Bipartite Graph Check in DSA Java - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Bipartite Check on Simple Graph

What is the output of the following Java code that checks if a graph is bipartite?

DSA Java

import java.util.*;

public class BipartiteCheck {
    public static boolean isBipartite(int[][] graph) {
        int n = graph.length;
        int[] colors = new int[n];
        Arrays.fill(colors, -1);

        for (int i = 0; i < n; i++) {
            if (colors[i] == -1) {
                Queue<Integer> queue = new LinkedList<>();
                queue.offer(i);
                colors[i] = 0;

                while (!queue.isEmpty()) {
                    int node = queue.poll();
                    for (int neighbor : graph[node]) {
                        if (colors[neighbor] == -1) {
                            colors[neighbor] = 1 - colors[node];
                            queue.offer(neighbor);
                        } else if (colors[neighbor] == colors[node]) {
                            return false;
                        }
                    }
                }
            }
        }
        return true;
    }

    public static void main(String[] args) {
        int[][] graph = {
            {1,3},
            {0,2},
            {1,3},
            {0,2}
        };
        System.out.println(isBipartite(graph));
    }
}

A Compilation error

B true

C false

D Runtime exception

💡 Hint

Think about whether the graph can be colored using two colors without adjacent nodes sharing the same color.

The graph is a square cycle (4 nodes connected in a cycle). Such a cycle with even length is bipartite, so the function returns true.

❓ Predict Output

intermediate

Output of Bipartite Check on Odd Cycle Graph

What is the output of the following Java code that checks if a graph is bipartite?

DSA Java

import java.util.*;

public class BipartiteCheck {
    public static boolean isBipartite(int[][] graph) {
        int n = graph.length;
        int[] colors = new int[n];
        Arrays.fill(colors, -1);

        for (int i = 0; i < n; i++) {
            if (colors[i] == -1) {
                Queue<Integer> queue = new LinkedList<>();
                queue.offer(i);
                colors[i] = 0;

                while (!queue.isEmpty()) {
                    int node = queue.poll();
                    for (int neighbor : graph[node]) {
                        if (colors[neighbor] == -1) {
                            colors[neighbor] = 1 - colors[node];
                            queue.offer(neighbor);
                        } else if (colors[neighbor] == colors[node]) {
                            return false;
                        }
                    }
                }
            }
        }
        return true;
    }

    public static void main(String[] args) {
        int[][] graph = {
            {1,2},
            {0,2},
            {0,1}
        };
        System.out.println(isBipartite(graph));
    }
}

A false

B Compilation error

C true

D Runtime exception

💡 Hint

Consider if a triangle graph can be colored with two colors without conflict.

The graph is a triangle (3 nodes all connected). Odd cycles are not bipartite, so the function returns false.

🧠 Conceptual

advanced

Why BFS is used in Bipartite Check?

Why is Breadth-First Search (BFS) preferred over Depth-First Search (DFS) for checking if a graph is bipartite?

A BFS colors nodes level by level, making it easier to detect conflicts between adjacent nodes.

B DFS cannot be used for bipartite check because it does not visit all nodes.

C BFS uses less memory than DFS in all cases.

D DFS always produces incorrect coloring for bipartite graphs.

💡 Hint

Think about how BFS explores nodes in layers and how that helps in coloring.

BFS explores nodes level by level, so it assigns colors in alternating layers, making it straightforward to detect conflicts.

🔧 Debug

advanced

Identify the Bug in Bipartite Check Code

What is the bug in the following Java code snippet for bipartite check?

DSA Java

public static boolean isBipartite(int[][] graph) {
    int n = graph.length;
    int[] colors = new int[n];
    Arrays.fill(colors, 0);

    for (int i = 0; i < n; i++) {
        if (colors[i] == 0) {
            Queue<Integer> queue = new LinkedList<>();
            queue.offer(i);
            colors[i] = 1;

            while (!queue.isEmpty()) {
                int node = queue.poll();
                for (int neighbor : graph[node]) {
                    if (colors[neighbor] == 0) {
                        colors[neighbor] = 3 - colors[node];
                        queue.offer(neighbor);
                    } else if (colors[neighbor] == colors[node]) {
                        return false;
                    }
                }
            }
        }
    }
    return true;
}

A The for loop should start from 1 instead of 0.

B Using 3 - colors[node] for color assignment is incorrect arithmetic.

C Queue should be a Stack for correct traversal.

D Colors array initialized with 0 causes confusion between uncolored and color 0.

💡 Hint

Check how uncolored nodes are represented and how colors are assigned.

Initializing colors with 0 means uncolored nodes and color 0 are the same, causing logic errors. Using -1 for uncolored is better.

🚀 Application

expert

Maximum Bipartite Matching Application

Given a bipartite graph represented as adjacency lists, which algorithm is best suited to find the maximum matching?

A Dijkstra's algorithm for shortest paths

B Prim's algorithm for minimum spanning tree

C Ford-Fulkerson algorithm using DFS for augmenting paths

D Kruskal's algorithm for minimum spanning tree

💡 Hint

Maximum bipartite matching can be solved by finding augmenting paths in a flow network.

Ford-Fulkerson algorithm uses DFS to find augmenting paths in the flow network constructed from the bipartite graph, enabling maximum matching calculation.

Practice

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

easy

A. Its vertices can be divided into two sets with no edges inside the same set

B. It has an equal number of vertices and edges

C. It contains at least one cycle

D. All vertices have the same degree

Solution

1. Step 1: Understand bipartite graph definition

   A bipartite graph is one where vertices can be split into two groups so that no edge connects vertices within the same group.

2. Step 2: Compare options with definition

   Only Its vertices can be divided into two sets with no edges inside the same set matches this definition exactly. Other options describe unrelated properties.

