## # 134. Gas Station

There are N gas stations along a circular route, where the amount of gas at station i is gas[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). You begin the journey with an empty tank at one of the gas stations.

Return the starting gas station's index if you can travel around the circuit once in the clockwise direction, otherwise return -1.

Note:

• If there exists a solution, it is guaranteed to be unique.
• 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:
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.
``````

### # Solution

Approach 1: For a path from A to B, if A couldn't reach B, then any stations between A and B couldn't finish the circle. We next try starting from B+1.

Approach 2: For each potential starting station, check if it's enough to support the gas need of the whole route.

Approach 3: Two pointers. Starting from the last station (just to avoid overflow). If it can reach the next station, keep moving on; otherwise, start from the station one step ahead.

### # Code (Python)

Approach 2:

``````class Solution:
def canCompleteCircuit(self, gas: List[int], cost: List[int]) -> int:
# Greedily find the max continous subsequence by keep adding diffs for as long as the car can go on, otherwise reset.
# Another explanation: we essentially try to start at any index from 0 and on. Since the amt of gas we carried over from previous indices (variable named 'total') is always >= 0, we got extra help from the previous indices. With those help if it still can't reach the next station, we give up by a reset. And we end by trying all indices -- no need to wrap around, because it's proved that everything can work up till the index right before the new start.
diff = [g - c for g, c in zip(gas, cost)]
if sum(diff) < 0:
return -1
start = 0
total = 0
for i in range(len(diff)):
# keep on if we can
if total + diff[i] >= 0:
total += diff[i]
else:
total = 0
start = i + 1
return start if start < len(diff) else -1
``````

### # Code (C++)

Approach 1:

``````// For a path from A to B, if A couldn't reach B, then any stations
// between A and B couldn't finish the cycle. We next try starting
// from B+1.
class Solution {
public:
int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
int len = gas.size();
int tail = 0;
int tank = 0;
{
tank += gas[tail] - cost[tail];
tail = (tail + 1) % len;
if (tank < 0)
{
return -1;
tank = 0;
}
{
}
}
return -1;
}
};
``````

Approach 2:

``````// For each potential starting station, check if it's enough to
// support the gas need of the whole route.
class Solution {
public:
int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
int len = gas.size();
int tank = 0;
int tankBefore = 0;
for (int i = 0; i < len; ++i)
{
tank += gas[i] - cost[i];
if (tank < 0)
{
tankBefore += tank;
tank = 0;
}
}
if (tankBefore + tank >= 0)
return -1;
}
};

// Note that the below function doesn't work.
// We need to reset the tank for every new try.
class Solution {
public:
int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
int len = gas.size();
int tank = 0;
for (int i = 0; i < len; ++i)
{
int newTank = tank + gas[i] - cost[i];
if (tank < 0 && newTank >= 0)
tank = newTank;
}
if (tank >= 0)
return -1;
}
};
``````

Approach 3:

``````// Two pointers.
class Solution {
public:
int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
int len = gas.size();
int tail = 1 % len; // In case that the len is 1.
{
if (tank < 0)
{
}
else
{
tank += gas[tail] - cost[tail];
tail = (tail + 1) % len;
}
}
if (tank >= 0)
return -1;
}
};
``````
``````// Two pointers. Starting from the last station (just to avoid
// overflow). If it can reach the next station, keep moving on;
// otherwise, start from the station one step ahead.
class Solution {
public:
int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
int len = gas.size();
int head = len - 1;
int tail = 0;
{
if (tank < 0)
{