human
I found an interesting problem on https://leetcode.com/problems/numbers-at-most-n-given-digit-set/:
Given an array of digits, you can write numbers using each digits[i] as many times as we want.  For example, if digits = ['1','3','5'], we may write numbers such as '13', '551', and '1351315'.
Return the number of positive integers that can be generated that are less than or equal to a given integer n.
 
Example 1:
Input: digits = ["1","3","5","7"], n = 100
Output: 20
Explanation: 
The 20 numbers that can be written are:
1, 3, 5, 7, 11, 13, 15, 17, 31, 33, 35, 37, 51, 53, 55, 57, 71, 73, 75, 77.

Example 2:
Input: digits = ["1","4","9"], n = 1000000000
Output: 29523
Explanation: 
We can write 3 one digit numbers, 9 two digit numbers, 27 three digit numbers,
81 four digit numbers, 243 five digit numbers, 729 six digit numbers,
2187 seven digit numbers, 6561 eight digit numbers, and 19683 nine digit numbers.
In total, this is 29523 integers that can be written using the digits array.

Example 3:
Input: digits = ["7"], n = 8
Output: 1

 
Constraints:

1 <= digits.length <= 9
digits[i].length == 1
digits[i] is a digit from '1' to '9'.
All the values in digits are unique.
1 <= n <= 109

I tried it in Python, but could not do it. Can you solve it?

 What is a the Python code you requested?
assistant
#Here is the Python code you requested.

class Solution:
    def atMostNGivenDigitSet(self, digits: List[str], n: int) -> int:
        
        count = 0
        length = 1
        n_str = str(n)
        while length < len(n_str):
            count+= len(digits)**length
            length+=1

        digits_sorted = sorted(digits)

        ## now length should equal to len(n), we compare the number with same length
        current_digit = 0
        while current_digit < length:
            for digit in digits_sorted:
                next_round = False
                if digit < n_str[current_digit]:
                    count+=len(digits)**(length-current_digit-1)
                elif digit > n_str[current_digit]:
                    return count
                else:
                    if current_digit == length-1:
                        return count+1
                    else: 
                        current_digit+=1
                        next_round = True
                        break
            if not next_round:
                return count

        return count