<p>There are <code>n</code> gas stations along a circular route, where the amount of gas at the <code>i<sup>th</sup></code> station is <code>gas[i]</code>.</p>

<p>You have a car with an unlimited gas tank and it costs <code>cost[i]</code> of gas to travel from the <code>i<sup>th</sup></code> station to its next <code>(i + 1)<sup>th</sup></code> station. You begin the journey with an empty tank at one of the gas stations.</p>

<p>Given two integer arrays <code>gas</code> and <code>cost</code>, return <em>the starting gas station&#39;s index if you can travel around the circuit once in the clockwise direction, otherwise return</em> <code>-1</code>. If there exists a solution, it is <strong>guaranteed</strong> to be <strong>unique</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> gas = [1,2,3,4,5], cost = [3,4,5,1,2]
<strong>Output:</strong> 3
<strong>Explanation:</strong>
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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> gas = [2,3,4], cost = [3,4,3]
<strong>Output:</strong> -1
<strong>Explanation:</strong>
You can&#39;t start at station 0 or 1, as there is not enough gas to travel to the next station.
Let&#39;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&#39;t travel around the circuit once no matter where you start.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == gas.length == cost.length</code></li>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= gas[i], cost[i] &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution:
    # def canCompleteCircuit(self, gas: List[int], cost: List[int]) -> int:
    #     cur_gas, started_at = 0, 0
    #     if sum(gas) < sum(cost): return -1
    #     for i in range(len(gas)):
    #         cur_gas += (gas[i] - cost[i]) # Assume that we reached the ith station from some station j (j < i)
    #         if cur_gas < 0: # if net gas falls below zero, we know that cost to reach ith station has to be so large that it emptied whole gas tank i.e brought cur_gas below 0
    #             # this station cant be our answer as we are losing a lot of gas
    #             cur_gas = 0
    #             started_at = i + 1  # so we pick next station in the hope that, that would be our answer
    #             # The statement 'if sum(gas) < sum(cost): return -1' ensures that we would always have an answer
    #             # since i can't be it, so our answer certainly lies ahead
    #     return started_at

    def canCompleteCircuit(self, gas: List[int], cost: List[int]) -> int:
        # Initialize the length of the gas and cost lists
        num_stations = len(gas)
        # Initialize pointers for traversing the gas stations
        start_index = end_index = num_stations - 1
        # Initialize counter for stations visited and total balance of gas
        stations_visited = total_gas_balance = 0
        # Loop until all stations have been visited
        while stations_visited < num_stations:
            # Updatethe total balance by adding current gas and subtracting current cost
            total_gas_balance += gas[end_index] - cost[end_index]
            # Increment the number of stations visited
            stations_visited += 1
            # Move to the next station, wrapping around if necessary
            end_index = (end_index + 1) % num_stations
            # While the total balance is negative and we haven't visited all stations
            # move the start index backwards and adjust the balance.
            while total_gas_balance < 0 and stations_visited < num_stations:
                print(f"At index {end_index}")
                # Move the start index to the previous station
                start_index = (start_index - 1) % num_stations
                # Update the total balance by adding gas and subtracting cost at the new start
                total_gas_balance += gas[start_index] - cost[start_index]
                # Increment the number of stations visited
                stations_visited += 1
        # If the total balance is negative, return -1 indicating the circuit cannot be completed,
        # otherwise return the start index
        return -1 if total_gas_balance < 0 else start_index

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