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`

Difficulty: MEDIUM_HARD
Tags: Mathematics

Starter code:
def almost_everywhere_zero(n, k):
	

Solutions:
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)


def almost_everywhere_zero(S, k):
	S = [int(c) for c in str(S)]
	(D, n) = ({}, len(S))

	def F(i, k, L):
		if i == n:
			return k == 0
		if k == 0:
			return F(i + 1, k, L)
		if (i, k, L) in D:
			return D[i, k, L]
		if i == 0 or L:
			D[i, k, L] = F(i + 1, k, S[i] == 0) + (S[i] - 1) * F(i + 1, k - 1, False) + F(i + 1, k - 1, S[i] != 0)
		else:
			D[i, k, L] = F(i + 1, k, False) + 9 * F(i + 1, k - 1, False)
		return D[i, k, L]
	return F(0, k, 0)


def fact(n):
	if n == 1 or n == 0:
		return 1
	else:
		return n * fact(n - 1)

def C(n, k):
	if n < k:
		return 0
	return fact(n) // (fact(k) * fact(n - k))

def almost_everywhere_zero(n, k):
	if n == 0 and k > 0:
		return 0
	if k == 0:
		return 1
	ntoString = str(n)
	digits = len(ntoString)
	first_digit = int(ntoString[0])
	tight = 0 if len(ntoString) < 2 else int(ntoString[1:])
	return (first_digit - 1) * 9 ** (k - 1) * C(digits - 1, k - 1) + almost_everywhere_zero(tight, k - 1) + 9 ** k * C(digits - 1, k)


R = lambda a, b: range(a, b)

def almost_everywhere_zero(S, k):
	S = [int(c) for c in str(S)]
	(D, n) = ({}, len(S))

	def F(i, k, L):
		if i == n:
			return k == 0
		if k == 0:
			return F(i + 1, k, L)
		if (i, k, L) in D:
			return D[i, k, L]
		if i == 0 or L:
			D[i, k, L] = sum((F(i + 1, k - (j > 0), j == S[i]) for j in R(0, S[i] + 1)))
		else:
			D[i, k, L] = sum((F(i + 1, k - (j > 0), 0) for j in R(0, 10)))
		return D[i, k, L]
	return F(0, k, 0)


from functools import lru_cache

@lru_cache(maxsize=None)
def calc(d, k, is_first_digit=True):
	if d < 0 or k < 0 or d < k:
		return 0
	elif d == k:
		return 9 ** k
	return calc(d - 1, k - 1, False) * 9 + calc(d - 1, k, False) * (not is_first_digit)

def almost_everywhere_zero(n, k):
	if n < 10:
		return {0: 1, 1: n}.get(k, 0)
	ans = sum((calc(d, k) for d in range(k, len(str(n)))))
	for d in range(1, int(str(n)[0])):
		ans += calc(len(str(n)) - 1, k - 1, False)
	return ans + almost_everywhere_zero(int(str(n)[1:]), k - 1)


import functools
import sys

@functools.lru_cache(None)
def almost_everywhere_zero(n, k):
	n = str(n)
	if len(n) == 0 or k < 0:
		return k == 0
	d = int(n[0])
	return sum((almost_everywhere_zero(n[1:] if i == d else '9' * (len(n) - 1), k - 1 if i > 0 else k) for i in range(d + 1)))
sys.setrecursionlimit(100000)


try:
	from math import comb
except ImportError:
	from math import factorial

	def comb(n, k):
		if k < 0 or k > n:
			return 0
		return factorial(n) // (factorial(k) * factorial(n - k))

def almost_everywhere_zero(n, k):
	digits = list(reversed([int(c) for c in str(n + 1)]))
	aez = 0
	for i in range(len(digits) - 1, -1, -1):
		d = digits[i]
		if d == 0:
			continue
		aez += comb(i, k) * 9 ** k
		k -= 1
		if k < 0:
			break
		aez += (d - 1) * comb(i, k) * 9 ** k
	return aez