach C but cannot reach B. Since A cannot reach B, the gas collected between A and B is short of the cost. Starting from A, at the time when the car reaches C, it brings in gas >= 0, and the car still cannot reach B. Thus if the car just starts from C, it definitely cannot reach B.

Proof for the second point:

• If there is only one gas station, it’s true.
• If there are two gas stations a and b, and gas(a) cannot afford cost(a), i.e., gas(a) < cost(a), then gas(b) must be greater than cost(b), i.e., gas(b) > cost(b), since gas(a) + gas(b) > cost(a) + cost(b); so there must be a way too.
• If there are three gas stations a, b, and c, where gas(a) < cost(a), i.e., we cannot travel from a to b directly, then:
• either if gas(b) < cost(b), i.e., we cannot travel from b to c directly, then cost(c) > cost(c), so we can start at c and travel to a; since gas(b) < cost(b), gas(c) + gas(a) must be greater than cost(c) + cost(a), so we can continue traveling from a to b. Key Point: this can be considered as there is one station at c’ with gas(c’) = gas(c) + gas(a) and the cost from c’ to b is cost(c’) = cost(c) + cost(a), and the problem reduces to a problem with two stations. This in turn becomes the problem with two stations above.
• or if gas(b) >= cost(b), we can travel from b to c directly. Similar to the case above, this problem can reduce to a problem with two stations b’ and a, where gas(b’) = gas(b) + gas(c) and cost(b’) = cost(b) + cost(c). Since gas(a) < cost(a), gas(b’) must be greater than cost(b’), so it’s solved too.
• For problems with more stations, we can reduce them in a similar way. In fact, as seen above for the example of three stations, the problem of two stations can also reduce to the initial problem with one station.