Ari - ✂️ - Bi-partition-Graph

.gitignore

@@ -142,3 +142,6 @@ dmypy.json
 # Cython debug symbols
 cython_debug/
 
+# scratch file
+scratch.py
+scratch_2.py

graphs/possible_bipartition.py

@@ -1,12 +1,62 @@
 # Can be used for BFS
 from collections import deque 
+from pprint import pprint
 
 def possible_bipartition(dislikes):
     """ Will return True or False if the given graph
-        can be bipartitioned without neighboring nodes put
+        can be bi-partitioned without neighboring nodes put
         into the same partition.
-        Time Complexity: ?
-        Space Complexity: ?
+        Time Complexity: O(nm)
+        Space Complexity: O(n)
     """
-    pass
+    if not dislikes:
+        return True
 
+    dislikes_graph = {}
+    for index, value in enumerate(dislikes):
+        if index not in dislikes_graph:
+            dislikes_graph[index] = value
+        else:
+            dislikes_graph[index] += value     
+    pprint(dislikes_graph)
+
+    visited = set()
+    red = set()
+    green = set()
+
+    dogs_queue = deque()
+
+    for i in range(len(dislikes_graph)):  # if graph nodes are not connected
+        if i in visited:
+            continue
+
+        dogs_queue.append((i, "red"))   # target color/ add it to queue
+
+        while dogs_queue:
+            dog, pen = dogs_queue.popleft()  # process dog
+            visited.add(i)   # once dog processed
+
+            # if we've seen this dog before and it's in the opposite pen already
+            if dog in red and pen == "green":
+                return False
+            elif dog in green and pen == "red":
+                return False
+
+            if dog not in red and dog not in green:  # haven't separated dog
+                if pen == "green":
+                    green.add(dog)  
+                    visited.add(dog)
+
+                elif pen == "red":         
+                    red.add(dog)   
+                    visited.add(dog)
+
+                # process neighboors of dog / putting dislike dogs in opposite pen   
+                for neighbor in dislikes_graph[dog]:  
+                    if neighbor not in red and neighbor not in green:
+                        if pen == "red":  # opposite pen!
+                            dogs_queue.append((neighbor, "green"))
+                        elif pen == "green":
+                            dogs_queue.append((neighbor, "red"))
+
+    return True

tests/test_possible_bipartition.py

@@ -9,10 +9,13 @@ def test_example_1():
       [1],
       [2]
     ]
+    # {0, 2, 1 , 1, 2}
+    # {3, 4}
+    # TRUE
 
     # Act
     answer = possible_bipartition(dislikes)
-
+    print(f' FIRST TEST ANSWER {answer}')
     # Assert
     assert answer
 
@@ -32,7 +35,7 @@ def test_example_2():
 
 def test_example_r():
     # Arrange
-    dislikes = [ [],
+    dislikes = [[],
       [2, 5],
       [1, 3],
       [2, 4],
@@ -80,21 +83,20 @@ def test_will_return_false_for_graph_which_cannot_be_bipartitioned():
     # Assert
     assert not answer
 
-
 def test_will_return_true_for_empty_graph():
     assert possible_bipartition([])
 
 def test_will_return_false_for_another_graph_which_cannot_be_bipartitioned():
     # Arrange
-    dislikes = [ [3, 6],
-      [2, 5],
-      [1, 3],
-      [0, 2, 4],
-      [3, 5],
-      [1, 4],
-      [0],
-      [8],
-      [7]
+    dislikes = [[3, 6],  # [0]
+      [2, 5],             # [1]
+      [1, 3],             # [2]
+      [0, 2, 4],          # [3]
+      [3, 5],             # [4]
+      [1, 4],             # [5]
+      [0],                # [6]
+      [8],                # [7]
+      [7]                 # [8]
     ]
 
     # Act