# LC 134. Gas Station

### [Problem link](https://leetcode.com/problems/gas-station/)

###### tags: `leedcode` `python` `c++` `medium` `Greedy`

There are <code>n</code> gas stations along a circular route, where the amount of gas at the <code>i^th</code> station is <code>gas[i]</code>.

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

Given two integer arrays <code>gas</code> and <code>cost</code>, return the starting gas station's index if you can travel around the circuit once in the clockwise direction, otherwise return <code>-1</code>. 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:**

- <code>n == gas.length == cost.length</code>
- <code>1 <= n <= 10⁵</code>
- <code>0 <= gas[i], cost[i] <= 10⁴</code>

## Solution 1 - Greedy
#### Python

class Solution:
    def canCompleteCircuit(self, gas: List[int], cost: List[int]) -> int:
        if sum(gas) < sum(cost):
            return -1

        start = 0
        total = 0

        for i in range(len(gas)):
            total += gas[i] - cost[i]
            if total < 0:
                total = 0
                start = i + 1

        return start

#### C++

class Solution {
public:
    int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
        int curSum = 0;
        int totalSum = 0;
        int start = 0;

        for (int i = 0; i < gas.size(); i++) {
            curSum += gas[i] - cost[i];
            totalSum += gas[i] - cost[i];
            if (curSum < 0) {
                start = i + 1;
                curSum = 0;
            }
        }
        if (totalSum < 0) {
            return -1;
        }
        return start;
    }
};

>### Complexity
>| | Time Complexity | Space Complexity |
>| ----------- | --------------- | ---------------- |
>| Solution 1 | O(n) | O(1) |

## Note
[referece](https://github.com/youngyangyang04/leetcode-master/blob/master/problems/0134.%E5%8A%A0%E6%B2%B9%E7%AB%99.md)