Bipartite Graph Check in DSA Python - Time & Space Complexity

Time Complexity: Bipartite Graph Check

O(n + m)

Understanding Time Complexity

Checking if a graph is bipartite means verifying if we can split nodes into two groups without edges inside the same group.

We want to know how the time needed grows as the graph gets bigger.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

from collections import deque

def is_bipartite(graph):
    color = {}
    for node in graph:
        if node not in color:
            queue = deque([node])
            color[node] = 0
            while queue:
                current = queue.popleft()
                for neighbor in graph[current]:
                    if neighbor not in color:
                        color[neighbor] = 1 - color[current]
                        queue.append(neighbor)
                    elif color[neighbor] == color[current]:
                        return False
    return True
    

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

Identify Repeating Operations

• Primary operation: Visiting each node and its neighbors once using BFS.
• How many times: Each node and edge is processed at most once.

How Execution Grows With Input

As the graph grows, the time to check grows roughly with the number of nodes plus edges.

Input Size (n nodes, m edges)	Approx. Operations
10 nodes, 15 edges	About 25 operations
100 nodes, 200 edges	About 300 operations
1000 nodes, 5000 edges	About 6000 operations

Pattern observation: Operations grow roughly in a straight line with total nodes plus edges.

Final Time Complexity

Time Complexity: O(n + m)

This means the time grows linearly with the number of nodes and edges in the graph.

Common Mistake

[X] Wrong: "The algorithm checks every possible pair of nodes, so it must be O(n²)."

[OK] Correct: The algorithm only visits each edge once, not all pairs, so it runs faster than checking all pairs.

Interview Connect

Understanding this time complexity helps you explain how graph algorithms scale and shows you can analyze BFS-based solutions clearly.

Self-Check

"What if the graph was represented as 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 has an equal number of vertices and edges

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

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 means its vertices can be split into two sets where no edge connects vertices within the same set.

2. Step 2: Compare options with definition

   Only Its vertices can be divided into two groups with no edges inside the same group matches this definition exactly. Other options describe different graph properties.

3. Final Answer:

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

4. Quick Check:

   Bipartite = Two groups, no internal edges [OK]

Hint: Bipartite means two groups with edges only between groups [OK]

Common Mistakes:

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

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

easy

A. graph = {0: 1, 1: 2, 2: 3, 3: 0}

B. graph = [[0, 1], [1, 2], [2, 3], [3, 0]]

C. graph = {0: [1, 3], 1: [0, 2], 2: [1, 3], 3: [0, 2]}

D. graph = [0, 1, 2, 3]

Solution

1. Step 1: Identify adjacency list format

   An adjacency list uses a dictionary mapping each vertex to a list of connected vertices.

2. Step 2: Check each option

   graph = {0: [1, 3], 1: [0, 2], 2: [1, 3], 3: [0, 2]} uses a dictionary with lists, correctly representing adjacency. graph = [[0, 1], [1, 2], [2, 3], [3, 0]] is a list of edges, not adjacency list. graph = {0: 1, 1: 2, 2: 3, 3: 0} maps to single vertex, not list. graph = [0, 1, 2, 3] is just a list of vertices.

3. Final Answer:

   graph = {0: [1, 3], 1: [0, 2], 2: [1, 3], 3: [0, 2]} -> Option C

4. Quick Check:

   Adjacency list = dict with lists [OK]

Hint: Adjacency list uses dict with vertex keys and list of neighbors [OK]

Common Mistakes:

• Using list of edges instead of adjacency list
• Mapping vertex to single vertex instead of list
• Using just a list of vertices without edges

3. What is the output of this Python code that checks if the graph is bipartite?

def is_bipartite(graph):
    color = {}
    def dfs(node, c):
        if node in color:
            return color[node] == c
        color[node] = c
        return all(dfs(nei, 1 - c) for nei in graph[node])
    return all(dfs(node, 0) for node in graph if node not in color)

graph = {0: [1], 1: [0, 2], 2: [1, 3], 3: [2]}
print(is_bipartite(graph))

medium

A. True

B. False

C. None

D. Error

Solution

1. Step 1: Understand DFS coloring logic

   The function colors nodes 0 or 1 alternately. If a conflict occurs, it returns False.

2. Step 2: Trace the graph coloring

   Start at node 0 with color 0, node 1 gets color 1, node 2 color 0, node 3 color 1. No conflicts found.

3. Final Answer:

   True -> Option A

4. Quick Check:

   Graph is bipartite, so output is True [OK]

Hint: DFS colors nodes alternately; no conflicts means bipartite [OK]

Common Mistakes:

• Assuming graph with chain edges is not bipartite
• Confusing color assignment logic
• Ignoring disconnected nodes

4. Find the error in this bipartite check code:

def is_bipartite(graph):
    color = {}
    def dfs(node, c):
        if node in color:
            return color[node] != c
        color[node] = c
        return all(dfs(nei, 1 - c) for nei in graph[node])
    return all(dfs(node, 0) for node in graph if node not in color)

medium

A. The graph parameter should be a list, not dict

B. The color dictionary should be a list

C. The dfs function should return True unconditionally

D. The condition should be color[node] == c, not != c

Solution

1. Step 1: Analyze the color check condition

   The code returns False if a node is already colored with the same color c, so it should check equality to detect conflict.

2. Step 2: Identify the incorrect operator

   The code uses != which returns True if colors differ, which is wrong. It should be == to detect conflict and return False.

3. Final Answer:

   The condition should be color[node] == c, not != c -> Option D

4. Quick Check:

   Conflict check needs equality, not inequality [OK]

Hint: Conflict if same color found, so use == not != [OK]

Common Mistakes:

• Using != instead of == in color conflict check
• Changing color dict to list unnecessarily
• Ignoring return values in dfs

5. Given a graph with edges [(0,1), (1,2), (2,3), (3,0), (2,4)], 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 it has an even cycle

D. The graph is not bipartite because it is disconnected

Solution

1. Step 1: Identify cycles in the graph

   The edges form a cycle 0-1-2-3-0 which has 4 edges (even), so no odd cycle exists.

2. Step 2: Check for odd cycles

   Since the only cycle is even length, the graph is bipartite.

3. Final Answer:

   The graph is bipartite because it has no odd cycles -> Option A

4. Quick Check:

   Odd cycle means not bipartite; no odd cycle means bipartite [OK]

Hint: Odd cycle means not bipartite; even cycle is allowed [OK]

Common Mistakes:

• Confusing even and odd cycle lengths
• Assuming disconnected means not bipartite
• Ignoring extra edges when checking cycles