possible bipartition

graphs/possible_bipartition.py

@@ -1,12 +1,57 @@
 # Can be used for BFS
-from collections import deque 
+from collections import deque
+import collections
+from email.policy import default 
 
 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(V+E)
+        Space Complexity: O(N)
     """
-    pass
+    # input = [ [], [2, 3], [1, 4], [1], [2]]
+    # output =  True 
 
+    #  input1 =  [ [],[2, 3],[1, 3],[1, 2]]
+    # output = False 
+
+    ### Note ###
+    # I had a hard time to come with a solution on my own
+    # After looking up online resources/videos, this solution made sense to me
+
+    start_node = 0
+    visited = [False] * len(dislikes)
+    q = deque()
+    q.append(start_node)
+
+    # initialize two groups for neighboring nodes 
+    group_a = set()
+    group_b = set()
+
+    # edge cases when dislikes is empty or the len is 1
+    if not dislikes or len(dislikes) == 1:
+        return True 
+
+    while q:
+        current = q.popleft() #start_node is 0, 
+        visited[current] = True
+
+        if not dislikes[current]:
+            q.append(current + 1)
+
+        # iterate through the nodes in dislikes 
+        for neighbor in dislikes[current]:
+            if not visited[neighbor]: # not visited is append to q 
+                q.append(neighbor)
+
+            if current not in group_a:
+                if neighbor in group_b:
+                    return False
+                group_a.add(neighbor)
+            else:
+                if neighbor in group_a:
+                    return False
+                group_b.add(neighbor)
+
+    return True