加油站

Problem

There are n gas stations along a circular route, where the amount of gas at the ith station is gas[i].

You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from the ith station to its next (i + 1)th station. You begin the journey with an empty tank at one of the gas stations.

Given two integer arrays gas and cost, return the starting gas station's index if you can travel around the circuit once in the clockwise direction, otherwise return -1. If there exists a solution, it is guaranteed to be unique.

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.

Constraints:

• n == gas.length == cost.length
• 1 <= n <= \(10^5\)
• 0 <= gas[i], cost[i] <= \(10^4\)
• The input is generated such that the answer is unique.

Idea

Use a greedy one-pass scan over the array.

If the total gas is less than the total cost, no start can work, so return -1. Otherwise a valid start always exists.

Scan the stations while keeping the current tank, which is the sum of gas[i] - cost[i] since the candidate start. If the tank becomes negative at station i, no station between the candidate start and i can be the answer, because each of them would reach i with even less gas. So reset the tank to 0 and move the candidate start to i + 1.

Time complexity: \(O(n)\), Space complexity: \(O(1)\)

Pseudo Code

total = 0;
tank = 0;
start = 0;

for (i = 0; i < n; i++) {
    diff = gas[i] - cost[i];
    total += diff;
    tank += diff;
    if (tank < 0) {
        start = i + 1;
        tank = 0;
    }
}

if (total < 0) {
    return -1;
}
return start;