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

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First solution in Clear category for Pawn Brotherhood by Corwine

def safe_pawns(pawns):
    pawns = {(ord(p[0]) - ord('a'), int(p[1])) for p in pawns}
    return len([p for p in pawns if (p[0] - 1, p[1] - 1) in pawns or (p[0] + 1, p[1] - 1) 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

Feb. 19, 2015

Comments:

cizeko on Feb. 19, 2015, 11:49 a.m.
<p>Nice!</p>

Look up in the Dict()

0 %

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