# Gas Station π

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:**

`n == gas.length == cost.length`

`1 <= n <= 105`

`0 <= gas[i], cost[i] <= 104`

# Solution

Time complexity

O(nΒ²)Space complexity

O(1)

Time complexity O(2n) ->

O(n)Space complexity

O(1)

**Do you see that green follow button? π**

**What if?**