Car Fueling Problem by Greedy Algorithm (getting list index out of range)

Question

I have a small problem solving the Car fueling problem using the Greedy Algorithm.

Problem Introduction

You are going to travel to another city that is located 𝑑 miles away from your home city. Your car can travel at most 𝑚 miles on a full tank and you start with a full tank. Along your way, there are gas stations at distances stop1, stop2, ... , stopN from your home city. What is the minimum number of refills needed?

Input:

950
400
4
200 375 550 750

Output:

2

What I've tried as of now

def car_fueling(dist,miles,n,gas_stations):
  num_refill, curr_refill, last_refill = 0,0,0
  while curr_refill <= n:
    last_refill = curr_refill
    while (curr_refill <= n-1) & (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

What is the problem I'm facing

In the statement

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

I am getting the error IndexError: list index out of range. It is because of gas_stations[curr_refill + 1]. So when I try to separate it as a while loop and an if statement as in

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

It is entering an infinite loop.

Can you kindly point out the mistake I'm facing?

Pentalpha

I am not sure if this is intended or not, but & operator is not AND in python. If you want to use logical and use keyword "and", not &. But to me it seem like you are trying to use short circuit evaluation so right condition is not being checked if left one is already false, thus preventing index out of range error. – David

Thanks for replying, Just to clarify, if you are saying that if I use and instead of & then the condition gas_stations[curr_refill + 1] - gas_stations[last_refill] <= miles will not be checked if the condition curr_refill <= n-1 this is false, then I tried that it is not working still getting the same list out of index error. – Pentalpha

Is your first index 0 or 1 in the array? – David

The array gas_stations starts from 0. – Pentalpha

Replace while (curr_refill <= n-1) with while (curr_refill < n-1). Because when you have equality sign and curr_refill = n-1, gas_stations[curr_refill + 1] will cause an index error because gas_stations[curr_refill + 1] ---> gas_stations[n], which is beyond last index. Remove equal sign in while condition too so it becomes while curr_refill < n: . Also keep using "and" instead of &. – David

Nope, Still not working. Same error – Pentalpha

I have implemented the same code in java in the following link: #63915761 – Led

Answer 1

A few issues:

• & is not the boolean and-operator. Use and
• curr_refill + 1 can be n, and hence produce the error you got. Note that the distance after the last gas station can be determined using dist
• The value of last_refill is wrong from the start: you did not refill (yet) at station 0, so it should not be initialised as 0. Instead use another variable that represents how far you can currently drive.

Corrected code:

def car_fueling(dist,miles,n,gas_stations):
    num_refill, curr_refill, limit = 0,0,miles
    while limit < dist:  # While the destination cannot be reached with current fuel
        if curr_refill >= n or gas_stations[curr_refill] > limit:
            # Cannot reach the destination nor the next gas station
            return -1
        # Find the furthest gas station we can reach
        while curr_refill < n-1 and gas_stations[curr_refill+1] <= limit:
            curr_refill += 1
        num_refill += 1  # Stop to tank
        limit = gas_stations[curr_refill] + miles  # Fill up the tank 
        curr_refill += 1
    return num_refill

# Test cases
print(car_fueling(950, 400, 4, [200, 375, 550, 750]))  # 2
print(car_fueling(10, 3, 4, [1, 2, 5, 9]))  # -1

Kowtow

Thanks for replying. The code is working perfectly for cases that have a solution but for cases where the output is to be -1, it is entering an infinite loop. Eg: Input car_fueling(10, 3, 4, [1, 2, 5, 9]) Output: -1 – Pentalpha

Added missing increment of curr_refill and added exit condition. – Kowtow

who can explain why @Kowtow used current_refill < n-1 instead of just current_refill < n – Negotiable

@Andrew, short answer: if we have current_refill < n , then the next condition, gas_stations[curr_refill+1], could lead to an out of range index, as that index could evaluate to n). Longer answer: We can stop looping when current_refill == n-1 because that means we are at the last fuel station. We already know that we don't have enough fuel to reach the destination without refuelling (outer loop condition), so we must consider refuelling here: it is our only remaining chance. – Kowtow

