Rock - Juliana's Solution. All tests passing.

graphs/possible_bipartition.py

@@ -5,8 +5,28 @@ 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 + E)
+        Space Complexity: O(n)
     """
-    pass
+    N = len(dislikes)
+    dogs = [None] * N
+    queued = deque()
 
+    for i in range(N):
+        if dogs[i] != None:
+            continue
+        queued.append((i, "Baker"))
+        while queued:
+            node, dog = queued.popleft()
+            puppy = "Baker"
+            if not dogs[node]:
+                dogs[node] = dog
+            for next in dislikes[node]:
+                if dogs[next] == dogs[node]:
+                    return False
+                elif dogs[next] and dogs[next] != dogs[node]:
+                    continue
+                elif not dogs[next] and dogs[node] == "Baker":
+                    puppy = "Jpug"
+                queued.append((next, puppy))
+    return True