Bipartite Graph Check in DSA C# - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a bipartite graph?

A bipartite graph is a graph whose nodes can be divided into two groups so that no two nodes in the same group are connected by an edge.

beginner

Which method is commonly used to check if a graph is bipartite?

We use graph coloring with two colors or Breadth-First Search (BFS) or Depth-First Search (DFS) to try to color the graph with two colors without conflicts.

beginner

What does it mean if during the bipartite check, a node is found with the same color as its neighbor?

It means the graph is not bipartite because two connected nodes have the same color, breaking the bipartite condition.

beginner

In C#, which data structure is useful to represent a graph for bipartite check?

An adjacency list using a List of Lists or Dictionary of Lists is useful to store graph edges efficiently.

intermediate

What is the time complexity of checking if a graph is bipartite using BFS or DFS?

The time complexity is O(V + E), where V is the number of vertices and E is the number of edges, because each node and edge is visited once.

What is the main property of a bipartite graph?

A Nodes can be split into two groups with no edges inside the same group

B Every node connects to every other node

C Graph has no edges

D Graph has only one node

A bipartite graph's nodes can be divided into two groups with no edges connecting nodes within the same group.

Which algorithm can be used to check if a graph is bipartite?

A Dijkstra's algorithm

B Floyd-Warshall algorithm

C Prim's algorithm

D BFS or DFS with two-coloring

BFS or DFS with two-coloring helps check if the graph can be colored with two colors without conflicts.

If a node and its neighbor have the same color during bipartite check, what does it mean?

A Graph is bipartite

B Graph is not bipartite

C Graph is disconnected

D Graph is complete

Same color on connected nodes means the graph cannot be bipartite.

What data structure is best to represent a graph for bipartite check in C#?

A Stack

B Array of integers

C Adjacency list using List<List<int>> or Dictionary<int, List<int>>

D Queue

Adjacency lists efficiently store graph edges and are easy to use with BFS or DFS.

What is the time complexity of bipartite graph check using BFS or DFS?

A O(V + E)

B O(E^2)

C O(V^2)

D O(1)

Each vertex and edge is visited once, so the complexity is O(V + E).

Explain how to check if a graph is bipartite using BFS or DFS.

Think about coloring nodes and checking neighbors.

Describe the meaning and importance of bipartite graphs in simple terms.

Imagine two teams with no internal conflicts.

Practice

1. What does it mean for a graph to be bipartite?

easy

A. All vertices have the same degree

B. It has an equal number of vertices and edges

C. It contains at least one cycle

D. Its vertices can be divided into two groups with no edges inside the same group

Solution

1. Step 1: Understand bipartite graph definition

   A bipartite graph is one where vertices can be split into two sets with edges only between sets, not inside a set.

2. Step 2: Compare options with definition

   Its vertices can be divided into two groups with no edges inside the same group matches this definition exactly, while others describe unrelated properties.

3. Final Answer:

   Its vertices can be divided into two groups with no edges inside the same group -> Option D

4. Quick Check:

   Bipartite graph = two groups, no internal edges [OK]

Hint: Bipartite means two groups with no edges inside groups [OK]

Common Mistakes:

• Confusing bipartite with equal vertices and edges
• Thinking bipartite means no cycles
• Assuming all vertices have same degree

2. Which C# data structure is best to represent a graph for bipartite check?

easy

A. List adjacency list

B. Dictionary adjacency list

C. int[,] adjacency matrix

D. HashSet<int> for vertices only

Solution

1. Step 1: Identify graph representation needs

   We need to store edges efficiently and access neighbors quickly for bipartite check.

2. Step 2: Evaluate options

   Dictionary allows flexible vertex keys and fast neighbor lookup, better than fixed-size arrays or sets.

