I'm trying to write a Python program to find the number of notes in an integer, but my code isn't working

Question

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This is what the python program should do. The notes to use are [500, 200, 100, 50, 20, 10]

Can someone explain what is wrong with my code? I've been sitting here and scratching my head for the longest time.

def note_calculator(num):
    notes = [500, 200, 100, 50, 20, 10]
    counter = 0
    mult = 0
    for x in range(6):
        enote = notes[x]
        if num <= enote:
            if num % enote == 0:
                mult = num / enote
                num = num - (mult * enote)
                counter += mult
                if num == 0:
                    break
            else:
                num = num - enote
                counter += 1
                if num == 0:
                    break
        else:
            continue
    return counter

Basically, my method is to put sample notes in a list and iterate through all of them. If the number you input is greater or equal to the first iteration (500) it will then check the divisibility, if there are no remainders, I calculate the multiple. Then I subtract the multiple * the sample note iteration (500) because the program should run until the input (num) is equal to zero. Then I add to the counter, which will tell me the number of notes. If it is not divisible with remainders I simply subtract the first iteration (500) and add it to the counter. This repeats until the last iteration (10)

I know there is an easier and shorter way to do this, but whatever.

Answer 1

Your algorithm works fine, but I noticed two bugs, first one:

if num <= enote: should be if num >= enote: You want to get enote when it's not bigger than rest amount, not otherwise.

Second, notice that when if num % enote == 0: you correctly take proper amount of enotes, but in else clause you take only one. You should take as much as you can, so do something similar as in if :

counter += num // enote
num = num % enote

And one small thing, your function now returns float (as / operator do so). You can use //, look:

print(type(10/2)) # <class 'float'>
print(type(10//2)) # <class 'int'>

Answer 2

It appears you are looking for the builtin divmod .

notes = [500, 200, 100, 50, 20, 10]
x = 880
s = 0
for n in notes: 
    q,r = divmod(x,n)
    x   = r
    s  += q
print(s)
    

Answer 3

Here is how I would implement it (Python 3):

def note_calculator(num):
    counter = 0
    for note in [500, 200, 100, 50, 20, 10]:
        counter += num // note
        num %= note
    return counter

In Python 2, you can just replace the // with a / .

Answer 4

You can simplify it like this...

from time import time

start = time()

def note_calculator(num):
    notes, count= [500, 200, 100, 50, 20, 10], 0

    for note in notes:
        if note <= num:
            multiplier = num // note
            num -= note*multiplier
            count += multiplier
    return count

print(note_calculator(880))
print(note_calculator(1000))

end = time()

print(f"Time Taken : {end-start}")

Output:-

6
2
Time Taken : 3.695487976074219e-05