C15 - Christian Spencer #13
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CheezItMan
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Jan 11, 2022
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| def possible_bipartition(dislikes): | ||
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| """ 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(Nodes + Edges) | ||
| - Every edge for a dog starting with the first dog and exploring all edges. Edges are only expored once and all nodes must be | ||
| colored. | ||
| Space Complexity: O(Nodes + Edges) | ||
| - The adjacency list has O + E space complexity. Each node and edge requires an array or list space, but it is | ||
| not squre when stored in 2 dimensions. | ||
| """ | ||
| pass |
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👍 Interesting solution using numbers (1 and 0) for node colors. Well done.
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