# Gas Station

Published: Oct 16, 2022

Medium Greedy Array

## Introduction

A naive idea is to start from every index and check all stations can be reached. However, a greedy approach makes it possible to do in one path. Just calculate gas - cost at each index. If the accumulation of the diff becomes negative, the start should be moved. This way, we can get the answer.

## Problem Description

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

You have a car with an unlimited gas tank and it costs `cost[i]` of gas to travel from the i-th 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

Constraints:

• `n == gas.length == cost.length`
• `1 <= n <= 10**5`
• `0 <= gas[i], cost[i] <= 10**4`

https://leetcode.com/problems/gas-station/

## Examples

``````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.
``````
``````Example 3
Input: gas = [3,1,1], cost = [1,2,2]
Output: 0
``````

## Analysis

At first, check there is enough amount of gas compared to cost. If it is, there should be an unique solution. Calculate gas - cost. If the accumulation of gas - cost becomes negative, it’s impossible to reach to the starting point. So, initialize the parameter and update start index. When the all stations are checked, we get the answer.

## Solution

``````class GasStation:
def canCompleteCircuit(self, gas: List[int], cost: List[int]) -> int:
if sum(gas) < sum(cost):
return -1
n = len(gas)
total, start = 0, 0
for i in range(n):
total += gas[i] - cost[i]
if total < 0:
total = 0
start = i + 1
return start
``````

## Complexities

• Time: `O(n)`
• Space: `O(1)`