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

**Problem**

There areÂ `n`

Â gas stations along a circular route, where the amount of gas at theÂ `i`

Â station isÂ ^{th}`gas[i]`

.

You have a car with an unlimited gas tank and it costsÂ `cost[i]`

Â of gas to travel from theÂ `i`

Â station to its nextÂ ^{th}`(i + 1)`

Â station. You begin the journey with an empty tank at one of the gas stations.^{th}

Given two integer arraysÂ `gas`

Â andÂ `cost`

, returnÂ *the starting gas stationâ€™s index if you can travel around the circuit once in the clockwise direction, otherwise return*Â `-1`

. If there exists a solution, it isÂ **guaranteed**Â to beÂ **unique**

**Example 1:**

Input:gas = [1,2,3,4,5], cost = [3,4,5,1,2]Output:3Explanation: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:-1Explanation: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}

Now, letâ€™s see the leetcode solution ofÂ **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:**Â This problem **Gas Station** is generated byÂ **LeetcodeÂ **but the solution is provided byÂ **Chase2learn **This tutorial is only forÂ EducationalÂ andÂ LearningÂ purpose.