Bipartite Graph Check in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the color array with -1 for all vertices.

DSA Typescript

const color: number[] = new Array(graph.length).fill([1]);

Drag options to blanks, or click blank then click option'

A-1

B0

C1

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 or 1 which are valid colors instead of -1.

Using null which is not a number.

We use -1 to mark all vertices as uncolored initially.

2fill in blank

medium

Complete the code to assign the first color (0) to the starting vertex.

DSA Typescript

color[start] = [1];

Drag options to blanks, or click blank then click option'

A1

B0

C-1

D2

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning -1 which means uncolored.

Assigning 1 which is the other color but not the start.

We start coloring the first vertex with color 0.

3fill in blank

hard

Fix the error in the condition to check if the neighbor has the same color as the current vertex.

DSA Typescript

if (color[neighbor] === color[[1]]) { return false; }

Drag options to blanks, or click blank then click option'

Acurrent

Bneighbor

Cstart

Dgraph

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing neighbor color with itself.

Using an unrelated variable like 'start' or 'graph'.

We compare the neighbor's color with the current vertex's color to detect conflict.

4fill in blank

hard

Fill both blanks to assign the opposite color to the neighbor and add it to the queue.

DSA Typescript

color[neighbor] = 1 - color[[1]];
queue.push([2]);

Drag options to blanks, or click blank then click option'

Acurrent

Bneighbor

Cstart

Dgraph

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'neighbor' instead of 'current' for color assignment.

Pushing 'current' instead of 'neighbor' to the queue.

We assign the opposite color of the current vertex to the neighbor and add the neighbor to the queue for BFS.

5fill in blank

hard

Fill all three blanks to complete the BFS loop for checking bipartite graph.

DSA Typescript

while (queue.length > 0) {
  const [1] = queue.shift()!;
  for (const [2] of graph[[3]]) {
    if (color[neighbor] === -1) {
      color[neighbor] = 1 - color[current];
      queue.push(neighbor);
    } else if (color[neighbor] === color[current]) {
      return false;
    }
  }
}

Drag options to blanks, or click blank then click option'

Acurrent

Bneighbor

Dstart

Attempts:

3 left

💡 Hint

Common Mistakes

Using inconsistent variable names causing confusion.

Using 'start' instead of 'current' in the loop.

We dequeue 'current', iterate over its neighbors, and check colors to verify bipartite property.

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