Brothers creative solution for How to Find Friends by spoty - python coding challenges

Enable Javascript in your browser and then refresh this page, for a much enhanced experience.

Brothers solution in Creative category for How to Find Friends by spoty

def check_connection(network, first, second):
    l = lambda m: sum([c.split('-') for c in network if m in c], [])
    d, f = l(first), [first]
    for n in d:
        if n not in f:
            d.extend(l(n))
            f.append(n)
    return second in set(d)


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."

July 7, 2014

Comments:

sevaader on Dec. 26, 2014, 8:52 p.m.
I understand your code, as f is the already visited list, and d the list of friends extended each time, but ... what's the idea with sum ??

spoty on Dec. 30, 2014, 9:54 p.m.
sum("list of lists", []) it's just a trick, how to flatten list of lists.

sevaader on Jan. 6, 2015, 3:15 p.m.
ah, good to know. It's like itertools.chain

GitHub

0 %

You ClassRoom Demo is Ready With the new tool CheckiO ClassRoom, you will be able to analyze your students' progress, have your own leaderboard, change missions order on the map, and so on.