First clear solution for Pawn Brotherhood by wakka - python coding challenges

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First solution in Clear category for Pawn Brotherhood by wakka

def safe_pawns(pawns):
    p = {(ord("8")-ord(x[1]), ord(x[0])-ord("a")) for x in pawns}
    s1 = {(x[0]-1,x[1]-1) for x in p}
    s2 = {(x[0]-1,x[1]+1) for x in p}
    return len(p & s1 | p & s2)

if __name__ == '__main__':
    #These "asserts" using only for self-checking and not necessary for auto-testing
    assert safe_pawns({"b4", "d4", "f4", "c3", "e3", "g5", "d2"}) == 6
    assert safe_pawns({"b4", "c4", "d4", "e4", "f4", "g4", "e5"}) == 1

July 26, 2015

Comments:

bsquare on Sept. 13, 2019, 9:21 p.m.
<p>Nice solution.</p>

ShelbyJ on Oct. 2, 2019, 1:40 p.m.
<p>Nice Answer, Its easy to understand while still being concise</p>

Look up in the Dict()

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