Ada Barries - C17 Otters

graphs/possible_bipartition.py

@@ -5,8 +5,39 @@ 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) (N is dogs, E is edges (dogs who got beef))
+        Space Complexity: O(N) (nodes (dogs) reliant on amount of dogs)
     """
-    pass
+    # return true if dogs can be split into two groups
+    # and no fighting will occur. otherwise return false.
+    if not dislikes:
+        return True
+
+    # dict for dog groups. default is 0, other group is 1.
+    kennel = { dog:0 for dog in dislikes }
+
+    first_dog = list(dislikes.keys())[0]
+    kennel[first_dog] = 1
+
+    #BFS
+    queue = [first_dog]
+    visited = [first_dog]
+
+    for dog in dislikes:
+        while queue:
+            current_dog = queue.pop(0)
+            visited.append(current_dog)
+
+            if dislikes[current_dog]:
+                for angy_dog in dislikes[current_dog]:
+                    if kennel[angy_dog] == 0:
+                        kennel[angy_dog] = kennel[current_dog] + 1
+                        queue.append(angy_dog)
+                    else:
+                        if kennel[current_dog] == kennel[angy_dog]:
+                            return False
+        if dog not in visited:
+            queue.append(dog)
+
+    return True