7-liner: simple clear solution for New Cities by przemyslaw.daniel - python coding challenges

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7-liner: simple solution in Clear category for New Cities by przemyslaw.daniel

def subnetworks(net, crushes):
    result = {x: {x} for x in sum(net, []) if x not in crushes}
    for a, b in net*len(net):
        if set(crushes) & {a, b}: continue
        result[a] |= result[b] | {a, b}
        result[b] |= result[a] | {a, b}
    return len({frozenset(x) for x in result.values()})

Oct. 25, 2017

Comments:

martin.pilka on Jan. 31, 2019, 6:54 p.m.
<p>Indeed this code is short, however I believe it needs some explanation, could you please update it with comments?</p>

przemyslaw.daniel on Feb. 1, 2019, 5:25 a.m.
<pre class='brush: python'>def subnetworks(net, crushes): result = {x: {x} for x in # create dict for all not sum(net, []) if x not in crushes} # crushed nodes for a, b in net*len(net): # iterate len(net)**2 times if set(crushes) & {a, b}: continue # skip if not valid result[a] |= result[b] | {a, b} # exchange result[b] |= result[a] | {a, b} # exchange return len({frozenset(x) for x in result.values()}) # count unique </pre> <p>The idea here is pretty simple. For every valid pair of neighbors, we exchange information about the neighborhood as long as the entire network is aware. At the end just count unique subnetworks. This is rather inefficient, but clear way to do it.</p>

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