Day014 - 134. Gas Station

134. Gas Station

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

내 풀이

class Solution:
    def canCompleteCircuit(self, gas: List[int], cost: List[int]) -> int:
        total_gas = 0 
        current_gas = 0
        start_index = 0
        
        # 전체 순회를 할 수 있는지 확인하기 위해 total_gas를 계산
        for i in range(len(gas)):
            total_gas += gas[i] - cost[i]
            current_gas += gas[i] - cost[i]
            
            # current_gas가 음수가 되는 경우 출발할 수 없으므로 다음 지점으로 이동
            if current_gas < 0:
                # 현재 인덱스 + 1에서 다시 출발
                start_index = i + 1
                current_gas = 0

        # total_gas가 음수라면 순회를 할 수 없으므로 -1 반환
        if total_gas < 0:
            return -1
        else:
            return start_index

이

# Time Complexity : \(O(N)\)