How curious that all these people put that redundant code at the end...

Confused as to why this solution is considered best practice and clever. It's a fairly simple solution that is at least O(n^2) for a problem that can be solved in O(n). The (!!) operator is also fairly inefficient.

Duplicate?
https://www.codewars.com/kata/5728a1bc0838ffea270018b2

At year 1: f0=10000
At year 2: f1=10000-10000=0

Therefore at the end of year 2 John still has a non-negative amount on money.

This would occur if n is a single digit number, because the product of all the digis in a single digit number is that number.

Instead of this:

def strip_url_params(url, strip=None):
    if not strip: strip = []

perhaps use this:

def strip_url_params(url, strip=[]):

This is a 6 kyu? You are basically given the program in the description.

check_prime function could be improved vastly, check out Blind4Basics solution for inspiration.

I would not consider this the best solution since this problem can be solved in O(1).

I agree, field.index(subfield) adds an extra loop that is not necessarily needed.

Counter(arr).items() returns a list of [(key1, value1), (key2, value2), ...] where the keys are the distinct numbers in arr and the values are the occurrences of that key in arr. You can pass this into the max function with a custom key to retrieve the max value. The key takes in a (key, value) instance and compares it against other elements by first comparing x[1] then x[0] where x[1] is the occurrence of that number and x[0] is the number as stated before. You need to apply [0] to the outputted (key,value) of the max function to obtain the distinct number.

Better safe than sorry I guess. In this kata it is not needed however.

Nice but you are counting every digit regardless if it is in digits_list. Not too important but would speed up your solution for large integers_list and small digits_list.

This didn't age well.