Gas Station gas station problem

2019-06-01 17:09:30

Problem Description:

Problem Solving:

In fact, this is a math problem on nature.

【theorem】

For a loop through the array, if the whole array and SUM> = 0, then there must be found such an element in the array: From this array element, the array around the circle, and has been able to accumulate guarantee for non-negative status.

【prove】

Start from the first and the figures for the middle and there must be a minimum cumulative point, we set x. The final sum> = 0.

Now we are back from the x accumulate, then there must be a non-negative in the state, and to the last site there will be some surplus, and then bound to a negative number does not appear to start from the lowest point x marching.

【Leetcode Discuss】

If sum of all gas[i]-cost[i] is greater than or equal to 0, then there is a start position you can travel the whole circle.
Let i be the index such that the the partial sum

gas[0]-cost[0]+gas[1]-cost[1]+...+gas[i]-cost[i]

is the smallest, then the start position should be start=i+1 ( start=0 if i=n-1). Consider any other partial sum, for example,

gas[0]-cost[0]+gas[1]-cost[1]+...+gas[i]-cost[i]+gas[i+1]-cost[i+1]

Since gas[0]-cost[0]+gas[1]-cost[1]+...+gas[i]-cost[i] is the smallest, we must have

gas[i+1]-cost[i+1]>=0

in order for gas[0]-cost[0]+gas[1]-cost[1]+...+gas[i]-cost[i]+gas[i+1]-cost[i+1] to be greater.
The same reasoning gives that

 gas[i+1]-cost[i+1]>=0
 gas[i+1]-cost[i+1]+gas[i+2]-cost[i+2]>=0 ....... gas[i+1]-cost[i+1]+gas[i+2]-cost[i+2]+...+gas[n-1]-cost[n-1]>=0 

What about for the partial sums that wraps around?

gas[0]-cost[0]+gas[1]-cost[1]+...+gas[j]-cost[j] + gas[i+1]-cost[i+1]+...+gas[n-1]-cost[n-1] >= gas[0]-cost[0]+gas[1]-cost[1]+...+gas[i]-cost[i] + gas[i+1]-cost[i+1]+...+gas[n-1]-cost[n-1] >=0

The last inequality is due to the assumption that the entire sum of gas[k]-cost[k] is greater than or equal to 0.
So we have that all the partial sums

gas[i+1]-cost[i+1]>=0,
gas[i+1]-cost[i+1]+gas[i+2]-cost[i+2]>=0, gas[i+1]-cost[i+1]+gas[i+2]-cost[i+2]+...+gas[n-1]-cost[n-1]>=0, ... gas[i+1]-cost[i+1]+...+gas[n-1]-cost[n-1] + gas[0]-cost[0]+gas[1]-cost[1]+...+gas[j]-cost[j]>=0, ...

Thus i+1 is the position to start.

Therefore, for this question, we can calculate whether the total of gas -? Cost> = 0 If yes, then there must be a solution, in this context, only need to iterate over the array, if you encounter a situation can not be reached, then the it can be calculated from the first can not reach the start again.

    public int canCompleteCircuit(int[] gas, int[] cost) {
        int n = gas.length;
        int sum = 0;
        for (int i = 0; i < n; i++) sum += gas[i] - cost[i];
        if (sum < 0) return -1;
        int start = 0;
        int tank = 0;
        for (int i = 0; i < n; i++) {
            tank += gas[i];
            if (tank < cost[i]) {
                start = i + 1;
                tank = 0;
            }
            else {
                tank -= cost[i];
            }
        }
        return start;
    }