[LintCode] Gas Station

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.

Example

Given 4 gas stations with gas[i]=[1,1,3,1], and the cost[i]=[2,2,1,1]. The starting gas station's index is 2.

Analysis: This problem can be transformed to maximum subarray. If there is a station from where we can go through all the stations, and given the solution is unique, it must be the starting of the maximum subarray.

class Solution {
public:
    int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
        int n = gas.size();
        if(0 == n) return -1;
        int lastMax = gas[0]-cost[0];
        int totalMax = lastMax;
        int start = 0, maxStart = 0;
        for(int i=1;i<2*n-1;i++)
        {
            int j = i % n;
            int suf = gas[j] - cost[j];
            if(lastMax < 0)
            {
                if(i>=n) break;
                lastMax = suf;
                start = i;
            }
            else lastMax += suf;
            if(lastMax > totalMax)
            {
                totalMax = lastMax;
                maxStart = start;
            }
        }
        int sum = 0;
        for(int i=maxStart;i<n+maxStart;i++)
        {
            int j = i%n;
            sum += gas[j]-cost[j];
            if(sum < 0) return -1;
        }
        return maxStart;
    }
};