diff --git a/graphs/possible_bipartition.py b/graphs/possible_bipartition.py
index ca55677..9f0de0c 100644
--- a/graphs/possible_bipartition.py
+++ b/graphs/possible_bipartition.py
@@ -1,12 +1,56 @@
 # Can be used for BFS
-from collections import deque 
+from collections import deque
 
 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) number of nodes in the graph and e is the numbner of edges
+        Space Complexity: O(n) storing 4 variables? Checked, G1, G2, queue. drop the constant
     """
     pass
+    
+    if not dislikes:
+        return True
 
+    # keep track of nodes checked
+    # Create empty queue
+    # add start_node (in this case 0) to queue
+    checked = [None] * len(dislikes)
+    queue = deque()
+    queue.append(0)
+    group1 = set()
+    group2 = set()
+
+    # So long as the queue is not empty, remove current dog
+    # Mark that current dog has been checked
+    # if dog is not matched to a another hating ass dog, add the next dog to the queue
+    while queue:
+        current = queue.popleft()
+        checked[current] = True
+        if not dislikes[current]:
+            queue.append(current + 1)
+
+        # Loope through list of dogs. 
+        # If the next dog has NOT been checked
+        # Add it to the queue to be assigned
+        for dog in dislikes[current]:
+            if not checked[dog]:
+                queue.append(dog)
+            
+            # If current dog is not in group 1 and if the next dog is in group 2 
+            # Then return false becuase there is no peaceful way to group these doggs.
+            # But if the current dog is not in group 1 and the next dog is not in group 2
+            # Then add the current dog to group 1
+            if current not in group1:
+                if dog in group2:
+                    return False
+                group1.add(dog)
+        
+            # Additionally, if the next dog is NOT group 1, add the current dog to group 2
+            # But if the next dog is NOT in group 1 then add it to group 2
+            else:
+                if dog in group1:
+                    return False
+                group2.add(dog)
+    return True