Gas Station

• leetcode 134
• Medium

Description

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.

Hide Tags: Greedy

Java Solution

public class Solution {
    public int canCompleteCircuit(int[] gas, int[] cost) {
        int i, j, n = gas.length;


        for(i = 0; i < n; i += j + 1){
            int gasLeft = 0;
            for( j = 0; j < n; j++){
                int k = (i + j) % n;
                gasLeft += gas[k] - cost[k];
                if(gasLeft < 0) break;
            }
            if(j == n) return i;
        }
        return -1;
    }
}

C++ Solution

There is another solution using two variables.

class Solution {
public:
    int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
        int total = 0;
        int sum = 0;

        int j = -1;

        for(int i = 0;i < gas.size();i++){
            sum += gas[i] - cost[i];
            total += gas[i] - cost[i];

            if(sum < 0){
                j = i;
                sum = 0;
            }
        }

        return total >= 0 ? j + 1 : -1;
    }
};

Comment

Compared with the second solution, I prefer to the first solution which clarify the think of "Greedy" more clearly.