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Solve the following problem using Python, implementing the functions described below, one line at a time
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Description:
 def flatten_dict_join_keys(dct, join_symbol=" "):
 """ Flatten dict with defined key join symbol. :param dct: dict to flatten :param join_symbol: default value is " " :return: """

    return dict( flatten_dict(dct, join=lambda a,b:a+join_symbol+b) )