My solution in O(n) time

  • 5

    The idea is to keep track of the sum of all (gas[i] - cost[i]) (a random start would work), in the end, if sum >= 0, we can complete the circle. The position to start is the one next of the minimum of the sum. I actually don't know how to prove this, but it is accepted.

    class Solution {
        int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
            int len = gas.size();
            if(len == 0) return -1;
            int min = gas[0] - cost[0];
            int sum = min;
            int pos = 1%len;
            for(int i = 1; i < len; ++i){
                sum += gas[i] - cost[i];
                if(min > sum){
                    min = sum;
                    pos = (i+1)%len;
            if(sum < 0) return -1;
            return pos;

  • 0
    This post is deleted!

  • 1

    Your algorithm is pretty good! It could be proved.

    1. Suppose there are N gas stations, with gas[i] and cost[i];

    2. Let x[i] = gas[i] - cost[i].

    3. Let k be the index, where its accumulative sum (x[1] + ... + x[k]) reaches the minimum. We let (x[1] + ... + x[k]) be A;

    4. If the problem has a solution, then the remaining accumulative sum B, which is [x[k+1] + ... + x[N]) should be subjective to A + B > 0;

    5. The index of the optimal starting station is the (k + 1); We consider two types of indices, before and after the k, to prove the solution satisfies all requirement. Finally we show the solution actually is not unique.

    6. Regarding any index m, where m < k, let its accumulative sum (x[1] + ... + x[m]) be C. From definition of the index k in (3), we known C > A. Combing (4), we know B + C > B + A > 0, which means any gas station prior to k suffices to provide gas for a car.

    7. Regarding any index n, where k < n <= N, let its accumulative sum(x[k+1] + ... + x[m]) be D. Easy to know D must be none-negative, otherwise by definition of k in (3), the best k should be n, as A + D < A. This contradicts with our assumption of k in (3). So any gas station after k suffices to provide gas for a car.

    8. Proof is done.

    We show a case where the optimal solution is not unique.

    Let x = -4, 5, -6, 1, 5, where x is defined in (2). Its optimal solution is starting from 1 or 5. So the testing data of this problem might be combing any sequential x[i], which have the same sign. In this case, we just combine 1 and 5 to obtain unique solution.

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