Minimum Number of Refills for a Car Journey

Minimum Number of Refills Problem Explained Using Greedy Algorithms

A complete educational guide covering intuition, mathematics, algorithm design, complexity analysis, code implementation, examples, dry runs, CLI demonstrations, FAQs, and interview preparation.

Table of Contents

• Introduction
• Problem Statement
• Real World Analogy
• Mathematical Foundation
• Why Greedy Works
• Algorithm Breakdown
• Example Walkthrough
• Code Example
• CLI Demonstration
• Complexity Analysis
• Edge Cases
• FAQ
• Key Takeaways

Introduction

The Minimum Number of Refills problem is one of the most famous examples used to teach Greedy Algorithms. It appears in algorithm courses, coding interviews, competitive programming contests, and university assignments.

At first glance, the problem appears simple. You have a car. You know the distance to your destination. You know how far the car can travel on a full tank. You also know where gas stations are located.

The challenge is determining the minimum number of times you need to stop for fuel.

Stopping too early can increase the number of refills. Stopping too late may leave you stranded. Therefore, a strategy is needed that guarantees the smallest number of refueling stops.

Key Insight: Always drive to the farthest gas station that is still reachable with your current fuel.

Problem Statement

Given:

• Total distance to destination
• Maximum distance possible on a full tank
• Number of gas stations
• Locations of gas stations

Determine the minimum number of refills needed to reach the destination. Return -1 if the destination cannot be reached.

Real World Understanding

Imagine driving from one city to another. Your fuel tank allows you to travel only 400 kilometers. Along the highway, fuel stations are placed at different intervals.

Should you stop at every station? No.

Should you skip all stations? Also no.

The optimal strategy is stopping only when necessary and always maximizing the distance covered before each refill.

This mirrors many real-world optimization problems:

• Battery charging for electric vehicles
• Drone charging stations
• Network packet routing
• Resource allocation systems
• Supply chain logistics

Mathematical Foundation

Let's define:

• D = destination distance
• M = fuel tank capacity
• S = set of station positions

The car can move at most:

Distance Reachable = Current Position + M

If:

Next Station > Reachable Distance

Then:

Journey becomes impossible.

The optimization objective is:

Minimize:

Number of Refills

subject to:

Every movement between consecutive stops must be less than or equal to fuel capacity.

Why Farthest Reachable Station?

Suppose two stations are reachable:

• Station A at 200
• Station B at 350

Fuel capacity = 400

Stopping at 200 provides no advantage over stopping at 350.

In fact:

350 dominates 200 because it extends future reach further.

This is the mathematical reason Greedy works.

Why Greedy Algorithm Works

Greedy algorithms make the best local decision at every step.

The local decision here is:

Choose the farthest reachable gas station.

The remarkable property is that this local choice also produces the globally optimal solution.

Proof Idea

Assume an optimal solution stops earlier than the farthest reachable station.

Replacing that stop with the farther station cannot reduce future options. It only increases reach.

Therefore stopping earlier can never be better.

Thus choosing the farthest reachable station is always safe.

Algorithm Breakdown

1. Start with a full tank.
2. Compute current reachable distance.
3. Find the farthest station within range.
4. If no station exists, return -1.
5. Refill.
6. Update reach.
7. Repeat until destination is reachable.

Detailed Example Walkthrough

Input

Distance = 950
Fuel Capacity = 400

Stations:
200
375
550
750

Step 1

Initial reach:

0 + 400 = 400

Reachable stations:

• 200
• 375

Choose 375.

Refills = 1

Step 2

375 + 400 = 775

Reachable stations:

• 550
• 750

Choose 750.

Refills = 2

Step 3

750 + 400 = 1150

1150 is greater than destination 950.

Destination reached.

Answer

2

Python Implementation

def car_fueling(dist, miles, n, gas_stations):

    num_refill = 0
    curr_refill = 0

    gas_stations = [0] + gas_stations + [dist]

    while curr_refill <= n:

        last_refill = curr_refill

        while (
            curr_refill <= n and
            gas_stations[curr_refill + 1] -
            gas_stations[last_refill] <= miles
        ):
            curr_refill += 1

        if curr_refill == last_refill:
            return -1

        if curr_refill <= n:
            num_refill += 1

    return num_refill

CLI Demonstration

Example 1

$ python fueling.py

Distance: 950
Fuel: 400

Stations:
200
375
550
750

Output:
2

Example 2

$ python fueling.py

Distance: 10
Fuel: 3

Stations:
1
2
5
9

Output:
-1

Execution Trace

Current Reach = 400

Reachable:
200
375

Choosing:
375

Refill Count = 1

Current Reach = 775

Reachable:
550
750

Choosing:
750

Refill Count = 2

Destination Reachable

Important Edge Cases

• No gas stations
• Destination already reachable
• First station unreachable
• Last station too far from destination
• Multiple stations at same location
• Very large input sizes

Case: No Stations

If fuel capacity is enough to reach destination, answer is 0. Otherwise answer is -1.

Case: First Station Too Far

If first station exceeds fuel range, journey is impossible immediately.

Case: Destination Reachable Initially

If capacity is greater than destination distance, no refill is needed.

Complexity Analysis

Metric	Value
Time Complexity	O(n)
Space Complexity	O(1)
Optimal	Yes

Each gas station is processed at most once. Therefore runtime remains linear.

Common Mistakes

• Stopping at nearest station instead of farthest.
• Forgetting destination acts like a final station.
• Ignoring unreachable gaps.
• Miscounting initial tank as a refill.
• Using unnecessary dynamic programming.

Interview Questions

Why is Greedy better than Dynamic Programming here?

Because local optimal choices always lead to the global optimum. No overlapping subproblems exist.

Can this be solved in O(n)?

Yes. Every station is visited at most once.

What makes Greedy valid?

The farthest reachable station always dominates earlier reachable stations.

Frequently Asked Questions

What algorithm category does this belong to?

Greedy Algorithms.

Can stations be unsorted?

Usually no. If unsorted, sort first.

Can there be multiple optimal solutions?

Possibly, but the greedy solution still achieves the minimum refill count.

Why not stop earlier?

Stopping earlier never increases future reach. Therefore it cannot improve the solution.

Key Takeaways

• Always choose the farthest reachable station.
• Greedy strategy guarantees optimality.
• Time complexity is O(n).
• Space complexity is O(1).
• If no station is reachable, return -1.
• The destination can be treated as a virtual gas station.
• This is a classic interview and competitive programming problem.
• The problem demonstrates how local optimization can lead to a globally optimal result.
• Understanding the proof is as important as understanding the implementation.
• The technique appears in logistics, routing, networking, and EV charging systems.

Final Thoughts

The Minimum Number of Refills problem is one of the most elegant demonstrations of Greedy Algorithm design. Instead of exploring every possible route, we make a single intelligent decision at every step: travel as far as possible before refueling.

This seemingly simple idea dramatically reduces complexity while still guaranteeing an optimal answer. The lesson extends beyond programming. Many optimization problems in transportation, networking, scheduling, and resource allocation can be solved by identifying a locally optimal decision that remains globally valid.

Mastering this problem provides a strong foundation for understanding more advanced greedy strategies such as interval scheduling, Huffman coding, activity selection, and minimum spanning tree algorithms.