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

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First solution in Clear category for Pawn Brotherhood by flex

def safe_pawns(pawns):
    count = 0
    for i in pawns:
        if ((chr(ord(i[0])-1)+chr(ord(i[1])-1)) in pawns) or ((chr(ord(i[0])+1)+chr(ord(i[1])-1)) in pawns):
            count+=1                
    return count

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

Oct. 15, 2014

Look up in the Dict()

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