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 such that no two nodes within 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 BFS/DFS to check if the graph can be colored without conflicts, meaning no two connected nodes share the same color.

intermediate

In the bipartite check algorithm, what does it mean if a conflict in coloring is found?

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

beginner

Why do we use two colors in bipartite graph checking?

Two colors represent the two groups of nodes. Assigning colors helps verify if nodes connected by edges belong to different groups.

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 does it mean if a graph is bipartite?

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 splits nodes into two groups with edges only between groups.

Which algorithm is commonly used to check bipartite property?

A Dijkstra's algorithm

B Floyd-Warshall algorithm

C BFS or DFS with two-coloring

D Prim's algorithm

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

If two connected nodes have the same color during bipartite check, what does it indicate?

A Graph is not bipartite

B Graph is disconnected

C Graph is bipartite

D Graph is complete

Same color on connected nodes means the graph is not bipartite.

How many colors are used in bipartite graph checking?

A Four

B One

C Three

D Two

Two colors represent the two groups in a bipartite graph.

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

A O(E^2)

B O(V + E)

C O(V^2)

D O(log V)

Each vertex and edge is visited once, so 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.

What does it mean for a graph to be bipartite and why is it useful?

Consider the groups and edge connections.

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