Gas Station Leetcode Solution

In this post, we are going to solve the Gas Station Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

Table of Contents

Problem

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:

n == gas.length == cost.length
1 <= n <= 10⁵

0 <= gas[i], cost[i] <= 10⁴

Now, let’s see the leetcode solution of Gas Station Leetcode Solution.

Gas Station Leetcode Solution in Python

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

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

    return -1 if net < 0 else ans

Gas Station Leetcode Solution in CPP

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 Leetcode 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;
  }
}

Note: This problem Gas Station is generated by Leetcode but the solution is provided by Chase2learn This tutorial is only for Educational and Learning purpose.