Minimum Number of Refills for a Car Journey

You are driving to a destination that is a certain distance away, and your car has a limited fuel tank capacity. Along the way, there are gas stations at specific distances from your starting point. The task is to determine the **minimum number of refills** you will need to reach the destination, if possible, based on the given car fuel capacity and the locations of the gas stations.

### Inputs:

1. **Distance to destination** (`dist`): The total distance you need to travel.

2. **Fuel tank capacity** (`miles`): The maximum number of miles your car can travel on a full tank.

3. **Number of gas stations** (`n`): The number of gas stations along the route.

4. **Gas station locations** (`gas_stations`): A list of distances where gas stations are located from your starting point.

### Objective:

Determine the minimum number of refills required to reach the destination. If it’s not possible to reach the destination, return `-1`.

### Solution

1. **Initial Setup:**

   - You start with a full tank of gas, so the car can travel `miles` distance before needing a refill.

   - The variable `num_refill` keeps track of the total number of refills.

   - `curr_refill` tracks your current position at a gas station, and `limit` keeps track of how far you can travel with the current fuel level.

2. **Loop Until the Destination is Reachable:**

   - While the destination is beyond your current limit (i.e., `limit < dist`), you need to check whether you can reach the destination or the next gas station.

   - If you can't reach the next gas station (or there are no more stations), return `-1` to indicate that the trip is impossible.

3. **Find the Furthest Reachable Gas Station:**

   - The algorithm moves to the furthest reachable gas station that is within the current fuel limit.

   - It iterates through the list of gas stations and moves `curr_refill` to the furthest station within range.

   

4. **Refill at the Furthest Station:**

   - After finding the furthest gas station within reach, refill the tank, increase the refill count, and update the `limit` to reflect how far you can now travel after refilling.

5. **Continue Until the Destination is Reachable:**

   - The loop continues, finding the next furthest reachable station and refilling until you can reach the destination.

6. **Return the Number of Refills:**

   - If you reach the destination, return the number of refills made during the journey.

### Step-by-Step Breakdown

1. **Check Feasibility**:

   - Start with a full tank, and as long as the destination cannot be reached with the current fuel (`limit < dist`), look for gas stations within range.

2. **Handle Edge Cases**:

   - If the gas station is too far away from the current limit or there are no more gas stations within reach, return `-1` to indicate that the destination is unreachable.

3. **Optimize Refills**:

   - Always stop at the furthest reachable gas station to minimize the number of refills. This prevents unnecessary stops at closer stations.

### Example Walkthrough:

#### Test Case 1:

- **Input:** `car_fueling(950, 400, 4, [200, 375, 550, 750])`

  - **Explanation:**

    - Start with a full tank (can travel 400 miles).

    - Reach the first station at 200 miles, but continue to the furthest station at 375 miles.

    - Refill at 375 miles and continue.

    - Reach the next station at 550 miles but continue to the furthest one at 750 miles.

    - Refill at 750 miles and then reach the destination at 950 miles.

  - **Result:** 2 refills are needed.

#### Test Case 2:

- **Input:** `car_fueling(10, 3, 4, [1, 2, 5, 9])`

  - **Explanation:**

    - Start with a full tank (can travel 3 miles).

    - The first station is at 1 mile, but you cannot reach the next station at 5 miles with the current fuel capacity.

  - **Result:** The trip is impossible, so return `-1`.

When Dynamic Programming May Not Be the Best Choice

🚫 When Dynamic Programming is NOT the Right Choice

Dynamic Programming (DP) is often seen as a “go-to” technique for optimization problems. But here’s the truth:

Using DP blindly can make your solution slower, more complex, and harder to maintain.

This guide helps you understand when NOT to use DP—and what to use instead.

---

📚 Table of Contents

• Quick Refresher
• When DP Fails
• Time & Space Trade-off (Math)
• Comparison Code
• CLI Output
• Better Alternatives
• Key Takeaways
• Related Articles

---

⚡ Quick Refresher

DP works best when two conditions are met:

• Overlapping subproblems
• Optimal substructure

Mathematically:

\[ DP(n) = \min / \max (DP(n-1), DP(n-2), ...) \]

If your problem doesn’t follow this pattern, DP may not help.

---

🚨 When Dynamic Programming is NOT Ideal

1. Simple Problems

If a problem can be solved in one pass:

\[ Time = O(n) \]

Using DP may unnecessarily increase complexity.

---

2. Small Input Size

For small \( n \), even brute force works:

\[ O(n^2) \approx O(n) \quad \text{(for small n)} \]

👉 Optimization doesn’t matter when input is tiny.

---

3. No Overlapping Subproblems

If each subproblem is unique:

\[ Total\ Work = Unique\ Subproblems \]

No repetition → No benefit from memoization.

---

4. Greedy Works Better

Some problems follow:

\[ Global\ Optimum = Local\ Optimum \]

In such cases, greedy is faster and simpler.

---

5. Memory Constraints

DP space usage:

\[ Space = O(n) \text{ or } O(n^2) \]

If memory is limited, this is a problem.

---

6. Changing Inputs

If inputs keep changing:

\[ Recompute\ Cost = O(n) \]

DP tables must be rebuilt → inefficient.

---

7. Low Recursion Depth

If recursion is shallow:

\[ Depth \approx 1 \text{ or } 2 \]

No repeated work → DP unnecessary.

---

8. Hard to Define State

If you cannot define:

\[ State = f(parameters) \]

DP becomes messy and impractical.

---

📐 Time vs Space Trade-off

DP improves time but increases space:

\[ Time_{DP} < Time_{Brute} \]

\[ Space_{DP} > Space_{Brute} \]

👉 You are trading memory for speed.

---

💻 Code Comparison

Without DP (Simple Loop)

def sum_array(arr): total = 0 for x in arr: total += x return total

With DP (Unnecessary)

memo = {} def sum_dp(i, arr): if i == len(arr): return 0 if i in memo: return memo[i] memo[i] = arr[i] + sum_dp(i+1, arr) return memo[i]

---

🖥️ CLI Output

Click to Expand

Input: [1,2,3,4]
Simple Loop Output: 10
DP Output: 10
Time: Loop faster, DP slower due to overhead

---

🛠️ Better Alternatives

• Greedy Algorithms → Fast and simple
• Divide & Conquer → Efficient recursion
• Brute Force → Fine for small inputs
• Graph Algorithms → When structure matters

---

💡 Key Takeaways

• DP is powerful—but not always necessary
• Check problem structure before using DP
• Greedy and simple methods often win
• Always balance time vs space

---

🎯 Final Thought

Smart engineers don’t just solve problems—they choose the right tool.

DP is powerful… but knowing when NOT to use it is even more powerful.