Answer 2

probably this will fix the index error. The logic of your code is correct and that is the same as what I did in the code in the block of else statement. Add the start point and end point(total distance) can avoid index error. First check the total distance to see if it can be arrived in a full tank. If not do the else statement.

def compute_min_refills(distance, tank, stops):

    numrefill, currentrefill= 0,0
    stops = [0] + stops + [distance] #include the start and end points in the stops list   
    if distance <= tank:
        return 0
    else:
        while currentrefill < len(stops)-1:
            lastrefill = currentrefill
            #print(currentrefill, lastrefill, len(stops))
            while currentrefill < len(stops)-1 and stops[currentrefill+1] - stops[lastrefill]<=tank:
           
                currentrefill += 1

            if currentrefill == lastrefill:
                return -1
            if currentrefill < len(stops)-1:
                numrefill +=1

        #print(numrefill)

        return numrefill

if __name__ == '__main__':
    
    #print(compute_min_refills(10, 3, [1,2,5,9]))

Cheka

Code dumps without any explanation are rarely helpful. Stack Overflow is about learning, not providing snippets to blindly copy and paste. Please edit your question and explain how it works better than what the OP provided. See How to Answer. – Jimmie

I thought the author who asked this question actually know the logic of the code and the only problem he/she had is the index. So I added the start point as 0 and also include the end point which will make this problem easy to understand. Besides, the names of the variables that I created are quite obvious. I thought that would be help for this question. – Cheka

Answer 3

I used this code and got the correct answer from a course in Coursera.

# python3
import sys

def compute_min_refills(distance, tank, stops):
    capacity_tank = tank
    refill = 0

    if capacity_tank >= distance:
        return 0
    if capacity_tank < stops[0] or (distance-stops[-1]) > capacity_tank:
        return -1

    for i in range(1, len(stops)):
        if (stops[i]-stops[i-1]) > capacity_tank:
            return -1
        if stops[i] > tank:
            tank = (stops[i-1] + capacity_tank)
            refill += 1
    if distance > tank:
        refill += 1

    return refill


    if __name__ == '__main__':
        d, m, _, *stops = map(int, sys.stdin.read().split())
        print(compute_min_refills(d, m, stops))

Minor

Answer 4

def car_refill(dist,cap,n,stops):
    stops.insert(0,0)
    stops.append(dist)
    num_refill,curr_refill = 0,0
    while curr_refill <= n:
        last_refill = curr_refill
        while (curr_refill <= n and stops[curr_refill + 1] - stops[last_refill] <= cap):
            curr_refill += 1
        if curr_refill == num_refill :
            return -1
        if curr_refill <= n:
            num_refill +=1
    return num_refill

try this ....

Criterion

Thx for your answer and welcome to stackoverflow! =) we're here to learn something, so comment your contributions an explain the reader what and why you've coded something – Deodar

sorry i am new so not really familiar with the norms, this works the same way as above code but it also considers that x is a 1-d matrix starting from zero to the destination and then we use the inner while loop to get to the farthest point and if there is no point the num_refill and curr_refill will remain same so -1 else we can go and increase the num_refill meaning we fill the tank – Criterion

Answer 5

I dont know why but answers seems overly complicated i just imagined myself driving and cameup with this simple solution

function minfill(distance, miles, n, stations) {

    //added final distance to last station for simplicity can simply push to array. 
    stations = [...stations, distance]

    let refill = 0,
        limit = miles,
        dt = 0, //distance travelled
        current = 0; //current station

    while (current <= n) {

        //check if next  or first station is unreachable
        if ((Math.abs(stations[current] - stations[current + 1]) > limit) || stations[0] > limit) return -1

        //check if we need to refuel or pass
        if (Math.abs(dt - stations[current]) <= limit) { 
            current++
        }

        //if next distance was over limit we set distance tavelled to previous station ,current station was already pointed to next in above block
        else {
            dt = stations[current - 1]

            refill++
        }
    }
    return refill
}

p.s-this code is written in node/javascript, though i have passed all test for this question but i know here there are smarter people who will help to improve/correct this code or provide some pointers.

