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  1. from itertools import*
  2. a='()[]{}<>'
  3. all(any(c[-1]!=d[0]for c,d in zip(q,q[1:]))or print(''.join(q))for p in permutations(open(0))for q in product(*[(c[:-1],a[a.find(c[-2])^1]+c[-3:0:-1]+a[a.find(c[0])^1])for c in p]))
Success #stdin #stdout 0.02s 9984KB
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[----->
)------------[
{--<
}---)
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<-----]]------------((---{{--<
https://ideone.com/d8s8Uy
http://c...content-available-to-author-only...e.com/questions/65892/can-i-string-all-my-cords-and-adapters-together
language:
Python 3 (python  3.9.5)
created:
8 years ago
visibility:
secret

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