4 lines creative solution for How to Find Friends by DiZ - python coding challenges

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4 lines solution in Creative category for How to Find Friends by DiZ

def check_connection(edges, source, destination):
    graph, edges = {source}, [set(e.split('-')) for e in edges]
    for _ in edges: graph.update(*filter(graph.intersection, edges))
    return destination in graph

Sept. 3, 2015

Comments:

StefanPochmann on Sept. 22, 2015, 12:26 p.m.
Very nice update command.

DiZ on Sept. 22, 2015, 12:38 p.m.
Thanks. Still not sure about its readibility, but it's a terrible shortcut!

veky on Sept. 28, 2015, 11:05 a.m.
It's completely readable. Add to graph points of those edges that intersect graph. :-) <pre class='brush: python'>add to : .update points of : * those that : filter </pre>

DiZ on Sept. 28, 2015, 12:09 p.m.
Thanks! this translation will help many others :-D

SergeyEremeykin on Nov. 16, 2015, 11:59 p.m.
Really smart solution!

hao.daniel on Feb. 16, 2020, 6:19 p.m.
it's like a magician playing some tricks... Wonder how long to take to come up this ingenious. More interesting is to share the "thought process", so we can apply the pattern to similar problems? Great Job!

CDG.Axel on Nov. 13, 2021, 9:01 p.m.
Great idea with filter(set.intersection,...)! I used [e for e in network if e & friends] for this :( Btw it helps to avoid PEP8 violation: <pre class='brush: python'>[graph.update(*filter(graph.intersection, edges)) for _ in edges] </pre>

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