**Difficulty: **Medium

**Asked in: **Amazon, Google

**Understanding The Problem**

**Problem Description**

There are `N`

gas stations along a circular route, where the amount of gas at station ** i** is

`arr[i]`

. You have a car with an unlimited gas tank and it costs `cost[i]`

of gas to travel from station `i`

to its next station `(i+1)`

. At the beginning of the journey, the tank is empty at one of the gas stations.Return the **minimum starting gas station’s index** if you need to travel around the circuit once, otherwise return -1.

**Problem Note**

- Completing the circuit means starting at
`i`

and ending up at`i`

again. - Both input arrays are non-empty and have the same length.
- Each element in the input arrays is a non-negative integer.

**Example 1**

```
Input:
arr[] = [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. Gas available in Tank = 0 + 4 = 4
Travel to station 4.Gas available in tank = 4 - 1 + 5 = 8
Travel to station 0.Gas available in tank = 8 - 2 + 1 = 7
Travel to station 1.Gas available in tank = 7 - 3 + 2 = 6
Travel to station 2.Gas available in 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:
arr[] = [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. Gas available in tank = 0 + 4 = 4
Travel to station 0.Gas available in tank = 4 - 3 + 2 = 3
Travel to station 1.Gas available in 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.
```

#### Solutions

We will be discussing a brute force and a greedy solution for the problem

**Brute Force Solution:**Check if you can cover all the gas stations starting from any station. If that’s not possible, repeat it by choosing another starting station.**Greedy Solution**: Optimize the brute force by concluding that if you cannot cover station B from station A, then any station between them cannot lead to an answer.

#### 2. Greedy Solution

The problem enables you to have these two ideas:

- If the car starts at A and can not reach B. Any station between A and B can not reach B.(B is the first station that A can not reach.)
- If the total number of gas is bigger than the total number of cost. There must be a solution.

**The proof of the above two ideas:**

- In any station between A and B, let’s say C. C will have gas left in our tank, if we go from A to that station. We can’t reach B from A with some gas(maybe 0) left in the tank in C, so we can’t reach B from C with an empty tank.
- If the gas is more than the cost in total, there must be some stations we have enough gas to go through them. Let’s say they are
**green**stations. So the other stations are**red**. The**adjacent stations**with the**same color**can be joined together as one. Then there must be a red station that can be joined into a precedent green station unless there isn’t any red station because the total gas is more than the total cost. In other words, all of the stations will join into a green station at last.

**This Concludes:**

- If the sum of gas is more than the sum of cost, then there must be a solution. And the question guaranteed that the solution is unique(The first one you found will the right one).
- The tank should never be negative, so restart whenever there is a negative number.

**Solution Steps**

- Create a
`start`

to store the valid starting index from where the car could touch all the stations. - For each station
`i`

, fill the fuel tank with`gas[i]`

and burn the fuel by`cost[i]`

- If at any point the tank is
`< 0`

then, choose the next index as the starting point. - At last, check if the total fuel available at the gas stations is greater than the total fuel burnt during the travel.
- Return the
`start`

**Pseudo Code**

```
int canCompleteCircuit(int[] gas, int[] cost) {
start = 0
total = 0
tank = 0
for( i = 0 to i < gas.size() ) {
tank = tank + gas[i] - cost[i]
if(tank < 0) {
start = i+1
total= total + tank
tank=0
}
}
if(total + tank < 0) {
return -1
} else {
return start
}
}
```

**Complexity Analysis**

Time Complexity: O(n)

Space Complexity: O(1)

**Critical Ideas To Think**

- Whenever
`tank`

becomes`< 0`

we set the next index as the`start`

. Why? - If the
`tank`

becomes`< 0`

, We then set it to`0`

. Why? - Why did we return -1 if
`total + tank < 0`

? - How did we ensure that whether or not a solution exists for the input?