Shubha 🧙♀️ #9
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Shubha 🧙♀️ #9
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CheezItMan
reviewed
Nov 5, 2019
| @@ -1,4 +1,43 @@ | |||
| # I looked at the leetcode solution in Python | |||
| # (https://leetcode.com/problems/possible-bipartition/solution/) | |||
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Ok great, it's been a while since you saw graphs in class anyway.
lib/possible_bipartition.rb
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| def build_graph(dislikes) | ||
| nodes = Hash.new {|h,k| h[k] = [] } #had to look up how to do this on StackOverflow | ||
| dislikes.each do |dislike| |
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Minor thing: You don't need to indent here.
lib/possible_bipartition.rb
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| if queue.empty? && node < graph.keys.max | ||
| queue.push(node + 1) | ||
| end |
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This doesn't seem like the most efficient way to handle a disconnected graph. Maybe also check to ensure that grouped[node].nil?
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I updated this to seek the next ungrouped node. It gets rid of a lot of the aimless wandering of the last approach, but still doesn't seem very efficient. Thoughts?:
if queue.empty?
graph.keys.each do |key|
if grouped[key].nil?
queue.push(key)
break
end
end
end
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