C16 - Pine - Ainur #44
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kyra-patton
reviewed
Jul 25, 2022
| Time Complexity: ? | ||
| Space Complexity: ? | ||
| Time Complexity: O(N+E) N-number of nodes, E -number of edges | ||
| Space Complexity: O(N)?? |
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🪐 Yes! Space complexity will be O(N) because the data structures you create, queue, visited, group1, and group2 can each hold at most N nodes (aka dogs).
| @@ -5,8 +5,33 @@ 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+E) N-number of nodes, E -number of edges | |||
| group1 = set() | ||
| group2 = set() | ||
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| queue.append(0) |
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Your two test cases are failing because you need a guard clause to find the first node in the list that is connected to other nodes in the graph.
Say you have a graph such as this:
dislikes = [ [], [2, 3], [1, 3], [1, 2] ]
If you draw out the graph, you will see that Node 0 is disconnected from the rest of the graph - it has no edges. Because it has no edges/neighbors, nothing else will be added to your queue after you pop Node 0 off the queue and you will return True by default.
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