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@CheezItMan CheezItMan left a comment

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Nice work, I like the Set solution, and this works and works well. You hit the learning goals here. If you're curious about my solution (more DFS), you can see it here.

Comment on lines +8 to +9
dogz = Set.new()
doggos = Set.new()

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Nice use of a Set.

dogz1.add(i)
doggos.add(i)

return bipartition_helper(i, dislikes, dogz1, doggos1) || bipartition_helper(i, dislikes, dogz, doggos)

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Originally seeing this pair of recursive calls in this method set off alarm bells, but then I traced through and it doesn't actually make many more recursive calls, it's just a little different from my solution. It does however mean you have do the nested if-else blocks below.

I instead assigned the current "color" for the current node and assigned it's neighbors the alternating color, and then continue in a depth-first-search, and I repeat until the stack is empty. If they already have an incompatible color I return false. This results in a bit simpler code.

This does work however and work well.

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2 participants