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.

#include<stdio.h>
#include<vector>
using namespace std;

int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
    int i, j ,k ,index = 0, sum = 0, sumold;
    int n = cost.size();
    for(i = 0; i < n; i++) {
        for(j = index; j < n + i; j++) {
            if(j >= n) index = j - n;
            else index = j;
            sumold = sum;
            sum = sum + gas[index] - cost[index];
            if(sum < 0) {
                index = j;
                sum = sumold - gas[i] + cost[i];
                break;
            }
        }
        if(j >= n + i && sum >= 0) return i;
    }
    if(i == n) return -1;
}

int main() {
    int gas1[] = {67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66};
    int cost1[] = {27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26};
    vector<int> gas;
    vector<int> cost;
    gas.push_back(2);
    gas.push_back(4);
    cost.push_back(3);
    cost.push_back(4);
    printf("%d\n", canCompleteCircuit(gas, cost));
    return 1;
}

Gas Station