# Gas Station LeetCode Solution | Easy Approach Share:

Gas Station 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

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.
```

Constraints:

• `gas.length == n`
• `cost.length == n`
• `1 <= n <= 105`
• `0 <= gas[i], cost[i] <= 104`

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
``````

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