# Task
 Two integer numbers are added using the column addition method. When using this method, some additions of digits produce non-zero carries to the next positions. Your task is to calculate the number of non-zero carries that will occur while adding the given numbers.

 The numbers are added in base 10.

# Example

 For `a = 543 and b = 3456,` the output should be `0`
 
 For `a = 1927 and b = 6426`, the output should be `2`
 
 For `a = 9999 and b = 1`, the output should be `4`
 
 
# Input/Output


 - `[input]` integer `a`

  A positive integer.

  Constraints: `1 ≤ a ≤ 10^7`


 - `[input]` integer `b`

  A positive integer.

  Constraints: `1 ≤ b ≤ 10^7`

 
 - `[output]` an integer

Difficulty: EASY
Tags: Ad-hoc

Starter code:
def number_of_carries(a, b):
	

Solutions:
def number_of_carries(a, b):
	(ans, carrie) = (0, 0)
	while a > 0 or b > 0:
		carrie = (a % 10 + b % 10 + carrie) // 10
		ans += [0, 1][carrie > 0]
		a //= 10
		b //= 10
	return ans


def number_of_carries(a, b):
	i = 1
	carry_over = 0
	counter = 0
	length1 = len(str(a))
	length2 = len(str(b))
	if length1 > length2:
		while length2 - i >= 0:
			if int(str(a)[length1 - i]) + int(str(b)[length2 - i]) + carry_over >= 10:
				counter += 1
				carry_over = 1
				i += 1
			else:
				carry_over = 0
				i += 1
		while length1 - i >= 0:
			if int(str(a)[length1 - i]) + carry_over >= 10:
				counter += 1
				carry_over = 1
				i += 1
			else:
				carry_over = 0
				i += 1
	elif length2 > length1:
		while length1 - i >= 0:
			if int(str(a)[length1 - i]) + int(str(b)[length2 - i]) + carry_over >= 10:
				counter += 1
				carry_over = 1
				i += 1
			else:
				carry_over = 0
				i += 1
		while length2 - i >= 0:
			if int(str(b)[length2 - i]) + carry_over >= 10:
				counter += 1
				carry_over = 1
				i += 1
			else:
				carry_over = 0
				i += 1
	elif length2 == length1:
		while length1 - i >= 0:
			if int(str(a)[length1 - i]) + int(str(b)[length2 - i]) + carry_over >= 10:
				counter += 1
				carry_over = 1
				i += 1
			else:
				carry_over = 0
				i += 1
	return counter


def number_of_carries(a, b):
	(a, b, c, r) = ([int(x) for x in f'000000{a}'], [int(x) for x in f'000000{b}'], 0, 0)
	for (x, y) in zip(reversed(a), reversed(b)):
		c = x + y + c > 9
		r += c
	return r


from itertools import zip_longest
help = lambda n: map(int, reversed(str(n)))

def number_of_carries(a, b):
	carry = res = 0
	for (x, y) in zip_longest(help(a), help(b), fillvalue=0):
		tmp = x + y + carry
		res += tmp > 9
		carry = tmp // 10
	return res


from itertools import zip_longest

def number_of_carries(a, b):
	(carry, li) = ([0], [])
	for (i, j) in zip_longest(str(a)[::-1], str(b)[::-1], fillvalue='0'):
		s = int(i) + int(j) + carry[-1]
		li.append(s % 10)
		if s > 9:
			carry.append(int(str(s)[0]))
			continue
		carry.append(0)
	return carry.count(1)


from itertools import zip_longest

def number_of_carries(a, b):
	(x, y) = (str(a)[::-1], str(b)[::-1])
	result = c = 0
	for (a, b) in zip_longest(map(int, x), map(int, y), fillvalue=0):
		c = a + b + c > 9
		result += c
	return result


def number_of_carries(a, b):
	(a, b) = (str(a)[::-1], str(b)[::-1])
	l = max(len(a), len(b))
	(carry, carries) = (0, 0)
	for col in [int(m) + int(n) for (m, n) in zip(f'{a:<0{l}s}', f'{b:<0{l}s}')]:
		carry = col + carry > 9
		carries += carry
	return carries


def number_of_carries(a, b):
	(a, b) = (str(a)[::-1], str(b)[::-1])
	while len(a) < len(b):
		a += '0'
	while len(a) > len(b):
		b += '0'
	(carry, carries) = (0, 0)
	for (ad, bd) in zip(a, b):
		(ad, bd) = (int(ad), int(bd))
		nd = ad + bd + carry
		if nd > 9:
			carry = (nd - nd % 10) // 10
			carries += 1
		else:
			carry = 0
	return carries


def number_of_carries(a, b):
	c = [0]
	while a or b:
		(c, a, b) = (c + [(c[-1] + a % 10 + b % 10) // 10], a // 10, b // 10)
	return sum(c)


number_of_carries = r = lambda a, b, c=0: a + b + c and c + r(a // 10, b // 10, a % 10 + b % 10 + c > 9)