134. Gas Station LeetCode Solution

Last updated: February 14, 2025

In this guide, you will get 134. Gas Station LeetCode Solution with the best time and space complexity. The solution to Gas Station problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Gas Station solution in C++
Gas Station solution in Java
Gas Station solution in Python
Additional Resources

134. Gas Station LeetCode Solution image

Problem Statement of 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

Complexity Analysis

Time Complexity: O(n)
Space Complexity: O(1)

134. Gas Station LeetCode Solution in C++

class Solution {
 public:
  int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
    const int gasSum = accumulate(gas.begin(), gas.end(), 0);
    const int costSum = accumulate(cost.begin(), cost.end(), 0);
    if (gasSum - costSum < 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 the next index.
      }
    }

    return ans;
  }
};
/* code provided by PROGIEZ */

134. Gas Station LeetCode Solution in Java

class Solution {
  public int canCompleteCircuit(int[] gas, int[] cost) {
    final int gasSum = Arrays.stream(gas).sum();
    final int costSum = Arrays.stream(cost).sum();
    if (gasSum - costSum < 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 the next index.
      }
    }

    return ans;
  }
}
// code provided by PROGIEZ

134. Gas Station LeetCode Solution in Python

class Solution:
  def canCompleteCircuit(self, gas: list[int], cost: list[int]) -> int:
    ans = 0
    net = 0
    summ = 0

    # Try to start from each index.
    for i in range(len(gas)):
      net += gas[i] - cost[i]
      summ += gas[i] - cost[i]
      if summ < 0:
        summ = 0
        ans = i + 1  # Start from the next index.

    return -1 if net < 0 else ans
 # code by PROGIEZ

Additional Resources

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