Bipartite Graph Check in DSA Typescript - 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.

function isBipartite(graph: number[][]): boolean {
  const n = graph.length;
  const colors = new Array(n).fill(-1);

  function dfs(node: number, color: number): boolean {
    if (colors[node] !== -1) return colors[node] === color;
    colors[node] = color;
    for (const neighbor of graph[node]) {
      if (!dfs(neighbor, 1 - color)) return false;
    }
    return true;
  }

  for (let i = 0; i < n; i++) {
    if (colors[i] === -1 && !dfs(i, 0)) return false;
  }
  return true;
}
    

This code checks if the graph is bipartite by coloring nodes using depth-first search.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The depth-first search (DFS) visits each node and its neighbors once.
• How many times: Each node and edge is checked once during the DFS calls.

How Execution Grows With Input

As the graph grows, the time to check 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 total nodes plus edges, so bigger graphs take proportionally more time.

Final Time Complexity

Time Complexity: O(n + e)

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

Common Mistake

[X] Wrong: "The DFS runs in O(n²) because it looks at all pairs of nodes."

[OK] Correct: The DFS only visits neighbors of each node, so it depends on actual edges, not all pairs.

Interview Connect

Understanding this time complexity helps you explain how graph traversal scales and shows you can analyze recursive algorithms clearly.

Self-Check

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

Practice

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

easy

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

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 can be split into two sets where no edges connect 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.

3. Final Answer:

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

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 means no cycles
• Assuming all vertices have same degree

2. Which TypeScript data structure is best to represent a graph for bipartite check?

easy

A. Adjacency list using Map

B. Array of numbers

C. String array

D. Single number variable

Solution

1. Step 1: Identify graph representation needs

   We need to store edges for each vertex efficiently to check bipartite property.

2. Step 2: Choose suitable data structure

   An adjacency list using Map stores neighbors for each vertex clearly and is efficient.

3. Final Answer:

   Adjacency list using Map<number, number[]> -> Option A

4. Quick Check:

   Graph as adjacency list = Map [OK]

Hint: Use adjacency list (Map) for graph edges [OK]

Common Mistakes:

• Using simple arrays without edge info
• Using strings which don't represent graph
• Using single variable instead of structure

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

const graph = new Map([
  [0, [1, 3]],
  [1, [0, 2]],
  [2, [1, 3]],
  [3, [0, 2]]
]);

function isBipartite(graph: Map): boolean {
  const color = new Map();
  for (const node of graph.keys()) {
    if (!color.has(node)) {
      const queue = [node];
      color.set(node, 0);
      while (queue.length) {
        const u = queue.shift()!;
        for (const v of graph.get(u)!) {
          if (!color.has(v)) {
            color.set(v, 1 - color.get(u)!);
            queue.push(v);
          } else if (color.get(v) === color.get(u)) {
            return false;
          }
        }
      }
    }
  }
  return true;
}

console.log(isBipartite(graph));

medium

A. false

B. true

C. undefined

D. SyntaxError

Solution

1. Step 1: Analyze graph structure

   The graph is a square cycle: 0-1-2-3-0, which is an even cycle and bipartite.

2. Step 2: Trace BFS coloring

   Starting from node 0 with color 0, neighbors get color 1, their neighbors color 0, no conflicts found.

3. Final Answer:

   true -> Option B

4. Quick Check:

   Even cycle graph is bipartite = true [OK]

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

Common Mistakes:

• Assuming all cycles are not bipartite
• Confusing color assignment logic
• Missing BFS queue processing

4. Find the bug in this TypeScript function that checks if a graph is bipartite:

function isBipartite(graph: Map): boolean {
  const color = new Map();
  for (const node of graph.keys()) {
    if (!color.has(node)) {
      const queue = [node];
      color.set(node, 0);
      while (queue.length) {
        const u = queue.shift()!;
        for (const v of graph.get(u)!) {
          if (!color.has(v)) {
            color.set(v, color.get(u)!);
            queue.push(v);
          } else if (color.get(v) === color.get(u)) {
            return false;
          }
        }
      }
    }
  }
  return true;
}

medium

A. Missing return type boolean

B. queue.shift()! should be queue.pop()

C. color.set(v, color.get(u)!) should be color.set(v, 1 - color.get(u)!)

D. graph.get(u)! should be graph.get(v)!

Solution

1. Step 1: Identify color assignment error

   The code assigns the same color to neighbors, which breaks bipartite logic.

2. Step 2: Correct color assignment

   Neighbors must get opposite color: 1 - color of current node.

3. Final Answer:

   color.set(v, 1 - color.get(u)!) -> Option C

4. Quick Check:

   Neighbor color = opposite of current [OK]

Hint: Neighbors must have opposite colors [OK]

Common Mistakes:

• Assigning same color to neighbors
• Using pop instead of shift in BFS
• Incorrect graph access

5. Given a graph represented as adjacency list:

const graph = new Map([
  [0, [1, 2]],
  [1, [0, 3]],
  [2, [0, 3]],
  [3, [1, 2, 4]],
  [4, [3]]
]);

Is this graph bipartite? Choose the correct reasoning.

hard

A. Yes, because it has no odd-length cycles

B. No, because node 3 connects to nodes in the same set

C. Yes, because all nodes have even degree

D. No, because it contains a triangle (3-cycle)

Solution

1. Step 1: Identify cycles in the graph

   There is a triangle cycle: nodes 1-3-2-1 (edges 1-3, 3-2, 2-1), which is an odd-length cycle.

2. Step 2: Recall bipartite property about cycles

   Graphs containing odd-length cycles are not bipartite.

3. Final Answer:

   No, because it contains a triangle (3-cycle) -> Option D

4. Quick Check:

   Odd cycle present = not bipartite [OK]

Hint: Odd cycles mean graph is not bipartite [OK]

Common Mistakes:

• Falsely identifying triangle in 1-2-3 (actually 1-3-2 forms triangle)
• Assuming even degree means bipartite
• Confusing connectivity with bipartite property