Ruthie Newman, C.15, Paper, Bipartition-Graph

graphs/possible_bipartition.py

@@ -1,12 +1,72 @@
 # Can be used for BFS
-from collections import deque 
+from collections import deque
+from pickle import FALSE 
 
 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(1) to access the adjacency list, o(1) to iterate through the dog_dict hash map used to 
+        organize the dislikes data into a more accessible format(w/out empty indices), o(n) to iterate through each value list of the 
+        dictionary, which is dependent on the size of the input. 
+
+        Space Complexity: O(n) for new varaibles created (visited list, dictionary) b/c the space occupied is dependent on the input. 
+        O(E) for the adjacency list because each node in the dislikes array has a list of edges. 
     """
-    pass
+
+    #number of nodes/dogs
+    num_dogs = len(dislikes)
 
+    #if adjacency list is empty == True
+    if num_dogs == 0:
+        return True
+
+    # # #You could put dislikes into an adjacecny dictionary to get rid of all empty values....
+    dog_dict = { i : dislikes[i] for i in range(0, len(dislikes)) if len(dislikes[i]) != 0}
+
+    #Set a value for a start node/dog to 1 b/c there is not always a dog 0
+    current_dog = 0
+
+    #track whether node has been vistited
+    dogs_visited = [False for i in range(num_dogs)]
+
+    #set start dog as visited in dog_visited
+    dogs_visited[current_dog] = True
+
+    # create a queue to do BFS and enqueue source vertex
+    dogs_q = deque()
+    dogs_q.append(current_dog)
+
+    groupA =[]
+    groupB =[]
+    groupA.append(current_dog)
+
+    while dogs_q:
+
+        current_dog = dogs_q.pop()
+
+        for dog, dislike in dog_dict.items():
+
+            for dis in dislike:
+
+                if dis not in dogs_visited:
+                    dogs_visited[dis] = True
+                    dogs_q.append(dis)
+
+                if dog in groupA and dog in groupB:
+                    return False
+
+                if dog not in groupA and dog not in groupB:
+                    groupA.append(dis)
+
+                elif dog in groupA and dog not in groupB:
+                    groupB.append(dis)
+
+
+                elif dog in groupB and dog not in groupA:
+                    groupA.append(dis)   
+
+        return True
+
+