3. Final Answer:

   Dictionary adjacency list -> Option B

4. Quick Check:

   Dictionary adjacency list = best for flexible graphs [OK]

Hint: Use Dictionary for flexible graph edges [OK]

Common Mistakes:

• Using HashSet only stores vertices, no edges
• Adjacency matrix wastes space for sparse graphs
• List lacks vertex key mapping

3. What is the output of this C# code snippet checking bipartite for a graph with edges: 0-1, 1-2, 2-3, 3-0?

var graph = new Dictionary {
  {0, new List<int>{1,3}},
  {1, new List<int>{0,2}},
  {2, new List<int>{1,3}},
  {3, new List<int>{2,0}}
};
bool IsBipartite(Dictionary g) {
  var color = new Dictionary();
  foreach(var node in g.Keys) {
    if(!color.ContainsKey(node)) {
      var queue = new Queue<int>();
      queue.Enqueue(node);
      color[node] = 0;
      while(queue.Count > 0) {
        var u = queue.Dequeue();
        foreach(var v in g[u]) {
          if(!color.ContainsKey(v)) {
            color[v] = 1 - color[u];
            queue.Enqueue(v);
          } else if(color[v] == color[u]) {
            return false;
          }
        }
      }
    }
  }
  return true;
}
Console.WriteLine(IsBipartite(graph));

medium

A. True

B. False

C. Compilation error

D. Runtime exception

Solution

1. Step 1: Analyze the graph structure

   The graph forms a cycle of length 4: 0-1-2-3-0.

2. Step 2: Check bipartite condition on even cycle

   Even cycles are bipartite, but here the cycle is length 4 (even), so it should be bipartite. However, the code returns false if adjacent nodes have same color.

3. Step 3: Trace coloring

   Coloring alternates 0 and 1 around the cycle, no conflict occurs, so function returns true.

4. Final Answer:

   True -> Option A

5. Quick Check:

   Even cycle graph is bipartite = True [OK]

Hint: Even cycles are bipartite; odd cycles are not [OK]

Common Mistakes:

• Assuming all cycles are not bipartite
• Confusing color assignment logic
• Ignoring queue BFS traversal

4. Identify the bug in this bipartite check code snippet:

bool CheckBipartite(Dictionary graph) {
  var color = new Dictionary();
  foreach(var node in graph.Keys) {
    if(!color.ContainsKey(node)) {
      color[node] = 0;
      var stack = new Stack<int>();
      stack.Push(node);
      while(stack.Count > 0) {
        var u = stack.Pop();
        foreach(var v in graph[u]) {
          if(!color.ContainsKey(v)) {
            color[v] = color[u];
            stack.Push(v);
          } else if(color[v] == color[u]) {
            return false;
          }
        }
      }
    }
  }
  return true;
}

medium

A. Colors assigned to neighbors are same as current node, should be opposite

B. Stack should be replaced with queue for BFS

C. Missing initialization of color dictionary

D. No return statement for true case

Solution

1. Step 1: Review color assignment logic

   Neighbors must have opposite colors to current node for bipartite check.

2. Step 2: Identify incorrect assignment

   Code assigns neighbor color same as current node, causing false negatives.

3. Final Answer:

   Colors assigned to neighbors are same as current node, should be opposite -> Option A

4. Quick Check:

   Neighbor colors must differ [OK]

Hint: Neighbors must have opposite colors, not same [OK]

Common Mistakes:

• Assigning same color to neighbors
• Confusing stack vs queue (both work but BFS preferred)
• Forgetting to initialize colors

5. Given a graph with vertices 0 to 5 and edges: 0-1, 1-2, 2-3, 3-4, 4-5, 5-0, and 1-4, is the graph bipartite? Explain why.

hard

A. Yes, because all cycles are even length

B. No, because it has a self-loop

C. No, because it contains an odd cycle including edges 1-2-3-4-1

D. Yes, because it has no cycles

Solution

1. Step 1: Identify cycles in the graph

   The graph has a large cycle 0-1-2-3-4-5-0 (length 6, even), and with edge 1-4, cycles 1-2-3-4-1 (length 4, even) and 0-1-4-5-0 (length 4, even).

2. Step 2: Check for odd-length cycles

   No odd-length cycles exist. A graph is bipartite if and only if it has no odd-length cycles.

3. Final Answer:

   No, because it contains an odd cycle including edges 1-2-3-4-1 -> Option C

4. Quick Check:

   Odd cycle means not bipartite [OK]

Hint: Check for odd cycles to decide bipartite [OK]

Common Mistakes:

• Miscounting cycle lengths
• Assuming any cycle means not bipartite
• Ignoring extra edges creating odd cycles