Bipartite Graph Check in DSA C++ - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Bipartite Check on Simple Graph

What is the output of the following C++ code that checks if the graph is bipartite?

DSA C++

#include <iostream>
#include <vector>
#include <queue>
using namespace std;

bool isBipartite(int V, vector<vector<int>>& adj) {
    vector<int> color(V, -1);
    for (int i = 0; i < V; i++) {
        if (color[i] == -1) {
            queue<int> q;
            q.push(i);
            color[i] = 0;
            while (!q.empty()) {
                int u = q.front(); q.pop();
                for (int v : adj[u]) {
                    if (color[v] == -1) {
                        color[v] = 1 - color[u];
                        q.push(v);
                    } else if (color[v] == color[u]) {
                        return false;
                    }
                }
            }
        }
    }
    return true;
}

int main() {
    int V = 4;
    vector<vector<int>> adj = {
        {1, 3},
        {0, 2},
        {1, 3},
        {0, 2}
    };
    cout << (isBipartite(V, adj) ? "Yes" : "No") << endl;
    return 0;
}

A "Yes"

B "No"

C Compilation error

D Runtime error

💡 Hint

Think about whether the graph can be colored with 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. The code colors nodes alternately and finds no conflict, so returns true, printing "Yes".

❓ Predict Output

intermediate

Output of Bipartite Check on Odd Cycle Graph

What is the output of this C++ code checking bipartite property on a triangle graph?

DSA C++

#include <iostream>
#include <vector>
#include <queue>
using namespace std;

bool isBipartite(int V, vector<vector<int>>& adj) {
    vector<int> color(V, -1);
    for (int i = 0; i < V; i++) {
        if (color[i] == -1) {
            queue<int> q;
            q.push(i);
            color[i] = 0;
            while (!q.empty()) {
                int u = q.front(); q.pop();
                for (int v : adj[u]) {
                    if (color[v] == -1) {
                        color[v] = 1 - color[u];
                        q.push(v);
                    } else if (color[v] == color[u]) {
                        return false;
                    }
                }
            }
        }
    }
    return true;
}

int main() {
    int V = 3;
    vector<vector<int>> adj = {
        {1, 2},
        {0, 2},
        {0, 1}
    };
    cout << (isBipartite(V, adj) ? "Yes" : "No") << endl;
    return 0;
}

A Runtime error

B "Yes"

C "No"

D Compilation error

💡 Hint

A triangle (3-node cycle) is not bipartite because it has an odd cycle.

The graph is a triangle (3 nodes all connected). Odd cycles cannot be bipartite. The code detects a conflict in coloring and returns false, printing "No".

🧠 Conceptual

advanced

Why BFS is used in Bipartite Check?

Why is BFS (Breadth-First Search) commonly used to check if a graph is bipartite?

A Because BFS always finds the shortest path between nodes.

B Because BFS can detect cycles faster than DFS.

C Because BFS uses less memory than DFS.

D Because BFS colors nodes level by level, ensuring adjacent nodes get alternate colors.

💡 Hint

Think about how BFS explores nodes in layers.

BFS explores nodes in layers or levels. Assigning colors alternately per level ensures adjacent nodes have different colors, which is key to bipartite checking.

🔧 Debug

advanced

Identify the Bug in Bipartite Check Code

What is the bug in this bipartite check code snippet?

DSA C++

bool isBipartite(int V, vector<vector<int>>& adj) {
    vector<int> color(V, -1);
    for (int i = 0; i < V; i++) {
        if (color[i] == -1) {
            queue<int> q;
            q.push(i);
            color[i] = 0;
            while (!q.empty()) {
                int u = q.front(); q.pop();
                for (int v : adj[u]) {
                    if (color[v] == -1) {
                        color[v] = color[u];
                        q.push(v);
                    } else if (color[v] == color[u]) {
                        return false;
                    }
                }
            }
        }
    }
    return true;
}

A color[v] should be assigned 1 - color[u] instead of color[u]

B Queue should be replaced with stack for correct traversal

C The initial color vector should be initialized with 0 instead of -1

D The function should return true immediately after coloring the first node

💡 Hint

Check how colors are assigned to neighbors.

Assigning color[v] = color[u] means neighbors get the same color, which breaks bipartite property. It should be color[v] = 1 - color[u] to alternate colors.

🚀 Application

expert

Number of Connected Components in Bipartite Graph

Given a graph with 6 nodes and edges: (0-1), (1-2), (3-4), (4-5), (5-3), how many connected components are bipartite?

A 3

B 1

C 2

D 0

💡 Hint

Check bipartite property for each connected component separately.

There are two connected components: one with nodes {0,1,2} which is a chain (bipartite), and one with nodes {3,4,5} forming a triangle (not bipartite). So 1 component is bipartite, 1 is not. Total bipartite components = 1.

