**Gas Station **There are `n`

gas stations along a circular route, where the amount of gas at the `i`

station is ^{th}`gas[i]`

.

You have a car with an unlimited gas tank and it costs `cost[i]`

of gas to travel from the `i`

station to its next ^{th}`(i + 1)`

station. You begin the journey with an empty tank at one of the gas stations.^{th}

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**

**Example 1:**

Input:gas = [1,2,3,4,5], cost = [3,4,5,1,2]Output:3Explanation: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:-1Explanation: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.

**Constraints:**

`gas.length == n`

`cost.length == n`

`1 <= n <= 10`

^{5}`0 <= gas[i], cost[i] <= 10`

^{4}

### Gas Station Solutions

✅**Time:** O(n)

✅**Space:** O(1)

**C**++

```
class Solution {
public:
int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
const int gasses = accumulate(begin(gas), end(gas), 0);
const int costs = accumulate(begin(cost), end(cost), 0);
if (gasses - costs < 0)
return -1;
int ans = 0;
int sum = 0;
// try to start from each index
for (int i = 0; i < gas.size(); ++i) {
sum += gas[i] - cost[i];
if (sum < 0) {
sum = 0;
ans = i + 1; // start from next index
}
}
return ans;
}
};
```

**Java**

```
class Solution {
public int canCompleteCircuit(int[] gas, int[] cost) {
final int gasses = Arrays.stream(gas).sum();
final int costs = Arrays.stream(cost).sum();
if (gasses - costs < 0)
return -1;
int ans = 0;
int sum = 0;
// try to start from each index
for (int i = 0; i < gas.length; ++i) {
sum += gas[i] - cost[i];
if (sum < 0) {
sum = 0;
ans = i + 1; // start from next index
}
}
return ans;
}
}
```

**Python**

```
class Solution:
def canCompleteCircuit(self, gas: List[int], cost: List[int]) -> int:
ans = 0
net = 0
sum = 0
for i in range(len(gas)):
net += gas[i] - cost[i]
sum += gas[i] - cost[i]
if sum < 0:
sum = 0
ans = i + 1
return -1 if net < 0 else ans
```

#### Watch Tutorial

**Checkout more Solutions here**