Bipartite Graph Check in DSA Typescript - Build from Scratch

Bipartite Graph Check

📖 Scenario: You are working on a social networking app. You want to check if the users can be split into two groups so that no two friends are in the same group. This helps in organizing events where friends from different groups meet.

🎯 Goal: Build a program that checks if a given graph of users and their friendships is bipartite. A bipartite graph means you can divide users into two groups with no friendships inside the same group.

📋 What You'll Learn

Create a graph using an adjacency list with exact user connections

Add a color array to track group assignments

Implement a function to check if the graph is bipartite using BFS

Print the result as true or false

💡 Why This Matters

🌍 Real World

Checking bipartite graphs helps in scheduling, grouping people, and solving problems where two distinct sets must be formed without internal conflicts.

💼 Career

Understanding bipartite graphs is useful for software engineers working on social networks, recommendation systems, and network analysis.

1

Create the graph adjacency list

Create a variable called graph as an array of arrays representing the adjacency list for 4 users with these exact connections: user 0 connected to 1 and 3, user 1 connected to 0 and 2, user 2 connected to 1 and 3, user 3 connected to 0 and 2.

DSA Typescript

# Your code here

Hint

Use a constant named graph and assign it an array of arrays with the exact connections.

2

Add a color array to track groups

Create a variable called color as an array of length 4 filled with -1 to represent uncolored users.

DSA Typescript

const graph: number[][] = [[1, 3], [0, 2], [1, 3], [0, 2]];
# Your code here

Hint

Use new Array(4).fill(-1) to create the color array.

3

Implement the bipartite check function

Write a function called isBipartite that takes graph and color as parameters. Use a for loop with variable start from 0 to graph.length. Inside, if color[start] is -1, assign color[start] = 0 and use a queue to perform BFS. For each node, check neighbors: if uncolored, assign opposite color; if same color, return false. Return true if no conflicts.

DSA Typescript

const graph: number[][] = [[1, 3], [0, 2], [1, 3], [0, 2]];
const color: number[] = new Array(4).fill(-1);
# Your code here

Hint

Use BFS with a queue and assign colors 0 and 1 to neighbors. Return false if a conflict is found.

4

Print the bipartite check result

Write a console.log statement to print the result of calling isBipartite(graph, color).

DSA Typescript

const graph: number[][] = [[1, 3], [0, 2], [1, 3], [0, 2]];
const color: number[] = new Array(4).fill(-1);

function isBipartite(graph: number[][], color: number[]): boolean {
    for (let start = 0; start  0) {
                const node = queue.shift()!;
                for (const neighbor of graph[node]) {
                    if (color[neighbor] === -1) {
                        color[neighbor] = 1 - color[node];
                        queue.push(neighbor);
                    } else if (color[neighbor] === color[node]) {
                        return false;
                    }
                }
            }
        }
    }
    return true;
}
# Your code here

Hint

Call console.log(isBipartite(graph, color)) to print the result.

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