Problem Description
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
Difficulty: Medium
Tags: array, greedy
Rating: 90.91%
Solution Approach
The solution uses two key insights:
- Total Gas vs Total Cost: If the total gas available is less than the total cost needed, it’s impossible to complete the circuit. This is our first check.
- Greedy Approach: If we can’t reach a station
j
from stationi
, then we can’t reach stationj
from any station betweeni
andj-1
.
Solution
Here’s my Python solution to this problem:
class Solution:
def canCompleteCircuit(self, gas: List[int], cost: List[int]) -> int:
if sum(gas) < sum(cost): return -1
curr_g = 0
start = 0
for i in range(len(gas)):
curr_g += gas[i] - cost[i]
if curr_g < 0:
start = i + 1
curr_g = 0
return start
Complexity Analysis
The solution has the following complexity characteristics:
- Time Complexity:
- Space Complexity: