Gas Station

There are N gas stations along a circular route, where the amount of gas stored 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:

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.

If the total sum of gas stored in all the gas stations is bigger than the sum of cost. Then, there must be a solution that exists. So, the following formula must be true for a valid solution for a given cost[n] and the gas[n].

The Formula should be satisfied

If a car starts at point A and can reach point B with the available gas, then the move from point A to point B is part of a valid route. The tank should never be negative in a valid route, so restart the trip whenever the car tank becomes negative.

Gas Stations

Implementation:

public int canCompleteCircuit(int[] gas, int[] cost) {
    int len = gas.length;
    int totalTank = 0,currentTank = 0, startStation = 0;
    for (int i = 0; i < len; i++) {
        totalTank = totalTank + (gas[i] - cost[i]);
        currentTank = currentTank + (gas[i] - cost[i]);
        if (currentTank < 0) {
            currentTank = 0;
            startStation = i + 1;
        }
    }
    return totalTank >= 0 ? startStation : -1;
}

Complexity Analysis

  • Time Complexity : O(N) since there is only one loop over all stations here.
  • Space Complexity : O(1) since it’s a constant space solution.

Categories: Dynamic Programming

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