Remove empty line before a pattern using sed - sed

Question

Context
E.g. I have this test file foo.py:
#!/usr/bin/env python3
'''foo'''
# comment
import ...
# [END import]
import ...
# [END import]
import ...
# [END import]
# [END import]
import even...
# [END import]
# [END import]
import odd...
# [END import]
# [END import]
Expected
I would like to remove blank line before # [END import
#!/usr/bin/env python3
'''foo'''
# comment
import ...
# [END import]
import ...
# [END import]
import ...
# [END import]
# [END import]
import even...
# [END import]
# [END import]
import odd...
# [END import]
# [END import]
Can someone give me a working version using sed and/or explain why the following didn't work
Test 0
sed '$!N;s/^\n\(# \[END\)/\1/g' foo.py
Observed
#!/usr/bin/env python3
'''foo'''
# comment
import ...
# [END import]
import ...
# [END import]
import ...
# [END import]
# [END import]
import even...
# [END import]
# [END import]
import odd...
# [END import]
# [END import]
Here only "even" line changed since here we "consume" two line at a time
using N; without coming back...
Test 1
sed ':r;$!{N;br};s/^\n\(# \[END\)/\1/g' foo.py
Observed
nothing change, here I don't understand why it is not working (i.e. why the pattern is not matched)...
Test 2
without the ^ anchor.
sed ':r;$!{N;br};s/\n\(# \[END\)/\1/g' foo.py
Observed
#!/usr/bin/env python3
'''foo'''
# comment
import ...
# [END import]
import ...
# [END import]
import ...# [END import]# [END import]
import even...
# [END import]
# [END import]
import odd...
# [END import]
# [END import]
notice the double #[END on the same line as expected by the command but not the result expected.
Test 3
sed ':r;$!{N;br};s/\n\(\n# \[END\)/\1/g' foo.py
Observed
WORKING as expected, BUT I can't figure out why it can match \n\n i.e. two consecutive returns

Answer 1

You'll need to add m flag for Test 1, so that ^ and $ anchors will match every line's start and end locations, otherwise they'll match start/end of entire string. This assumes m flag is supported by your implementation, like GNU sed does.
sed ':r;$!{N;br};s/^\n\(# \[END\)/\1/mg'
Test 3 works because there's a newline just before the empty line as part of that previous line. The below example might help you better visualize it:
$ printf 'a\nb\nc\n'
a
b
c
$ printf 'a\nb\n\nc\n'
a
b
c
With perl:
perl -0777 -pe 's/\n\K\n(?=# \[END)//g'
-0777 will slurp the entire input as single string
\n\K\n(?=# \[END) will match a newline provided there's a newline character before and # \[END after that newline
Another alternative with GNU sed, doesn't need to read whole file in one go.
sed '/^$/{N; s/\n\(# \[END\)/\1/; P; D}'
/^$/ will match an empty line
N add next line to pattern space
s/\n\(# \[END\)/\1/ remove the newline if required regexp matches
P and D are crucial here, so I'll quote the manual:
P Print out the portion of the pattern space up to the first newline.
D If pattern space contains no newline, start a normal new cycle as if the d command was issued. Otherwise, delete text in the pattern space up to the first newline, and restart cycle with the resultant pattern space, without reading a new line of input.

Answer 2

This might work for you (GNU sed):
sed 'N;/^\n# \[END import\]/!P;D' file
Open a window of 2 lines throughout the file and if the first line is empty and the second line is # [END import] do not print the first line.
N.B. The idiom N;...;P;D prints all lines in a file but allows the programmer to reason about 2 lines at a time.