First clear solution for Pawn Brotherhood by Lachesis_132296 - python coding challenges

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First solution in Clear category for Pawn Brotherhood by Lachesis_132296

#str(chr(97)) == 'a'

def safe_pawns(pawns):
    #row - o 1 mniejsze
    #column - o 1 mniejsze lub o 1 większe
    matrix = {}
    for i in pawns:
        matrix[i[0], int(i[1])] = 1
    safe = 0
   # print(str(chr(ord('b') - 1))) == 'a'
    for key in matrix.keys():
        #jeżeli w matrycy element o linijkę niżej i kolumnę o 1 w przód/w tył jest różny od zera, to pionek jest bezpieczny:
        if matrix.get((chr((ord(key[0])+1)), key[1]-1), 0) != 0 or matrix.get((chr((ord(key[0])-1)), key[1]-1), 0) != 0: 
            safe += 1
    return safe

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

Nov. 19, 2016

Look up in the Dict()

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