Possibly the MOST easiest approach, O(N), one variable, Python

  • 14
    def canCompleteCircuit(self, gas, cost):
        :type gas: List[int]
        :type cost: List[int]
        :rtype: int
        if len(gas) == 0 or len(cost) == 0 or sum(gas) < sum(cost):
            return -1
        position = 0
        balance = 0 # current tank balance
        for i in range(len(gas)):
            balance += gas[i] - cost[i] # update balance
            if balance < 0: # balance drops to negative, reset the start position
                balance = 0
                position = i+1
        return position

  • 3

    What do you mean with "one variable"? I count three.

  • 0

    Hi Stefan, sorry for the ambiguity.
    By "one variable", I mean only using one variable to store the tank balance state. No need to use additional variables to store previous total amount like others' approach.

  • 1

    That "additional" variable is hidden inside sum(gas) and sum(cost) making in a 3*n algorithm instead of 1*n. Granted it's still O(n), but usually I would choose one more variable and a condition over 2 full traversals.

  • 0

    @TWiStErRob You do realize that this approach can still be implemented in 2 traversals if we iterate over len(gas) to find sum(gas) and sum(cost) instead of using sum functions?

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