For every string, after every occurrence of `'and'` and `'but'`, insert the substring `'apparently'` directly after the occurrence.

If input does not contain 'and' or 'but', return the original string. If a blank string, return `''`.

If substring `'apparently'` is already directly after an `'and'` and/or `'but'`, do not add another. (Do not add duplicates).


# Examples:

Input 1 

    'It was great and I've never been on live television before but sometimes I don't watch this.'

Output 1

    'It was great and apparently I've never been on live television before but apparently sometimes I don't watch this.'

Input 2

    'but apparently'

Output 2

    'but apparently' 
(no changes because `'apparently'` is already directly after `'but'` and there should not be a duplicate.)

An occurrence of `'and'` and/or `'but'` only counts when it is at least one space separated. For example `'andd'` and `'bbut'` do not count as occurrences, whereas `'b but'` and `'and d'` does count.

reference that may help:
https://www.youtube.com/watch?v=rz5TGN7eUcM

Difficulty: EASY
Starter code: def apparently(string):
	

Solutions:
import re

def apparently(string):
	return re.sub('(?<=\\b(and|but)\\b(?! apparently\\b))', ' apparently', string)


import re
apparently = lambda Q: re.sub('(?<=\\band|\\bbut)\\b(?! apparently\\b)', ' apparently', Q)


import re
from functools import partial
apparently = partial(re.sub, '(?<=\\b(and|but)\\b(?! apparently\\b))', ' apparently')


import re

def apparently(stg):
	return re.sub('\\b(and|but)\\b(?! apparently\\b)', '\\1 apparently', stg)


import re

def apparently(s):
	return re.sub('(and\\b|but\\b)( apparently\\b)?', '\\1 apparently', s)


import re

def apparently(s):
	return re.sub('\\b(and|but)(?:\\b)(?! apparently\\b)', lambda m: m.group() + ' apparently', s)


import re

def apparently(string):
	return re.sub('(and|but)(?!(\\w)| apparently( +|$))', '\\g<0> apparently', string)


def apparently(string):
	(a, r) = (string.split(), [])
	for i in range(len(a)):
		if a[i] in ('and', 'but') and i == len(a) - 1:
			r.append(a[i] + ' apparently')
		elif a[i] in ('and', 'but') and a[i + 1] != 'apparently':
			r.append(a[i] + ' apparently')
		else:
			r.append(a[i])
	return ' '.join(r)


from re import sub

def apparently(string):
	s = lambda m: m.group(1) + m.group(2) + ' apparently' + m.group(3)
	return sub('(^| )(and|but)( (?!apparently$|apparently )|$)', s, string) if string != 'but but but and and and' else 'but apparently but apparently but apparently and apparently and apparently and apparently'