Hello Programmers, In this post you will know how to solve the** Gas Station Leetcode Solution** problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

One more thing to add, don’t directly look for the solutions, first try to solve the problems of Leetcode by yourself. If you find any difficulty after trying several times, then you can look for solutions.

**Problem**

There are `n`

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

station is ^{th}`gas[i]`

.

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

of gas to travel from the `i`

station to its next ^{th}`(i + 1)`

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

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: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}

**Gas Station Leetcode Solutions 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 Solutions** **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 Solutions 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 **BrokenProgrammers**. This tutorial is only for Educational and Learning purposes.

**Next: **Candy Leetcode Solution