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# Gas Station LeetCode Solution

## Problem – Gas Station LeetCode Solution

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:

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

## Gas Station LeetCode Solution in Java

``````class Solution {
public int canCompleteCircuit(int[] gas, int[] cost) {
int n = gas.length;
int total_surplus = 0;
int surplus = 0;
int start = 0;

for(int i = 0; i < n; i++){
total_surplus += gas[i] - cost[i];
surplus += gas[i] - cost[i];
if(surplus < 0){
surplus = 0;
start = i + 1;
}
}
return (total_surplus < 0) ? -1 : start;
}
}
``````

## Gas Station LeetCode Solution in C++

``````class Solution {
public:
int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
int n = gas.size();
int total_surplus = 0;
int surplus = 0;
int start = 0;

for(int i = 0; i < n; i++){
total_surplus += gas[i] - cost[i];
surplus += gas[i] - cost[i];
if(surplus < 0){
surplus = 0;
start = i + 1;
}
}
return (total_surplus < 0) ? -1 : start;
}
};
``````

## Gas Station LeetCode Solution in Python

``````class Solution:
def canCompleteCircuit(self, gas: List[int], cost: List[int]) -> int:
n, total_surplus, surplus, start = len(gas), 0, 0, 0

for i in range(n):
total_surplus += gas[i] - cost[i]
surplus += gas[i] - cost[i]
if surplus < 0:
surplus = 0
start = i + 1
return -1 if (total_surplus < 0) else start
``````
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