Gas Station - Practice Coding Problems

Gas Station - Problem

Array Greedy Medium

Imagine you're on a road trip around a circular highway with n gas stations positioned along the route. Each gas station i has a specific amount of fuel available: gas[i] gallons.

Your car has an unlimited fuel tank but starts completely empty. The catch? It costs cost[i] gallons of fuel to drive from station i to the next station (i+1). Since the route is circular, after the last station, you return to the first one.

Your mission: Find a starting gas station where you can complete the entire circular journey exactly once, or determine if it's impossible.

Input: Two arrays gas[] and cost[] of length n
Output: The index of the starting gas station, or -1 if no solution exists
Guarantee: If a solution exists, it's unique!

Input & Output

example_1.py — Standard Case

$ Input: gas = [1,2,3,4,5], cost = [3,4,5,1,2]

› Output: 3

💡 Note: Starting at station 3: fuel=4, can reach station 4 (cost=1). At station 4: fuel=3+5=8, can reach station 0 (cost=2). At station 0: fuel=6+1=7, can reach station 1 (cost=3). Continue this way to complete the circuit.

example_2.py — Impossible Case

$ Input: gas = [2,3,4], cost = [3,4,3]

› Output: -1

💡 Note: Total gas available is 2+3+4=9, but total cost needed is 3+4+3=10. Since total gas < total cost, it's impossible to complete the circuit from any starting point.

example_3.py — Edge Case Single Station

$ Input: gas = [5], cost = [4]

› Output: 0

💡 Note: With only one station, we start there, gain 5 fuel, and need 4 to return to the same station. Since 5 ≥ 4, we can complete the circuit starting from station 0.

Constraints

gas.length == n
cost.length == n
1 ≤ n ≤ 10⁵
0 ≤ gas[i], cost[i] ≤ 10⁴
The solution is guaranteed to be unique if it exists

Visualization

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Understanding the Visualization

Check Total Feasibility

First, verify if the total fuel available across all stations is enough to cover the total cost of the entire journey

Track Fuel Balance

Start from station 0 and keep track of your fuel balance as you visit each station

Reset on Fuel Shortage

Whenever you run out of fuel, you know that station (and all previous ones) cannot be the starting point

Find Optimal Start

The first station after your last fuel shortage becomes your candidate starting point

Guarantee Solution

If total fuel ≥ total cost, this candidate is guaranteed to work due to the circular nature

Key Takeaway

🎯 Key Insight: The optimal starting point is always right after the last station where we run out of fuel, because that's where the fuel surplus begins!

Asked in

a Amazon 45 G Google 38 f Meta 29 ⊞ Microsoft 22

The optimal solution uses a single-pass greedy algorithm with O(n) time complexity. The key insight is tracking both total fuel balance (for feasibility) and current fuel balance (for finding the start). When current fuel goes negative, we reset and try the next station as our new candidate start point.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Try Every Starting Point)	O(n²)	O(1)	Test each gas station as a starting point and simulate the complete journey
One-Pass Greedy (Optimal)	O(n)	O(1)	Single pass solution using greedy strategy with total fuel tracking
Two-Pass Greedy Approach	O(n)	O(1)	First check if solution exists, then find the optimal starting point using deficit tracking

Brute Force (Try Every Starting Point) — Algorithm Steps

For each station i from 0 to n-1, try it as starting point
Initialize current fuel to 0 and start the journey
At each station, add gas[i] and subtract cost[i] to travel to next station
If fuel becomes negative at any point, this starting point fails
If we complete the full circle, return this starting index
If no starting point works, return -1

Visualization

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Step-by-Step Walkthrough

Try Start Point 0

Begin at station 0, add gas[0], try to reach station 1

Continue or Fail

If fuel >= cost[0], continue to next station. Otherwise, try next start point

Complete Circuit

Keep going until either fuel runs out or we complete the full circle

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>

int canCompleteCircuit(int* gas, int* cost, int gasSize) {
    int n = gasSize;
    
    for (int start = 0; start < n; start++) {
        int currentFuel = 0;
        int stationsVisited = 0;
        int currentStation = start;
        
        while (stationsVisited < n) {
            // Add gas from current station
            currentFuel += gas[currentStation];
            // Check if we can reach next station
            if (currentFuel < cost[currentStation]) {
                break; // Can't continue from this starting point
            }
            // Travel to next station
            currentFuel -= cost[currentStation];
            currentStation = (currentStation + 1) % n;
            stationsVisited++;
        }
        
        if (stationsVisited == n) {
            return start; // Successfully completed the circuit
        }
    }
    
    return -1; // No valid starting point found
}

int main() {
    int gas[] = {1, 2, 3, 4, 5};
    int cost[] = {3, 4, 5, 1, 2};
    printf("%d\n", canCompleteCircuit(gas, cost, 5));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each of n starting points, we simulate n stations in the worst case

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a few variables to track current fuel and position

✓ Linear Space

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High Frequency

~18 min Avg. Time

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Try Every Starting Point) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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