Shubha 🧙‍♀️

lib/possible_bipartition.rb

@@ -1,4 +1,48 @@
+# I looked at the leetcode solution in Python 
+# (https://leetcode.com/problems/possible-bipartition/solution/) 
+# to get an idea of how to turn the list of dislikes into a graph, 
+# but chose to implement a breadth-first solution instead
 
 def possible_bipartition(dislikes)
-  raise NotImplementedError, "possible_bipartition isn't implemented yet"
+    return true if dislikes.empty?
+    graph = build_graph(dislikes)
+    # start with an arbitrary node of the graph, 
+    # assuming all nodes have at least one edge
+    grouped = {graph.keys[0] => true}
+    queue = [graph.keys[0]]   
+    # using a bfs approach, set all neighbor nodes to the opposite label
+    # and add them to the queue if they haven't been visited
+    # return false if a neighbor has the same label as the current node
+    while !queue.empty?
+      node = queue.shift()
+      graph[node].each do |neighbor|
+        if grouped[neighbor] == nil
+          grouped[neighbor] = !grouped[node]
+          queue.push(neighbor)
+        elsif grouped[neighbor] == grouped[node]
+          return false
+        end  
+      end     
+      if queue.empty?
+         graph.keys.each do |key|
+            if grouped[key].nil?
+                queue.push(key)
+                break
+            end
+        end
+      end
+    end  
+    return true
 end
+
+
+def build_graph(dislikes)
+  nodes = Hash.new {|h,k| h[k] = [] } #had to look up how to do this on StackOverflow
+  dislikes.each do |dislike|
+      dislike.each do |node|
+        nodes[node] += dislike - [node] 
+      end
+  end
+  return nodes
+end
+

test/possible_bipartition_test.rb

@@ -17,7 +17,7 @@
     expect(answer).must_equal true
   end
 
-  it "will work for example 2" do
+  it "will work for example 2" do 
     # Arrange
     dislikes =  [ [],
       [2, 3],
@@ -49,7 +49,7 @@
     expect(answer).must_equal false
   end
 
-  it "will return true for a graph which can be bipartitioned" do
+  it "will return true for a graph which can be bipartitioned" do 
     # Arrange
     dislikes = [ [3, 6],
       [2, 5],
@@ -68,7 +68,7 @@
   end
 
 
-  it "will return false for a graph which cannot be bipartitioned" do
+  it "will return false for a graph which cannot be bipartitioned"  do
     # Arrange
     dislikes = [ [3, 6],
       [2, 5],
@@ -89,4 +89,39 @@
   it "will work for an empty graph" do
     expect(possible_bipartition([])).must_equal true
   end
+end
+
+describe "edge cases" do
+  it "will work for a graph where all nodes are not interconnected" do
+    # Arrange
+    dislikes = [
+      [1,2],
+      [3,4],
+      [4,5],
+      [3,5]]
+
+    # Act
+    answer = possible_bipartition(dislikes)
+
+    # Assert
+    expect(answer).must_equal false
+
+  end
+
+  it "will work for a graph where all nodes don't have dislikes" do
+    # Arrange
+    dislikes = [
+      [1,2],
+      [4,5],
+      [5,6],
+      [4,6]]
+
+    # Act
+    answer = possible_bipartition(dislikes)
+
+    # Assert
+    expect(answer).must_equal false
+
+  end
+
 end