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.length
  • 1 <= n <= 105
  • 0 <= 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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Input

x
+
cmd
gas = [1,2,3,4,5], cost = [3,4,5,1,2]

Output

x
+
cmd
3

#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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Input

x
+
cmd
gas = [1,2,3,4,5], cost = [3,4,5,1,2]

Output

x
+
cmd
3

#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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Input

x
+
cmd
gas = [2,3,4], cost = [3,4,3]

Output

x
+
cmd
-1

#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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Input

x
+
cmd
gas = [2,3,4], cost = [3,4,3]

Output

x
+
cmd
-1

#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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Input

x
+
cmd
gas = [1,2,3,4,5], cost = [3,4,5,1,2]

Output

x
+
cmd
3

#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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Input

x
+
cmd
gas = [1,2,3,4,5], cost = [3,4,5,1,2]

Output

x
+
cmd
3
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