134. Gas Station

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

Array Greedy

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 <= 10^5
0 <= gas[i], cost[i] <= 10^4
The input is generated such that the answer is unique.

Solution

class Solution:
    def canCompleteCircuit(self, gas: list[int], cost: list[int]) -> int:
        start, tank, total = 0, 0, 0

        for i in range(len(gas)):
            drive = gas[i] - cost[i]
            tank = tank + drive

            if tank < 0:
                tank = 0
                start = i + 1
            total = total + drive

        return start if total >= 0 else -1