The Gas Station problem is a classic algorithmic challenge that tests one's ability to think about circular data structures and optimize for efficiency. we'll dive deep into three different approaches to solve this problem, discussing the logic, implementation, and time complexity of each.

The Problem Statement:

Imagine you’re on a circular road trip with a bunch of gas stations along the way. Each station has a certain amount of gas (let’s say in liters), and it takes a certain amount of gas to travel to the next station (let’s say in kilometers).

The question is: Can you complete the entire trip starting from any gas station, and if so, from which one?

means....

Your task is to find the starting station index from which you can complete a full circle of the gas stations. If no such station exists, return -1.

Given two arrays:

• gas[i] = Amount of gas at the ith station.
• cost[i] = Gas needed to get from the ith station to the (i+1)th station.

Here’s how it works:

• You start at one of the gas stations with an empty tank.
• At each station, you fill your tank with the gas available at that station.
• You then use some of that gas to drive to the next station.
• The goal is to find out if you can start from a particular station, travel all the way around the circuit, and end up back at the same station without running out of gas.

Key Points:

• The gas stations are arranged in a circle, so after the last station, you go back to the first one.
• The amount of gas at each station and the cost (gas required) to travel to the next station are given in two arrays, gas and cost.
• You need to find the starting station’s index from which you can complete the entire circuit. If it’s not possible to complete the trip from any station, you return -1.

Understanding the Example:

Let’s break down the example to understand what’s happening.

Example:

• gas =  [1, 2, 3, 4, 5]    indicates the gas remaining 
• cost = [3, 4, 5, 1, 2]    indicates cost ! 

This means:

• At station 0, you have 1 liter of gas,   but it costs 3 liters to reach station 1.
• At station 1, you have 2 liters of gas, but it costs 4 liters to reach station 2.
• And so on...

Now, the problem asks us to find out from which station you should start so that you can travel the entire circuit and end up back at the starting station.

Explanation:

Let's walk through the example:

1. Start at station 3 (index 3):
   • You fill up with 4 liters of gas (gas[3] = 4).
   • Tank status: 0 + 4 = 4 liters.
2. Travel to station 4:
   • It costs you 1 liter (cost[3] = 1) to reach station 4.
   • You fill up with 5 liters of gas at station 4.
   • Tank status: 4 - 1 + 5 = 8 liters.
3. Travel to station 0:
   • It costs you 2 liters (cost[4] = 2) to reach station 0.
   • You fill up with 1 liter of gas at station 0.
   • Tank status: 8 - 2 + 1 = 7 liters.
4. Travel to station 1:
   • It costs you 3 liters (cost[0] = 3) to reach station 1.
   • You fill up with 2 liters of gas at station 1.
   • Tank status: 7 - 3 + 2 = 6 liters.
5. Travel to station 2:
   • It costs you 4 liters (cost[1] = 4) to reach station 2.
   • You fill up with 3 liters of gas at station 2.
   • Tank status: 6 - 4 + 3 = 5 liters.
6. Travel to station 3:
   • It costs you 5 liters (cost[2] = 5) to reach station 3.
   • Now, your tank is just enough to return to station 3.
   • Tank status: 5 - 5 = 0 liters.

Lets Solve ! 

Approach 1: Brute Force (O(n^2) Complexity)

Explanation:

In this approach, we attempt to start the journey from each gas station and check if we can complete the circuit.

• For each station, calculate whether the car can move to the next station.
• Continue until either the car cannot move to the next station or it completes the full circle.

If the car successfully completes the circuit starting from a station, return that station's index. If none of the stations work, return -1.

#include <vector>
using namespace std;

class Solution {
public:
    int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
        int n = gas.size();
        
        // Try starting from each station
        for (int start = 0; start < n; start++) {
            int tank = 0;  // Initialize the tank with 0 fuel
            bool canComplete = true;
            
            // Check if we can complete the circle starting from station `start`
            for (int i = 0; i < n; i++) {
                int current = (start + i) % n;  // Current station
                tank += gas[current] - cost[current];  // Update tank
                
                // If at any point tank goes negative, we cannot continue
                if (tank < 0) {
                    canComplete = false;
                    break;
                }
            }
            
            // If we successfully complete the circle, return the start index
            if (canComplete) {
                return start;
            }
        }
        
        // If no start point is found
        return -1;
    }
};

Time Complexity:

• Worst Case: O(n2)O(n^2)O(n2), because for each station, we might traverse all the stations to check if we can complete the circuit.

