Bipartite Graph Check in DSA Java - Time & Space Complexity

Time Complexity: Bipartite Graph Check

O(n + 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.

public boolean isBipartite(int[][] graph) {
    int n = graph.length;
    int[] colors = new int[n];
    for (int i = 0; i < n; i++) {
        if (colors[i] == 0 && !dfs(graph, colors, i, 1)) {
            return false;
        }
    }
    return true;
}

private boolean dfs(int[][] graph, int[] colors, int node, int color) {
    if (colors[node] != 0) {
        return colors[node] == color;
    }
    colors[node] = color;
    for (int neighbor : graph[node]) {
        if (!dfs(graph, colors, neighbor, -color)) {
            return false;
        }
    }
    return true;
}

This code uses depth-first search to color nodes and check if the graph is bipartite.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: DFS visits each node and its neighbors once.
• How many times: Each node and edge is checked once during DFS calls.

How Execution Grows With Input

As the graph grows, the time to check colors grows with the number of nodes and edges.

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

Pattern observation: The work grows roughly with the sum of nodes and edges.

Final Time Complexity

Time Complexity: O(n + e)

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

Common Mistake

[X] Wrong: "The time depends only on the number of nodes, not edges."

[OK] Correct: Checking neighbors means edges matter too, so edges add to the work done.

Interview Connect

Understanding this helps you explain how graph traversal scales and why both nodes and edges matter in real problems.

Self-Check

"What if we used a queue and breadth-first search instead of recursion? How would the time complexity change?"

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