Gigi is a clever monkey, living in the zoo, his teacher (animal keeper) recently taught him some knowledge of "0".

In Gigi's eyes, "0" is a character contains some circle(maybe one, maybe two).

So, a is a "0",b is a "0",6 is also a "0"，and 8 have two "0" ,etc...

Now, write some code to count how many "0"s in the text.

Let us see who is smarter? You ? or monkey?

Input always be a string(including words numbers and symbols)，You don't need to verify it, but pay attention to the difference between uppercase and lowercase letters.

Here is a table of characters：

one zeroabdegopq069DOPQR         () <-- A pair of braces as a zerotwo zero%&B8

Output will be a number of "0".

Source: codewars | Difficulty: EASY
Tags: Ad-hoc

Starter code: def countzero(string):
	

Solutions:
def countzero(s):
	return sum((1 if c in 'abdegopq069DOPQR' else 2 if c in '%&B8' else 0 for c in s.replace('()', '0')))


def countzero(string):
	return sum((1 if c in 'abdegopq069DOPQR' else 2 if c in '%&B8' else 0 for c in string)) + string.count('()')


def countzero(string):
	return sum(map('abdegopq069DOPQR%&B8%&B8'.count, string.replace('()', '0')))


D = {**{c: 1 for c in 'abdegopq069DOPQR'}, **{c: 2 for c in '%&B8'}}

def countzero(string):
	return sum((D.get(c, string[i:i + 2] == '()') for (i, c) in enumerate(string)))


two = '%&B8'
some = f'abdegopq069DOPQR{two}'

def countzero(string):
	return sum((2 if c in two else 1 for c in string.replace('()', '0') if c in some))


ONE_ZERO = 'abdegopq069DOPQR'
TWO_ZERO = '%&B8'

def countzero(string):
	string = string.replace('()', '0')
	return sum((1 if c in ONE_ZERO else 2 if c in TWO_ZERO else 0 for c in string))


zeroes = {}
for c in 'abdegopq069DOPQR':
	zeroes[c] = 1
for c in '%&B8':
	zeroes[c] = 2

def countzero(string):
	string = string.replace('()', '0')
	return sum(map(lambda c: zeroes.get(c, 0), string))


def countzero(string):
	n = 0
	for i in string:
		if i in '069abdeopqDOPQRg(':
			n += 1
		elif i in '8%&B':
			n += 2
	return n


countzero = lambda s: sum(((c in '%&B8') * 2 + (c in 'abdegopq069DOPQR') * 1 for c in s)) + s.count('()')


def countzero(string):
	counter = 0
	for x in string:
		if x in 'abdegopq069DOPQR':
			counter += 1
		elif x in '%&B8':
			counter += 2
		else:
			pass
	return counter + string.count('()')