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Mary Tian CS Fun Bipartition Graph #49
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@@ -8,5 +8,40 @@ def possible_bipartition(dislikes): | |
Time Complexity: ? | ||
Space Complexity: ? | ||
""" | ||
pass | ||
# this is a guard clause | ||
if not dislikes: | ||
return True | ||
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queue = deque() | ||
visited = set() | ||
red = set() | ||
green = set() | ||
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if dislikes[0]: | ||
start_node = 0 | ||
else: | ||
start_node = 1 | ||
Comment on lines
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. Consider the edge case where there are multiple disconnected nodes indexed at the beginning of For example: `dislikes = [ [], [], [], [1,2] ] How might you refactor your code to account for this edge case? |
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queue.append(start_node) | ||
visited.add(start_node) | ||
red.add(start_node) | ||
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while len(queue) != 0: | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 🐩 Nice BFS approach |
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current = queue.popleft() | ||
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for neighbor in dislikes[current]: | ||
if neighbor not in visited: | ||
visited.add(neighbor) | ||
queue.append(neighbor) | ||
if current in red: | ||
green.add(neighbor) | ||
else: | ||
red.add(neighbor) | ||
elif neighbor in visited: | ||
if current in red and neighbor in red: | ||
return False | ||
if current in green and neighbor in green: | ||
return False | ||
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return True | ||
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There was a problem hiding this comment.
Choose a reason for hiding this comment
The reason will be displayed to describe this comment to others. Learn more.
⏱🪐 Time and space complexity?