You are given 2 numbers is `n` and `k`. You need to find the number of integers between 1 and n (inclusive) that contains exactly `k` non-zero digit.
Example1 
`
almost_everywhere_zero(100, 1) return 19`
by following condition we have 19 numbers that have k = 1 digits( not count zero ) 
` [1,2,3,4,5,6,7,8,9,10,20,30,40,50,60,70,80,90,100]`
Example2  
`
almost_everywhere_zero(11, 2) return 1`
we have only `11` that has 2 digits(ten not count because zero is not count) 
` 11`
constrains
`1≤n<pow(10,100)`
`1≤k≤100`
from scipy.special import comb

def almost_everywhere_zero(n, k):
    if k == 0: return 1
    first, *rest = str(n)
    l = len(rest)
    return 9**k*comb(l, k, exact=True) +\
           (int(first)-1)*9**(k-1)*comb(l, k-1, exact=True) +\
           almost_everywhere_zero(int("".join(rest) or 0), k-1)
from scipy.special import comb

def almost_everywhere_zero(n, k):
    if k == 0: return 1
    first, *rest = str(n)
    l = len(rest)
    return 9**k*comb(l, k, exact=True) +\
           (int(first)-1)*9**(k-1)*comb(l, k-1, exact=True) +\
           almost_everywhere_zero(int("".join(rest) or 0), k-1)