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This article mainly explains "how to solve the gas station problem with python". The content of the explanation is simple and clear, and it is easy to learn and understand. Please follow the editor's train of thought to study and learn "how to solve the gas station problem with python".

Topic: gas station

There are N gas stations on a ring road, of which the I gas station has gasoline gas [I] liters.

You have a car with unlimited fuel tank capacity, and it takes cost [I] liters of gasoline to drive from the I gas station to the first gas station. You start from one of the gas stations and the fuel tank is empty at first.

If you can drive around the loop, return to the number of the gas station at the time of departure, otherwise return to-1.

Description:

If there is a solution to the question, the answer is the only answer.

The input array is a non-empty array with the same length.

The elements in the input array are all non-negative.

Example 1:

Enter:

Gas = [1, 2, 3, 4, 5]

Cost = [3, 4, 5, 5, 1, 1, 2]

Output: 3

Explanation:

4 litres of gasoline can be obtained from the No. 3 gas station (index 3). At this time, the fuel tank has = 0 + 4 = 4 litres of gasoline.

Go to the No. 4 gas station, the fuel tank has 4-1 + 5 = 8 liters of gasoline.

Go to the No. 0 gas station, the fuel tank has 8-2 + 1 = 7 liters of gasoline.

Go to the No. 1 gas station, the fuel tank has 7-3 + 2 = 6 liters of gasoline.

Go to the No. 2 gas station, the fuel tank has 6-4 + 3 = 5 litres of gasoline.

To get to the No. 3 gas station, you need to consume 5 liters of gasoline, just enough to get you back to the No. 3 gas station.

Therefore, 3 can be the starting index.

Solve the problem:

Only when sum (gas)-sum (cost) > = 0 can there be a solution.

So, how to calculate the departure position?

As long as you start from one location, calculate the cumulative gasoline quantity and cumulative consumption to the next location, if the cumulative gasoline quantity

Int:

Residue = [gasI]-cost [I] for i in range (len (gas))]

If sum (residue) < 0:

Return-1

# and greater than 0, there must be a solution

Acc = 0

Start = 0

For i in range (len (residue)):

Acc + = residue [I]

If acc < 0:

Acc = 0

Start = I + 1

Return start

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