gas-station.cpp

// 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 in the clockwise direction, otherwise return -1.
//
// Note:
//
//
// 	If there exists a solution, it is guaranteed to be unique.
// 	Both input arrays are non-empty and have the same length.
// 	Each element in the input arrays is a non-negative integer.
//
//
// Example 1:
//
//
// Input: 
// gas  = [1,2,3,4,5]
// cost = [3,4,5,1,2]
//
// Output: 3
//
// Explanation:
// Start at station 3 (index 3) and fill up with 4 unit of gas. Your tank = 0 + 4 = 4
// Travel to station 4. Your tank = 4 - 1 + 5 = 8
// Travel to station 0. Your tank = 8 - 2 + 1 = 7
// Travel to station 1. Your tank = 7 - 3 + 2 = 6
// Travel to station 2. Your tank = 6 - 4 + 3 = 5
// Travel to station 3. The cost is 5. Your gas is just enough to travel back to station 3.
// Therefore, return 3 as the starting index.
//
//
// Example 2:
//
//
// Input: 
// gas  = [2,3,4]
// cost = [3,4,3]
//
// Output: -1
//
// Explanation:
// You can't start at station 0 or 1, as there is not enough gas to travel to the next station.
// Let's start at station 2 and fill up with 4 unit of gas. Your tank = 0 + 4 = 4
// Travel to station 0. Your tank = 4 - 3 + 2 = 3
// Travel to station 1. Your tank = 3 - 3 + 3 = 3
// You cannot travel back to station 2, as it requires 4 unit of gas but you only have 3.
// Therefore, you can't travel around the circuit once no matter where you start.
//
//


class Solution {
public:
    int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
        for (int i = 0; i < gas.size(); ++i) {
            gas[i] -= cost[i];
        }
        int ans = 0;
        int sum = 0;
        for (int i = 0; i < gas.size()*2-1; ++i) {
            sum += gas[i % gas.size()];
            if (sum < 0) {
                ans = i + 1;
                if (ans >= gas.size()) {
                    return -1;
                }
                sum = 0;
            }
        }
        return ans;
    }
};