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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0101_0200.s0134_gas_station;

// #Medium #Top_Interview_Questions #Array #Greedy
// #2022_06_24_Time_2_ms_(94.26%)_Space_62.5_MB_(87.11%)

/**
 * 134 - Gas Station\.
 *
 * Medium
 *
 * There are `n` gas stations along a circular route, where the amount of gas at the i^th station is `gas[i]`.
 *
 * You have a car with an unlimited gas tank and it costs `cost[i]` of gas to travel from the i^th 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:**
 *
 * *   `gas.length == n`
 * *   `cost.length == n`
 * *   1 <= n <= 10⁵
 * *   0 <= gas[i], cost[i] <= 10⁴
**/
public class Solution {
    public int canCompleteCircuit(int[] gas, int[] cost) {
        int sumGas = 0;
        int sumCost = 0;
        int curGas = 0;
        int result = -1;
        for (int i = 0; i < gas.length; i++) {
            curGas += gas[i] - cost[i];
            // re-calculate the starting point
            if (curGas < 0) {
                result = -1;
                curGas = 0;
            } else if (result == -1) {
                // set initial starting point
                result = i;
            }
            sumGas += gas[i];
            sumCost += cost[i];
        }
        if (sumGas < sumCost) {
            return -1;
        }
        return result;
    }
}