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all tests passing #25

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all tests passing #25

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FaithKauwe
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✨🐩 Nice work, Faith. I left a couple comments on time complexity and edge cases, but overall a very clean BFS solution. Let me know what questions you have.

🟢

@@ -5,8 +5,57 @@ def possible_bipartition(dislikes):
""" Will return True or False if the given graph
can be bipartitioned without neighboring nodes put
into the same partition.
Time Complexity: ?
Space Complexity: ?
Time Complexity: O(n)

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⏱ Time complexity will actually be O(N + E) where N is the number of nodes/dogs in dislikes and E is the number of edges or dislikes between two dogs. You loop through all of the nodes, then each of the nodes' edges.

Comment on lines +23 to +26
if dislikes[0]:
start = 0
else:
start = 1

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What if there are multiple dogs at the beginning of dislikes that don't dislike any other dogs (are disconnected from the rest of the graph)? How might you account for this edge case?

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2 participants