Explanation:

• Brute Force: We check each possible starting station one by one. If a station fails, we move to the next station and repeat the process. This approach can be slow for large input sizes due to the nested loops.

Approach 2: Optimized O(n) Solution Using Greedy Strategy

Explanation:

This approach is an optimized version of the first approach, where we minimize unnecessary checks:

• If a station fails to provide enough gas to move to the next station, it means all stations between the current starting point and this station are not viable starting points. Thus, we can directly move our starting point to the next station after the failed one.

• We use two pointers: front and rear. Initially, both start at 0, and we try to expand rear as long as we can move from one station to the next. If we find that a move is not possible, we shift the front to rear+1.

Code Implementation:

#include <vector>
using namespace std;

class Solution {
public:
    int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
        int n = gas.size();
        int total_tank = 0, curr_tank = 0;
        int start = 0;

        for (int i = 0; i < n; ++i) {
            total_tank += gas[i] - cost[i];
            curr_tank += gas[i] - cost[i];

            // If we cannot move to the next station from the current station
            if (curr_tank < 0) {
                start = i + 1;  // Move the starting point to the next station
                curr_tank = 0;  // Reset the current tank
            }
        }

        return total_tank >= 0 ? start : -1;
    }
};

Time Complexity:

• Worst Case: O(n)O(n)O(n), as we are making a single pass through the stations and adjusting our start point only when necessary.

Explanation:

• Greedy Strategy: This approach improves the efficiency by eliminating unnecessary checks. If we find that a station cannot be part of a viable solution, we skip all preceding stations.

Approach 3: Greedy with Deficit Calculation (Most Efficient O(n))

Explanation:

This approach further simplifies the problem by leveraging the idea that:

• If we sum up the total gas available and the total cost required to travel the circuit, we can determine if a solution exists globally.
• We keep track of the total balance (remaining gas) and deficit (required gas) as we iterate over the stations.
• If at any point, the balance becomes negative, we know that all previous stations cannot be the starting point, and we reset the starting point to the next station.

This approach is the most optimal because:

• We only need a single pass over the array.
• It minimizes unnecessary operations and directly identifies the valid starting point if one exists.

Code Implementation:

#include <vector>
using namespace std;

class Solution {
public:
    int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
        int total_balance = 0;  // Tracks total balance of gas over cost
        int deficit = 0;  // Tracks the total deficit encountered
        int start = 0;  // The starting index

        for (int i = 0; i < gas.size(); i++) {
            total_balance += gas[i] - cost[i];  // Update total balance

            // If balance is negative, current start is not possible
            if (total_balance < 0) {
                deficit += total_balance;  // Add to deficit
                start = i + 1;  // Move start to the next station
                total_balance = 0;  // Reset balance
            }
        }

        // Check if overall balance + deficit is non-negative
        return (total_balance + deficit >= 0) ? start : -1;
    }
};

Time Complexity:

• Worst Case: O(n)O(n)O(n), making it highly efficient.

Pros and Cons of Each Approach:

1. Approach 1 (Brute Force):

   • Pros: Easy to understand and implement.
   • Cons: Inefficient for large inputs due to O(n2)O(n^2)O(n2) time complexity.

2. Approach 2 (Greedy with Two Pointers):

   • Pros: Improved efficiency, reduces unnecessary checks.
   • Cons: Slightly more complex logic compared to brute force.

3. Approach 3 (Greedy with Deficit Calculation):

   • Pros: Most efficient, directly identifies the correct start point with minimal operations.
   • Cons: None significant; it's the optimal approach.

Why Approach 3 is Best:

• Optimal Time Complexity: O(n)O(n)O(n), making it suitable for large input sizes.
• Simple Logic: Only requires a single pass through the array, with no need for nested loops or complex pointer management.
• Global View: It effectively combines local decisions with a global understanding of the problem (through deficit and balance calculations), ensuring that we don't overlook any potential solutions.

This approach is the most efficient and cleanest solution for this problem.