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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 resource_string(package_or_requirement, resource_name):
 """ Similar to pkg_resources.resource_string but if the resource it not found via pkg_resources it also looks in a predefined list of paths in order to find the resource :param package_or_requirement: the module in which the resource resides :param resource_name: the name of the resource :return: the file content :rtype: str """

    with open(resource_filename(package_or_requirement, resource_name), 'r') as resource_file:
        return resource_file.read()