Resolved: how can I square the numbers between 0 and any input with extremely large numbers in python

Question:

I am in codewars and the challenge I have has one part I can’t get my head around this is what I am speaking of
You are given a number “n” (n >= 0) and a digit “d” (0 <= d <= 9).
Write a function nbDig(n, d) that finds the square of each integer from 0 to n, and returns the number of times that the digit d appears across all the squares. Note that d might appear multiple times in a single square.
Example:
n=12, d=1 Squares from 0 to n=12: 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144 The function returns 7 because the digit d=1 appears 7 times: in 1, 16, 81, 100, 121 (note: 1 appears twice in 121), and 144.
the part the I have trouble with is how do I get my function to get the square roots of everything between 0 and n and when the numbers I have to solve are large numbers like (5750)(195856) or (74747)

Answer:

If I am interpreting your question correctly, you simply need to devise a very simple list comprehension program.
Getting the square root of everything between zero and n is not necessary. Did you mean square?
Here’s my solution:

def nbDig(n,d):
    d = str(d)
    int_list = str([str(x*x) for x in range(0,n+1)])
    return_value_count = ((str(int_list.replace(",","")).replace("[","")).replace("]","")).count(d)
    print(return_value_count)
    return return_value_count

I can test it like so:

nbDig(12,1)

This returns 7.

If you have better answer, please add a comment about this, thank you!