Problem Description
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.
Examples
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.
Python Solution
def canCompleteCircuit(gas: List[int], cost: List[int]) -> int:
if sum(gas) < sum(cost):
return -1
start = 0
tank = 0
for i in range(len(gas)):
tank += gas[i] - cost[i]
if tank < 0:
start = i + 1
tank = 0
return start
Java Solution
class Solution {
public int canCompleteCircuit(int[] gas, int[] cost) {
int totalGas = 0;
int totalCost = 0;
for (int i = 0; i < gas.length; i++) {
totalGas += gas[i];
totalCost += cost[i];
}
if (totalGas < totalCost) {
return -1;
}
int start = 0;
int tank = 0;
for (int i = 0; i < gas.length; i++) {
tank += gas[i] - cost[i];
if (tank < 0) {
start = i + 1;
tank = 0;
}
}
return start;
}
}
C++ Solution
class Solution {
public:
int canCompleteCircuit(vector& gas, vector& cost) {
int totalGas = 0;
int totalCost = 0;
for (int i = 0; i < gas.size(); i++) {
totalGas += gas[i];
totalCost += cost[i];
}
if (totalGas < totalCost) {
return -1;
}
int start = 0;
int tank = 0;
for (int i = 0; i < gas.size(); i++) {
tank += gas[i] - cost[i];
if (tank < 0) {
start = i + 1;
tank = 0;
}
}
return start;
}
};
JavaScript Solution
/**
* @param {number[]} gas
* @param {number[]} cost
* @return {number}
*/
var canCompleteCircuit = function(gas, cost) {
const totalGas = gas.reduce((a, b) => a + b, 0);
const totalCost = cost.reduce((a, b) => a + b, 0);
if (totalGas < totalCost) {
return -1;
}
let start = 0;
let tank = 0;
for (let i = 0; i < gas.length; i++) {
tank += gas[i] - cost[i];
if (tank < 0) {
start = i + 1;
tank = 0;
}
}
return start;
};
C# Solution
public class Solution {
public int CanCompleteCircuit(int[] gas, int[] cost) {
int totalGas = 0;
int totalCost = 0;
for (int i = 0; i < gas.Length; i++) {
totalGas += gas[i];
totalCost += cost[i];
}
if (totalGas < totalCost) {
return -1;
}
int start = 0;
int tank = 0;
for (int i = 0; i < gas.Length; i++) {
tank += gas[i] - cost[i];
if (tank < 0) {
start = i + 1;
tank = 0;
}
}
return start;
}
}
Complexity Analysis
- Time Complexity: O(n) where n is the number of gas stations
- Space Complexity: O(1) as we only use constant extra space
Solution Explanation
This solution uses a greedy approach:
- Key concept:
- Total gas check
- Running sum
- Starting point
- Algorithm steps:
- Check feasibility
- Track gas tank
- Update start point
- Find solution
Key points:
- Early validation
- Greedy choice
- Optimal substructure
- Single pass