> ## Documentation Index
> Fetch the complete documentation index at: https://leetcode-py.wisl.dev/llms.txt
> Use this file to discover all available pages before exploring further.

> ## Agent Instructions
> leetcode-py is a Python LeetCode practice environment generator with one CLI: lcpy. It is not a service or platform.
> Each problem is a directory under leetcode/ with README.md, solution.py, test_solution.py, helpers.py, and playground.ipynb. lcpy gen creates them from JSON templates bundled with the package.
> Examples are backed by tests; copy them verbatim.

# Gas Station Python Solution with Tests

> Tested Python solution for LeetCode 134 with 12 pytest cases. Generate a practice environment with lcpy.

LeetCode 134, Medium. Topics: Array, Greedy. [View on LeetCode](https://leetcode.com/problems/gas-station/description/).

Generate this problem as a practice environment: tested reference solution, 12 [parametrized pytest cases](/practice/testing), and a playground notebook:

```bash theme={"theme":{"light":"github-light","dark":"github-dark"}}
lcpy gen -n 134   # by problem number
lcpy gen -s gas_station   # by problem name
```

## Problem

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**.

### Examples

```
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.
```

```
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

* `n == gas.length == cost.length`
* `1 <= n <= 10^5`
* `0 <= gas[i], cost[i] <= 10^4`
* The input is generated such that the answer is unique.

## Solution

Reference implementation from [solution.py on GitHub](https://github.com/wislertt/leetcode-py/blob/main/leetcode/gas_station/solution.py), full suite in [test\_solution.py](https://github.com/wislertt/leetcode-py/blob/main/leetcode/gas_station/test_solution.py):

```python theme={"theme":{"light":"github-light","dark":"github-dark"}}
class Solution:
    """
    Gas Station Circuit - Greedy Approach

    Visual Example: gas=[1,2,3,4,5], cost=[3,4,5,1,2]

         Station:  0   1   2   3   4
                  ┌─┐ ┌─┐ ┌─┐ ┌─┐ ┌─┐
         Gas:     │1│ │2│ │3│ │4│ │5│
                  └─┘ └─┘ └─┘ └─┘ └─┘
         Cost:     3   4   5   1   2
                   ↓   ↓   ↓   ↓   ↓
         Net:     -2  -2  -2  +3  +3

    Algorithm trace:
    i=0: tank=0+(-2)=-2 < 0 → reset tank=0, start=1
    i=1: tank=0+(-2)=-2 < 0 → reset tank=0, start=2
    i=2: tank=0+(-2)=-2 < 0 → reset tank=0, start=3
    i=3: tank=0+(+3)=+3 ≥ 0 → continue
    i=4: tank=3+(+3)=+6 ≥ 0 → return start=3

    Key insight: If total_gas ≥ total_cost, greedy start position works!
    """

    # Time: O(n)
    # Space: O(1)
    def can_complete_circuit(self, gas: list[int], cost: list[int]) -> int:
        if sum(gas) < sum(cost):
            return -1

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

## Complexity

| Time | Space |
| ---- | ----- |
| O(n) | O(1)  |

## Tags

[Grind](/catalog/grind), [NeetCode 150](/catalog/neetcode-150), [NeetCode 250](/catalog/neetcode-250), [NeetCode All](/catalog/neetcode).