Second speedy solution for Pawn Brotherhood by Gruggi - python coding challenges

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Second solution in Speedy category for Pawn Brotherhood by Gruggi

def safe_pawns(pawns):
    
    safe_positions = set();
    for item in pawns:
        safe_positions.add(chr(ord(item[0]) + 1) + chr(ord(item[1]) + 1));
        safe_positions.add(chr(ord(item[0]) - 1) + chr(ord(item[1]) + 1));
        
    return len(pawns) - len(pawns.difference(safe_positions)); 

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

Sept. 11, 2014

Comments:

tommycarstensen on April 24, 2015, 1:49 p.m.
<p>I like this one although I suspect it might not be that fast.</p>

tivvit on Jan. 1, 2016, 12:09 a.m.
<p>intersect is better than difference ;)</p>

Look up in the Dict()

0 %

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