134. Gas Station

Description

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

You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from the i^th 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
• The input is generated such that the answer is unique.

Submitted Code

class Solution:
    def canCompleteCircuit(self, gas: List[int], cost: List[int]) -> int:
        if sum(gas) < sum(cost): return -1          # 전체 가스량과 이동비용 비교
        
        tank = 0
        start = 0

        for station in range(len(gas)):             # 출발점 찾기
            tank += gas[station] - cost[station]
            if tank < 0:
                start = station + 1
                tank = 0
        
        return start

Runtime: 14 ms | Beats 94.66%
Memory: 25.94 MB | Beats 67.80%

탐욕법(Greedy)을 사용해서 해결할 수 있다.

Other Solutions

1st

class Solution:
    def canCompleteCircuit(self, gas: List[int], cost: List[int]) -> int:
        n, total_surplus, surplus, start = len(gas), 0, 0, 0
        
        for i in range(n):
            total_surplus += gas[i] - cost[i]
            surplus += gas[i] - cost[i]
            if surplus < 0:
                surplus = 0
                start = i + 1
        return -1 if (total_surplus < 0) else start

time complexity: 𝑂(𝑛)
space complexity: 𝑂(1)

sum() 없이 한 번의 순회만으로 해결할 수도 있다.