Regex: Remove all match plus one char before all

[buran]

so actually the requested result is "For each "|BS| group remove as many chars before that group as many times |BS| is present in that group"

#!/usr/bin/python3
import re

strings = ['it |BS||BS||BS|this is one|BS||BS||BS|an example',
           'it |BS|this is an example',
           'it |BS||BS|this is an example',
           'it |BS||BS||BS|this is an example',
           'it |BS||BS||BS||BS|this is an example']
ptrn = re.compile(r'[\w ]?\|BS\|')
for string in strings:
   print(string)
   while True:
       after_sub = ptrn.sub('', string, count=1)
       if string == after_sub:
           break
       else:
           string = after_sub
   print(string)
   print('\n')

Output:
it |BS||BS||BS|this is one|BS||BS||BS|an example
this is an example


it |BS|this is an example
itthis is an example


it |BS||BS|this is an example
ithis is an example


it |BS||BS||BS|this is an example
this is an example


it |BS||BS||BS||BS|this is an example
this is an example