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Ruiz #27
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -1,12 +1,44 @@ | ||
| # Can be used for BFS | ||
| from collections import deque | ||
| from collections import deque | ||
| # from curses import COLORS | ||
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| COLORS = ["blue", "green"] | ||
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| def dfs(dislikes, current_node, painted_graph, current_color): | ||
| neighbors = dislikes[current_node] | ||
| next_color = (current_color + 1) % len(COLORS) | ||
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| for neighbor in neighbors: | ||
| color = painted_graph.get(neighbor) | ||
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| if not color: | ||
| painted_graph[neighbor] = COLORS[next_color] | ||
| if not dfs(dislikes=dislikes, current_node=neighbor, painted_graph=painted_graph, current_color=next_color): | ||
| return False | ||
| elif color != COLORS[next_color]: | ||
| return False | ||
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| return True | ||
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| 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) | ||
| Space Complexity: o(n) | ||
| """ | ||
| pass | ||
| painted_graph = {} | ||
| current_color = 0 | ||
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| for node in range(len(dislikes)): | ||
| neighbors = dislikes[node] | ||
| if not painted_graph.get(node): | ||
| painted_graph[node] = COLORS[current_color] | ||
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| if not dfs(dislikes=dislikes, current_node=node, painted_graph=painted_graph, current_color=current_color): | ||
| return False | ||
| return True | ||
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| # struggled with an small syntax error and then looked at leetcode answers and solution and noticed i had set it up backwards. | ||
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⏱ Time complexity is actually O(N + E) here where N is the number of nodes in
dislikesand E is the number of edges in the graph. This is because depth first search will traverse each node and edge once.