From 0d23e069c4765430b59ca101aff806d1f7373525 Mon Sep 17 00:00:00 2001
From: xiomara <xrodri98@gmail.com>
Date: Tue, 17 Jan 2023 21:57:54 -0500
Subject: [PATCH] graphs exercise solution

---
 graphs/possible_bipartition.py | 34 +++++++++++++++++++++++++++++++---
 1 file changed, 31 insertions(+), 3 deletions(-)

diff --git a/graphs/possible_bipartition.py b/graphs/possible_bipartition.py
index ca55677..64d77de 100644
--- a/graphs/possible_bipartition.py
+++ b/graphs/possible_bipartition.py
@@ -1,12 +1,40 @@
 # Can be used for BFS
-from collections import deque 
+from collections import deque
 
-def possible_bipartition(dislikes):
+
+def possible_bipartition(dislikes_dict):
     """ Will return True or False if the given graph
         can be bipartitioned without neighboring nodes put
         into the same partition.
         Time Complexity: ?
         Space Complexity: ?
     """
-    pass
 
+    # return true if the graph is empty
+    if not dislikes_dict:
+        return True
+
+    graph = {}
+    for node in dislikes_dict:
+        graph[node] = dislikes_dict[node]
+
+    color = {}
+    for node in graph:
+        # check if the node has been visited
+        if node not in color:
+            # assign a color to the current node
+            color[node] = 0
+            queue = deque([node])
+            while queue:
+                # get the next node to visit
+                current = queue.popleft()
+                # visit the neighbors of the current node
+                for neighbor in graph[current]:
+                    if neighbor not in color:
+                        # assign the opposite color to the neighbor
+                        color[neighbor] = 1 - color[current]
+                        queue.append(neighbor)
+                    elif color[neighbor] == color[current]:
+                        # if the neighbor already has the same color as the current node, the graph is not bipartite
+                        return False
+    return True