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The solution for part two could now be used for part one as well but then I would have to rewrite part 1 .-.
I initially thought that, but now I reconsider I'm not so sure. Isn't it possible to have a 3-member clique overlapping two larger ones? In other words, there could be more than one way to partition the graph into completely connected components. Which means my solution to part 2 is technically incorrect. Bummer.
There probably are multiple ways to partition the graph.
I haven't applied any optimizations and my program checks members of already detected groups again, would that yield all possible partitions because I choose all the possible starting points for a k-clique?