Gas Station

Similar Problems:

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

You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from station i to its next station (i+1). You begin the journey with an empty tank at one of the gas stations.

Return the starting gas station’s index if you can travel around the circuit once, otherwise return -1.

Note:

The solution is guaranteed to be unique.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

## Blog link: https://code.dennyzhang.com/gas-station ## Basic Ideas: One pass ## If gas[i] - cost[i] < 0, this node can't be the target ## ## If total sum of gas-cost < 0, we can't find the solution. ## If total sum of gas-cost >=0, we must be able to find one solution ## ## Get accumulated sum of gas-cost ## If it turn to negative, this node and all previous nodes can't be the target ## ## Question: ## How to handle a circle, instead of an array? ## How to detect cases with no solution? ## ## Complexity: Time O(n), Space O(1) class Solution: def canCompleteCircuit(self, gas, cost): """ :type gas: List[int] :type cost: List[int] :rtype: int """ length = len(gas) res, curSum, totalSum = 0, 0, 0 for i in range(0, length): totalSum += gas[i] - cost[i] curSum += gas[i] - cost[i] if curSum < 0: res = i+1 curSum = 0 return res if totalSum >=0 else -1

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