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

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First solution in Clear category for Pawn Brotherhood by PythOff

def safe_pawns(pawns):
    ileBezp = 0

    lewy = []
    prawy = []
    
    for i in pawns:
        lewy.append(chr(ord(i[0])-1) + str(int(i[1]) - 1))
        prawy.append(chr(ord(i[0])+1) + str(int(i[1]) - 1))
        
    for j in range(0, len(pawns)):
        if lewy[j] in pawns or prawy[j] in pawns:
            ileBezp += 1
    
    return ileBezp

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. 28, 2016

Look up in the Dict()

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