Jourdan

Answer 6

import java.util.; import java.io.;

public class CarFueling {

static int compute_refills(int dist, int tank, int stops[], int n) {
    int current_refills=0;
    int num_refills=0;
    int last_refill=0;
    while(current_refills<=n) {
        last_refill = current_refills;
        while ((current_refills !=stops.length-1) && (stops[current_refills + 1] - stops[last_refill]) <= tank) {
            current_refills +=1 ;

        }

        if (current_refills == last_refill)
            return -1;
        if (current_refills <= n)
            num_refills +=1;


    }
    return num_refills;
}



    public static void main (String[]args){
        Scanner scanner = new Scanner(System.in);
        int dist = scanner.nextInt();
        int tank = scanner.nextInt();
        int n = scanner.nextInt();
        int stops[] = new int[n * n * n];// to solve array index out of bound exception increase the size of the array
        for (int i = 0; i < n; i++) {
            stops[i] = scanner.nextInt();
        }

        System.out.println(compute_refills(dist, tank, stops, n));

    }
}

Led

I have implemented in java . But, I think there is some issue with the iteration of while loop. It's not working well – Led

Answer 7

The answer that add [0] and [d] to stops should work, but current_refill comparison should be current_refill < len(stops)- 2 everywhere, because two stops are added. There's another way to solve this problem.

def compute_min_number_of_refills(d, m, stops):
    if d <= m:
        return 0
    total_refill = 0
    last_refill = -1
    limit = m 
    stops.append(d)
    i = 0
    while i < len(stops):
        if stops[i] >= limit: 
            current_refill = i - 1 if stops[i] > limit else i
            if current_refill == last_refill:
                return -1 
            last_refill = current_refill
            total_refill += 1
            limit = m + stops[current_refill]
            i = current_refill + 1
        else:
            i += 1
    return total_refill

Tooling

Answer 8

Logic

The idea is to loop through stops and check if each stop is less than the distance the car can go in one full tank. If yes then continue otherwise refuel at the previous stop and repeat the process with that stop as the new starting point.

Code

def min_refills(distance, tank, stops):
    stop_list = []
    stops.append(distance) # append the destination to the stop list
    # write your code here
    if distance <= tank: # if the travel distance <= distance travelled in one full tank
        return 0
    else:
        start = 0
        prev = 0
        for stop in stops:
            
            if stop - start < tank:     # if refueling stop is closer to the starting point than the car can travel in one full tank
                prev = stop     # move prev pointer to the refueling stop
            elif (stop - start == tank) and (stop != distance):     # don't consider destination as a refueling stop
                start = stop    # move the starting point to the current gas stop
                stop_list.append(stop)      # add the refueling stop to the stop list
                prev = stop     # move the prev pointer to the stop
            elif stop - start > tank:
                start = prev    # move the starting point to the prev gas stop
                if stop - start > tank:     # if next refuleing stop is farther than the dist. car can go in one full tank
                    return -1
                stop_list.append(prev)      # add the refueling stop to the list
                prev = stop     # move the prev pointer the stop

    return len(stop_list)

Redraft

Answer 9

It's in an infinite loop because n is not being incremented.

Increment n where it makes the most sense (for example, at the end of your while statement).

def car_fueling(dist,miles,n,gas_stations):
    num_refill, curr_refill, last_refill = 0,0,0

    while curr_refill <= n:
        last_refill = curr_refill

        while (curr_refill <= n-1) & (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

        n+=1 # Increment

  return num_refill

Duester

Thank you for the answer but this solution is providing a list index out of range error. – Pentalpha

My apologies. I thought your question was focused on solving the infinite loop. I now see that it's a question about implementing the greedy algo. – Duester