Practice

1. What does it mean if a graph is bipartite?

easy

A. It contains at least one cycle of odd length

B. It has an equal number of vertices and edges

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

D. All vertices have the same degree

Solution

1. Step 1: Understand bipartite graph definition

   A bipartite graph means vertices can be split into two sets where no edge connects vertices within the same 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 C

4. Quick Check:

   Bipartite = 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 graphs must have cycles
• Assuming all vertices have same degree

2. Which of the following C++ code snippets correctly initializes a graph as an adjacency list for bipartite check?

easy

A. vector graph(n);

B. vector<int> graph(n);

C. int graph[n][n];

D. map graph;

Solution

1. Step 1: Identify adjacency list structure

   An adjacency list for a graph with n vertices is best represented as vector of vectors: vector graph(n);

2. Step 2: Check other options

   vector<int> graph(n); is a vector of ints, not suitable for adjacency lists. int graph[n][n]; is a 2D array but less flexible. map graph; is a map of int to int, not adjacency list.

3. Final Answer:

   vector graph(n); -> Option A

4. Quick Check:

   Adjacency list = vector of vectors [OK]

Hint: Use vector of vectors for adjacency list in C++ [OK]

Common Mistakes:

• Using vector<int> instead of vector
• Using fixed size arrays limiting flexibility
• Using map which is not adjacency list

3. What is the output of the following code snippet that checks if a graph is bipartite?

int n = 4;
vector graph = {{1,3}, {0,2}, {1,3}, {0,2}};
vector<int> color(n, -1);

function dfs = [&](int node, int c) {
    color[node] = c;
    for (int neigh : graph[node]) {
        if (color[neigh] == -1) {
            if (!dfs(neigh, 1 - c)) return false;
        } else if (color[neigh] == c) {
            return false;
        }
    }
    return true;
};

bool result = true;
for (int i = 0; i < n; i++) {
    if (color[i] == -1) {
        if (!dfs(i, 0)) {
            result = false;
            break;
        }
    }
}
cout << (result ? "YES" : "NO") << endl;

medium

A. Runtime Error

B. NO

C. Compilation Error

D. YES

Solution

1. Step 1: Analyze graph structure

   The graph edges connect vertices 0-1, 0-3, 1-2, 2-3 forming a square (even cycle), which is bipartite.

2. Step 2: Trace DFS coloring

   DFS colors nodes alternatingly. No adjacent nodes share the same color, so dfs returns true for all nodes.

3. Final Answer:

   YES -> Option D

4. Quick Check:

   Even cycle graph is bipartite = YES [OK]

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

Common Mistakes:

• Assuming all cycles are non-bipartite
• Confusing color assignment logic
• Ignoring disconnected components

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

bool dfs(int node, int c, vector& graph, vector<int>& color) {
    color[node] = c;
    for (int neigh : graph[node]) {
        if (color[neigh] == -1) {
            dfs(neigh, 1 - c, graph, color);
        } else if (color[neigh] == c) {
            return false;
        }
    }
    return true;
}

medium

A. Missing return value check for recursive dfs call

B. Wrong color assignment to nodes

C. Using vector<int> instead of vector<bool>

D. Incorrect graph representation

Solution

1. Step 1: Check recursive call handling

   The recursive dfs call result is not checked. If it returns false, the function should also return false immediately.

2. Step 2: Explain fix

   Adding 'if (!dfs(...)) return false;' ensures early exit on conflict detection.

3. Final Answer:

   Missing return value check for recursive dfs call -> Option A

4. Quick Check:

   Always check recursive call results in DFS [OK]

Hint: Return false immediately if recursive DFS fails [OK]

Common Mistakes:

• Ignoring recursive call return values
• Assigning wrong colors
• Using wrong data structures for graph

5. Given a graph with 5 vertices and edges: (0-1), (1-2), (2-3), (3-4), (4-1), which statement is true about its bipartite property?

hard

A. The graph is bipartite because it has no odd cycles

B. The graph is not bipartite because it contains an odd cycle

C. The graph is bipartite because all vertices have degree 2

D. The graph is not bipartite because it is disconnected

Solution

1. Step 1: Identify cycles in the graph

   The edges form a cycle: 1-2-3-4-1 which has 4 edges (even length). However, the full cycle including vertices 0-1-4-3-2-1 forms a cycle of length 5 (odd cycle): 0-1-4-3-2-1 (0-1,1-4,4-3,3-2,2-1).

2. Step 2: Check bipartite condition

   Since the graph contains an odd cycle of length 5, it is not bipartite.

3. Step 3: Re-examine options

   The graph is bipartite because it has no odd cycles is false. The graph is not bipartite because it contains an odd cycle is true. The graph is bipartite because all vertices have degree 2 is incorrect reasoning. The graph is not bipartite because it is disconnected is false because the graph is connected.

4. Final Answer:

   The graph is not bipartite because it contains an odd cycle -> Option B

5. Quick Check:

   Odd cycle means not bipartite [OK]

Hint: Check cycle lengths: odd cycle means not bipartite [OK]

Common Mistakes:

• Miscounting cycle length
• Confusing connectivity with bipartite property
• Assuming degree affects bipartite