Amy W #8
Amy W #8
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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.
| dogz = Set.new() | ||
| doggos = Set.new() |
| dogz1.add(i) | ||
| doggos.add(i) | ||
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| 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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