or perhaps even:

for item in strs:
print(item.split(' '))

['1', '2']

['1', '', '2']

['1', '', '', '2']

['1', '', '', '', '2']

a zero length string, each time the dlimiter (a space) 'splits' other delimiters (other sapces)

Thank you!

Won't this unnecessarily perform calculations which could be bypassed by testing for 1 pillar?

I apologize in advance if I'm wrong, but as clever as this is I think it fails to fill the minimum quota of 10 by filling the deficit of 'Unpopular' with 'Popular'.
Am I missing something??
"If there are less than 7 popular projects, or less than 3 unpopular projects, fill the gaps from the other group of projects if possible."

Also:
"If there are "less than" 10 projects in total, just return all projects." You've got <= 10.

Can someone please tell me what I'm supposed to do with my solution?
Thanks