Numbers At Most N Given Digit Set

class Solution:
    def atMostNGivenDigitSet(self, digits: List[str], n: int) -> int:
        nums = list(map(int, str(n)))
        d = len(nums)
        sz = len(digits)

        #ps[i-1] = no.of digits in set less than digit i ( 1 - 9)
        #Thus, ps[i] = no. of digits in digits set less than or equal to digit i
        #Also, ps[i] - ps[i-1] indicates presence of digit i in given set
        #For digit 0 i-1 will be -1, thus for this adjustment ps list is made of 11 size with last element (i.e ps[-1]) = 0

        #prefix-sum like logic using two pointers
        ps = [0]*11
        digits = list(map(int, digits))
        i, j = 1, 0
        for i in range(10):
            if j < sz and i == digits[j]:
                ps[i] = ps[i-1] + 1
                j += 1
            else:
                ps[i] = ps[i-1]
            i += 1

        #count of 1-digit, 2-digit, ... , d-1 digit nos
        if sz > 1:
            res = sz * (sz ** (d - 1) - 1)//(sz - 1)
        else:
            res = (d - 1) * sz

        for idx, ele in enumerate(nums):
            res += ps[ele-1] * (sz ** (d-1-idx)) #with msd less than ele
            if ps[ele] - ps[ele-1] == 0: #if ele not present in set
                break
        res += ps[ele] - ps[ele-1] #can n be formed
        return res