Gas Station LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the Gas Station in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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Link for the ProblemGas Station– LeetCode Problem

Gas Station– LeetCode Problem

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:

  • gas.length == n
  • cost.length == n
  • 1 <= n <= 105
  • 0 <= gas[i], cost[i] <= 104
Gas Station– LeetCode Solutions
Gas Station Solution in 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;
  }
};
Gas Station Solution in 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;
  }
}
Gas Station Solution in 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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