Second creative solution for How to Find Friends by coells - python coding challenges

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Second solution in Creative category for How to Find Friends by coells

import re

def check_connection(net, drone1, drone2):
  t = {drone1}
  for _ in net: t |= {re.sub(r'(\A%s-)|(-%s\Z)'%(x,x), '', e) for x in t for e in net}
  return drone2 in t


if __name__ == '__main__':
    #These "asserts" using only for self-checking and not necessary for auto-testing
    assert check_connection(
        ("dr101-mr99", "mr99-out00", "dr101-out00", "scout1-scout2",
         "scout3-scout1", "scout1-scout4", "scout4-sscout", "sscout-super"),
        "scout2", "scout3") == True, "Scout Brotherhood"
    assert check_connection(
        ("dr101-mr99", "mr99-out00", "dr101-out00", "scout1-scout2",
         "scout3-scout1", "scout1-scout4", "scout4-sscout", "sscout-super"),
        "super", "scout2") == True, "Super Scout"
    assert check_connection(
        ("dr101-mr99", "mr99-out00", "dr101-out00", "scout1-scout2",
         "scout3-scout1", "scout1-scout4", "scout4-sscout", "sscout-super"),
        "dr101", "sscout") == False, "I don't know any scouts."

June 28, 2014

Comments:

gyahun_dash on July 7, 2014, 3:43 p.m.
Amazing! BTW, you can replace the expression in line 5 as follows (but not readable so much): <pre class='brush: python'>r'(-?%s-?)'%(x) </pre>

coells on July 7, 2014, 4:15 p.m.
Unfortunately, that wouldn't work for substrings, e.g. scout vs scout1

gyahun_dash on July 7, 2014, 5:01 p.m.
How about this ? but it looks less readable... <pre class='brush: python'>r'((\A|-)%s(-|\Z))'%(x) </pre>

coells on July 7, 2014, 5:21 p.m.
Yes, this should work. It's even more creative regex :-)

gyahun_dash on July 7, 2014, 4:49 p.m.
Oh, it can pass the tests but can't pass your test. Thanks.

xavierskip on Aug. 7, 2014, 1:40 a.m.
amazing re!

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