Gas Station LeetCode Solution | Easy Approach

Minimum Cost to Merge Stones
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Gas Station 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:

  • gas.length == n
  • cost.length == n
  • 1 <= n <= 105
  • 0 <= gas[i], cost[i] <= 104

Gas Station Solutions

Time: O(n)
Space: O(1)

C++

class Solution {
 public:
  int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
    const int gasses = accumulate(begin(gas), end(gas), 0);
    const int costs = accumulate(begin(cost), end(cost), 0);
    if (gasses - costs < 0)
      return -1;

    int ans = 0;
    int sum = 0;

    // try to start from each index
    for (int i = 0; i < gas.size(); ++i) {
      sum += gas[i] - cost[i];
      if (sum < 0) {
        sum = 0;
        ans = i + 1;  // start from next index
      }
    }

    return ans;
  }
};

Java

 class Solution {
  public int canCompleteCircuit(int[] gas, int[] cost) {
    final int gasses = Arrays.stream(gas).sum();
    final int costs = Arrays.stream(cost).sum();
    if (gasses - costs < 0)
      return -1;

    int ans = 0;
    int sum = 0;

    // try to start from each index
    for (int i = 0; i < gas.length; ++i) {
      sum += gas[i] - cost[i];
      if (sum < 0) {
        sum = 0;
        ans = i + 1; // start from next index
      }
    }

    return ans;
  }
}

Python

 class Solution:
  def canCompleteCircuit(self, gas: List[int], cost: List[int]) -> int:
    ans = 0
    net = 0
    sum = 0

    for i in range(len(gas)):
      net += gas[i] - cost[i]
      sum += gas[i] - cost[i]
      if sum < 0:
        sum = 0
        ans = i + 1

    return -1 if net < 0 else ans

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