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

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First solution in Clear category for Pawn Brotherhood by yukirin

def safe_pawns(pawns):
    barrier = map(
        lambda p: ((chr(ord(p[0]) - 1) + str(int(p[1]) - 1)),
                   (chr(ord(p[0]) + 1) + str(int(p[1]) - 1))),
        pawns)
    return len([True for b1, b2 in barrier if b1 in pawns or b2 in pawns])
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

March 30, 2015

Look up in the Dict()

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