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kyra-patton
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Jul 24, 2022
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| Time Complexity: ? | ||
| Space Complexity: ? |
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| group_1 = 0 | ||
| group_2 = 1 |
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Consider making all caps, as you use these as constants
Suggested change
| group_1 = 0 | |
| group_2 = 1 | |
| GROUP_1 = 0 | |
| GROUP_2 = 1 |
| groups[start_dog] = group_1 | ||
| queue = deque() | ||
| queue.append(start_dog) | ||
| while queue: |
kyra-patton
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Jul 25, 2022
kyra-patton
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kyra-patton
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🐾🐶 Overall nice implementation Mac. It looks like you got your logic a bit backwards at one point, but overall it seems like you have a good understanding of BFS. I'm marking this as a yellow since the time and space complexity are missing and not all of the tests are passing. Feel free to update to receive a green score.
Let me know what questions you have.
🟡
| for disliked_dog in dislikes[curr_dog]: | ||
| if groups[disliked_dog] is None: | ||
| groups[disliked_dog] = ( | ||
| group_1 if groups[curr_dog] == group_1 else group_2 |
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🤔 I think you mixed up your logic a bit here. If the current dog is in group_1 then you want the disliked dog to be in group_2.
Suggested change
| group_1 if groups[curr_dog] == group_1 else group_2 | |
| group_2 if groups[curr_dog] == group_1 else group_1 |
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