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khuddleup
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Jul 22, 2022
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| # Additionally, if the next dog is NOT group 1, add the current dog to group 2 | ||
| # But if the next dog is NOT in group 1 then add it to group 2 | ||
| else: |
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^^^ This shold actually say, if the current dog IS in group 1, then check to see if the next dog is also in group 1, and if it it is return false, cuz aint no way these dogs will get along. But if the next dog is NOT in group 1, then add it to group 2.
kyra-patton
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Jul 25, 2022
kyra-patton
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kyra-patton
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🐾🐶 Very nice BFS implementation, Kayla. Really appreciate you working through this! Let me know what questions you have.
🟢
| Time Complexity: ? | ||
| Space Complexity: ? | ||
| Time Complexity: O(n+e) number of nodes in the graph and e is the numbner of edges | ||
| Space Complexity: O(n) storing 4 variables? Checked, G1, G2, queue. drop the constant |
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✨ Yes, because queue, checked, group1, and group2 each can hold at most n elements, where n is the number of nodes (or dogs) in dislikes. Nice work!
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| if not dislikes[current]: | ||
| queue.append(current + 1) |
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Creative way to handle disconnected nodes in the graph!
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