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

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First solution in Clear category for Pawn Brotherhood by nalotjr

def safe_pawns(pawns):
    
    cnt = 0
    for pn in pawns:
        n1 = ""
        n2 = ""
        n1 = n1 + chr(ord(pn[0])-1) + chr(ord(pn[1])-1)
        n2 = n2 + chr(ord(pn[0])+1) + chr(ord(pn[1])-1)
        if n1 in pawns or n2 in pawns:
            cnt+=1
    return cnt

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

Dec. 17, 2015

Look up in the Dict()

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