Bipartite Graph Check in DSA C++ - Time & Space Complexity

Time Complexity: Bipartite Graph Check

O(V + E)

Understanding Time Complexity

We want to know how long it takes to check if a graph can be split into two groups without edges inside the same group.

How does the time needed grow as the graph gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

bool isBipartite(int V, 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 (auto 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;
}
    

This code checks if a graph is bipartite by coloring nodes using breadth-first search.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Visiting each vertex and its edges once during BFS.
• How many times: Each vertex is visited once, and each edge is checked once (in an undirected graph, edges appear twice in adjacency lists).

How Execution Grows With Input

As the number of vertices and edges grow, the time grows roughly in proportion to their sum.

Input Size (V, E)	Approx. Operations
10 vertices, 15 edges	About 25 checks (10 vertices + 15 edges)
100 vertices, 200 edges	About 300 checks (100 + 200)
1000 vertices, 5000 edges	About 6000 checks (1000 + 5000)

Pattern observation: The work grows linearly with the number of vertices plus edges.

Final Time Complexity

Time Complexity: O(V + E)

This means the time to check grows in a straight line with the total number of vertices and edges.

Common Mistake

[X] Wrong: "The check takes time proportional to the square of vertices because of nested loops."

[OK] Correct: The code visits each edge and vertex only once, not repeatedly for every vertex, so it grows linearly, not quadratically.

Interview Connect

Understanding this time complexity helps you explain how graph traversal algorithms scale, a key skill in many coding challenges and real-world problems.

Self-Check

"What if the graph was represented using an adjacency matrix instead of adjacency lists? How would the time complexity change?"

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