LeetCode 134: 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.
Note:
The solution is guaranteed to be unique.

public class Solution {
    public int canCompleteCircuit(int[] gas, int[] cost) {
        // O(n) time complexity
        // Let xi = gas[i] - cost[i]
        // A requirement is Σ xi ≥ 0 (enough gas to complete a full circle).
        // Consider Si = x1 + x2 + ... + xi
        // Note that Sn ≥ 0.
        // Now pick the smallest k such that Sk is the least and start at the station next to it.
        
        // Now for k < j ≤ n, we have the gas in tank = Sj - Sk ≥ 0.
        // e.g., j = k+1, Sj - Sk = (Sk + (gas[k+1]-cost[k+1])) - Sk = gas[k+1] - cost[k+1] > 0
        // So we should start at k+1
        
        // for 1 ≤ j ≤ k, we have gas in tank = xk+1 + .. + xn + x1 + x2 + .. + xj = Sn - Sk + Sj = Sn + (Sj-Sk) ≥ 0.
        // Thus starting at k+1 will ensure there is enough gas accumulated at each station to get to the next station.        
        int minS = Integer.MAX_VALUE, S = 0, position = 0;
        
        for (int i = 0; i < gas.length; i++)
        {
            S += gas[i] - cost[i];
            
            if (S < minS)
            {
                minS = S;
                position = (i+1) % gas.length;
            }
        }
        
        if (S >= 0)
            return position;
        else
            return -1;        
    }
}