3. Final Answer:

   Its vertices can be divided into two sets with no edges inside the same set -> Option A

4. Quick Check:

   Bipartite = Two sets, 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 Java code snippet correctly initializes a graph as an adjacency list for bipartite check?

easy

A. List graph = new ArrayList<>();

B. int graph = new int[5][5];

C. Map graph = new HashMap<>();

D. Set<Integer> graph = new HashSet<>();

Solution

1. Step 1: Identify adjacency list structure

   An adjacency list for a graph is usually a list of lists, where each inner list holds neighbors of a vertex.

2. Step 2: Match code snippets to adjacency list

   List<List<Integer>> graph = new ArrayList<>(); uses List of Lists, suitable for adjacency list. int graph = new int[5][5]; is a 2D array (adjacency matrix), not list. Options A and C use Set and Map, which are not standard adjacency lists.

3. Final Answer:

   List<List<Integer>> graph = new ArrayList<>(); -> Option A

4. Quick Check:

   Adjacency list = List of Lists [OK]

Hint: Adjacency list uses List of Lists in Java [OK]

Common Mistakes:

• Using 2D arrays instead of lists
• Using Map or Set incorrectly for adjacency
• Not initializing the list properly

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

List<List<Integer>> graph = new ArrayList<>();
for (int i = 0; i < 4; i++) graph.add(new ArrayList<>());
graph.get(0).add(1); graph.get(1).add(0);
graph.get(1).add(2); graph.get(2).add(1);
graph.get(2).add(3); graph.get(3).add(2);
graph.get(3).add(0); graph.get(0).add(3);

boolean result = isBipartite(graph);
System.out.println(result);

Assuming isBipartite correctly checks bipartite property.

medium

A. false

B. true

C. Compilation error

D. Runtime exception

Solution

1. Step 1: Analyze the graph structure

   The graph has edges forming a cycle of length 4: 0-1-2-3-0.

2. Step 2: Check bipartite condition for even cycle

   Even length cycles (4) are bipartite, so the graph should be bipartite. But the edges connect 0-3 and 3-2, forming a cycle of length 4.

3. Step 3: Re-examine edges carefully

   The cycle is 0-1-2-3-0, which is even length, so graph is bipartite. So output should be true.

4. Final Answer:

   true -> Option B

5. Quick Check:

   Even cycle graph is bipartite = true [OK]

Hint: Even length cycles are bipartite, odd length are not [OK]

Common Mistakes:

• Thinking all cycles are not bipartite
• Confusing cycle length parity
• Assuming runtime errors without cause

4. Identify the bug in this Java method for bipartite check:

boolean isBipartite(List<List<Integer>> graph) {
  int n = graph.size();
  int[] colors = new int[n];
  Arrays.fill(colors, -1);
  for (int i = 0; i < n; i++) {
    if (colors[i] == -1) {
      if (!dfsCheck(graph, i, colors, 0)) return false;
    }
  }
  return true;
}

boolean dfsCheck(List<List<Integer>> graph, int node, int[] colors, int color) {
  colors[node] = color;
  for (int neighbor : graph.get(node)) {
    if (colors[neighbor] == -1) {
      if (!dfsCheck(graph, neighbor, colors, 1 - color)) return false;
    } else if (colors[neighbor] == color) {
      return false;
    }
  }
  return true;
}

medium

A. Missing check for null graph

B. Colors array not initialized properly

C. Incorrect color flip in recursive call

D. DFS color assignment logic is correct, no bug

Solution

1. Step 1: Review color initialization

   Colors array is initialized with -1, meaning uncolored, which is correct.

2. Step 2: Check DFS color assignment and recursion

   DFS assigns current node color, then recursively assigns opposite color (1 - color) to neighbors. If neighbor has same color, returns false.

3. Step 3: Confirm logic correctness

   The logic correctly detects bipartite property using DFS and color flipping. No bug found.

4. Final Answer:

   DFS color assignment logic is correct, no bug -> Option D

5. Quick Check:

   DFS color flip and checks are correct [OK]

Hint: DFS flips color 0 and 1 to check bipartite [OK]

Common Mistakes:

• Forgetting to flip color in recursion
• Not initializing colors array
• Returning true too early

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 vertices can be colored the same

B. No, because it contains an odd length cycle

C. Yes, because it contains only even length cycles

D. No, because it has isolated vertices

Solution

1. Step 1: Identify cycles in the graph

   The graph edges form a 6-node cycle: 0-1-2-3-4-5-0 and an extra edge 1-4.

2. Step 2: Check for odd length cycles

   There exists an odd length cycle: 1-4-5-0-1 (length 4 is even, but 1-4-5-0-1 is actually length 4, so check carefully). Actually, the cycle 1-4-5-0-1 is length 4 (even). But consider cycle 1-2-3-4-1: edges 1-2,2-3,3-4,4-1 is length 4 (even). However, the presence of edge 1-4 creates a triangle cycle 1-4-5-0-1? No, 1-4-5-0-1 is length 4. But check for odd cycles: 1-4-5-0-1 is length 4, 0-1-4 is length 3? No edge 0-4. But edge 1-4 exists, 4-5, 5-0, 0-1. So cycle 1-4-5-0-1 is length 4. No odd cycles. But check cycle 1-4-5-0-1 is length 4. So no odd cycles. So graph is bipartite.

3. Step 3: Determine bipartite property

   Absence of odd length cycles means the graph is bipartite.

4. Final Answer:

   Yes, because it contains only even length cycles -> Option C

5. Quick Check:

   No odd cycles = bipartite [OK]

Hint: Odd cycles break bipartite property [OK]

Common Mistakes:

• Falsely assuming extra edge creates odd cycle (e.g., nonexistent 1-5)
• Miscalculating cycle lengths
• Confusing isolated vertices with bipartite condition