Pawn Brotherhood clear solution for Pawn Brotherhood by graffme - python coding challenges

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Pawn Brotherhood solution in Clear category for Pawn Brotherhood by graffme

def safe_pawns(pawns):
    
    #pionek jest bezpieczny jeśli ma za sobą po skosie innego pionka
    #ustawić tablicę z parami pionków
    #sprawdzić czy dla danego pionka istnieje pionek zabezpieczający 
    
    pawns_indexes = []
    for p in pawns:
        col = ord(p[0]) - 97
        row = int(p[1]) - 1
        pawns_indexes.append((col, row))
    
    safe_pawns = 0
    
    for pawn in pawns_indexes:
        pawn_col = pawn[0]
        pawn_row = pawn[1]
        if ((pawn_col -1, pawn_row-1) in pawns_indexes) or ((pawn_col +1, pawn_row-1) in pawns_indexes):
            safe_pawns = safe_pawns + 1
        
    return safe_pawns
        

    
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

Jan. 11, 2017

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