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
SOLUTION:
class Solution:
def canCompleteCircuit(self, gas: List[int], cost: List[int]) -> int:
n = len(gas)
total = 0
abstotal = 0
res = 0
for i in range(n):
diff = gas[i] - cost[i]
total += diff
abstotal += diff
if total < 0:
total = 0
res = i + 1
if abstotal < 0:
return -1
return res
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