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

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

Given two integer arrays gas and cost, return the starting gas station's index if you can travel around the circuit once in the clockwise direction, otherwise return -1. If there exists a solution, it is guaranteed to be unique

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Example 1:

Input: gas = [1,2,3,4,5], cost = [3,4,5,1,2]
Output: 3
Explanation:
Start at station 3 (index 3) and fill up with 4 unit of gas. Your tank = 0 + 4 = 4
Travel to station 4. Your tank = 4 - 1 + 5 = 8
Travel to station 0. Your tank = 8 - 2 + 1 = 7
Travel to station 1. Your tank = 7 - 3 + 2 = 6
Travel to station 2. Your tank = 6 - 4 + 3 = 5
Travel to station 3. The cost is 5. Your gas is just enough to travel back to station 3.
Therefore, return 3 as the starting index.


Example 2:

Input: gas = [2,3,4], cost = [3,4,3]
Output: -1
Explanation:
You can't start at station 0 or 1, as there is not enough gas to travel to the next station.
Let's start at station 2 and fill up with 4 unit of gas. Your tank = 0 + 4 = 4
Travel to station 0. Your tank = 4 - 3 + 2 = 3
Travel to station 1. Your tank = 3 - 3 + 3 = 3
You cannot travel back to station 2, as it requires 4 unit of gas but you only have 3.
Therefore, you can't travel around the circuit once no matter where you start.


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Constraints:


	n == gas.length == cost.length
	1 &lt;= n &lt;= 105
	0 &lt;= gas[i], cost[i] &lt;= 104


class Solution:
    def canCompleteCircuit(self, gas: List[int], cost: List[int]) -> int:
        deltas = [x-y for x, y in zip(gas, cost)]
        n = len(deltas)
        deltas = deltas + deltas
        cursum, curi = 0, 0
        maxsum, maxi = 0, 0
        for i, delta in enumerate(deltas):
            cursum = max(0, cursum + delta)
            if cursum == 0:
                curi = i+1
            if cursum > maxsum:
                maxi = curi
                maxsum = cursum
        cursum = 0
        for i in range(n):
            cursum += deltas[(maxi+i)%n]
            if cursum < 0: return -1
        return maxi
class Solution {
    public int canCompleteCircuit(int[] gas, int[] cost) {
       // *Upvote will be appreciated*
        int totalFuel = 0;
        int totalCost = 0;
        int n = gas.length;
        for(int i = 0; i < n; i++) {
            totalFuel += gas[i];
        }
        for(int i = 0; i < n; i++) {
            totalCost += cost[i];
        }
        // if totalfuel < totalCost then It is not possible to tavel
        if(totalFuel < totalCost) {
            return -1;
        }

        // It is greather then There may be an Answer
        int start = 0;
        int currFuel = 0;
        for(int i = 0; i < n; i++) {
            currFuel += (gas[i]-cost[i]);
            if(currFuel < 0) { // It Current Fuel is less than 0 mean we can't star from that index
                start = i+1; // so we start from next index
                currFuel = 0;
            }
        }
        return start;
    }
}
class Solution {
public:
    int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
        int n = gas.size();
        int start = -1;

        int sum = 0 , gastillnow = 0;
        for(int i = 0 ; i < 2*n ; i++){
            if(start == i%n){
                return i%n;
            }
            if(gas[i%n] + gastillnow >= cost[i%n]){ // we can start from this index
                if(start==-1) start = i;
                gastillnow += gas[i%n]-cost[i%n];
            }else if(gastillnow + gas[i%n] < cost[i%n]){ // previous start index was wrong we have to start from another
                start = -1;
                gastillnow = 0;
            }
        }
        return -1;
    }
};
var canCompleteCircuit = function(gas, cost) {
    const len = gas.length;
    // scan forward from the current index
    const scan = (i) => {
        let numTries = 0;
        let tank = 0;
        let c = 0;
        while (numTries <= len) {
            tank -= c;
            if (tank < 0) return -1;
            // if we made it back around, and we have gas, return the index, we made it!
            if (numTries === len && tank >= 0) {
                return i;
            }
            tank += gas[i];
            c = cost[i];
            i++;
            if (i === len) i = 0; // if we hit the end, bounce back to zero
            numTries++;
        }
        return -1;
    }

    for (let i = 0; i < len; i++) {
        if (!gas[i]) continue; // no gas / zero gas so let's just move on
        let index = scan(i);
        if (~index) return index; // if it's not -1, return it
    }

    return -1;
};
Gas Station