Algorithm
Problem Name: 134. Gas Station
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.length1 <= n <= 1050 <= gas[i], cost[i] <= 104
Code Examples
#1 Code Example with C Programming
Code -
C Programming
#include <stdio.h>
#include <assert.h>
int canCompleteCircuit(int* gas, int gasSize, int* cost, int costSize) {
if (gasSize != costSize) return -1;
int i;
int tank = 0, sum = 0;
int ans = 0;
for (i = 0; i < gasSize; i++) {
sum += gas[i] - cost[i];
tank += gas[i] - cost[i];
if (sum < 0) {
ans = i + 1;
sum = 0;
}
}
return tank >= 0 ? ans : -1;
}
int main() {
int gas[] = { 1, 2 };
int cost[] = { 2, 1 };
assert(canCompleteCircuit(gas, sizeof(gas) / sizeof(gas[0]),
cost, sizeof(cost) / sizeof(cost[0])) == 1);
printf("all tests passed!\n");
return 0;
}
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Output
#2 Code Example with C++ Programming
Code -
C++ Programming
class Solution {
public:
int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
int n(gas.size()), sum(0), total(0), res(0);
for (int i = 0; i < n; ++i) {
sum += gas[i] - cost[i];
if (sum < 0) {
res = i + 1;
sum = 0;
}
total += gas[i] - cost[i];
}
return total >= 0 ? res : -1;
}
};
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Output
#3 Code Example with Java Programming
Code -
Java Programming
class Solution {
public int canCompleteCircuit(int[] gas, int[] cost) {
int n = gas.length;
int tankCapacity = 0;
int currTankCapacity = 0;
int startingStation = 0;
for (int i = 0; i < n; i++) {
tankCapacity += gas[i] - cost[i];
currTankCapacity += gas[i] - cost[i];
if (currTankCapacity < 0) {
startingStation = i + 1;
currTankCapacity = 0;
}
}
return tankCapacity >= 0 ? startingStation : -1;
}
}
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Output
#4 Code Example with Javascript Programming
Code -
Javascript Programming
const canCompleteCircuit = function(gas, cost) {
let total = 0
let curLeft = 0
let curtIdx = 0
for (let i = 0; i < gas.length; i++) {
total += gas[i] - cost[i]
curLeft += gas[i] - cost[i]
if (curLeft < 0) {
curtIdx = i + 1
curLeft = 0
}
}
return total < 0 ? -1 : curtIdx
}
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Output
#5 Code Example with Python Programming
Code -
Python Programming
class Solution:
def canCompleteCircuit(self, gas, cost, cur = 0, index = 0):
for i in range(len(gas)):
cur += gas[i] - cost[i]
if cur < 0: cur, index = 0, i + 1
return index if index < len(gas) and sum(gas) >= sum(cost) else -1
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Output
#6 Code Example with C# Programming
Code -
C# Programming
namespace LeetCode
{
public class _0134_GasStation
{
public int CanCompleteCircuit(int[] gas, int[] cost)
{
int totalGas = 0, currentTank = 0;
var startIndex = 0;
for (int i = 0; i < gas.Length; i++)
{
totalGas += gas[i] - cost[i];
currentTank += gas[i] - cost[i];
if (currentTank < 0)
{
startIndex = i + 1;
currentTank = 0;
}
}
return totalGas >= 0 ? startIndex : -1;
}
}
}
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Output