Bipartite Graph Check in DSA Typescript

Mental Model

A bipartite graph can be split into two groups where no nodes inside the same group connect to each other.

Analogy: Imagine two teams playing a game where players only pass the ball to the other team, never to their own team members.

Group A: 1 -> 3 -> 5
Group B: 2 -> 4 -> 6
Edges only go between groups, never inside one group.

Dry Run Walkthrough

Input: Graph with edges: 1-2, 2-3, 3-4, 4-1

Goal: Check if this graph can be divided into two groups with no edges inside the same group

Step 1: Start coloring node 1 with color 0

Node colors: [1:0, 2:null, 3:null, 4:null]

Why: We pick a color for the first node to start the grouping

Step 2: Color neighbors of 1 (nodes 2 and 4) with color 1

Node colors: [1:0, 2:1, 3:null, 4:1]

Why: Neighbors must have opposite color to avoid same-group edges

Step 3: Color neighbor of 2 (node 3) with color 0

Node colors: [1:0, 2:1, 3:0, 4:1]

Why: Neighbors of node 2 get opposite color to 2

Step 4: Check neighbors of 3 (nodes 2 and 4) for color conflicts

Node colors: [1:0, 2:1, 3:0, 4:1]

Why: Neighbors have opposite colors, no conflict found

Step 5: Check neighbors of 4 (nodes 1 and 3) for color conflicts

Node colors: [1:0, 2:1, 3:0, 4:1]

Why: Neighbors have opposite colors, no conflict found

Result:

Node colors: [1:0, 2:1, 3:0, 4:1] -> Graph is bipartite

Annotated Code

DSA Typescript

class Graph {
  adjacencyList: Map<number, number[]>;

  constructor(edges: [number, number][]) {
    this.adjacencyList = new Map();
    for (const [u, v] of edges) {
      if (!this.adjacencyList.has(u)) this.adjacencyList.set(u, []);
      if (!this.adjacencyList.has(v)) this.adjacencyList.set(v, []);
      this.adjacencyList.get(u)!.push(v);
      this.adjacencyList.get(v)!.push(u);
    }
  }

  isBipartite(): boolean {
    const color = new Map<number, number>();

    for (const node of this.adjacencyList.keys()) {
      if (!color.has(node)) {
        if (!this.dfsCheck(node, color, 0)) return false;
      }
    }
    return true;
  }

  private dfsCheck(node: number, color: Map<number, number>, c: number): boolean {
    color.set(node, c); // assign color to current node

    for (const neighbor of this.adjacencyList.get(node)!) {
      if (!color.has(neighbor)) {
        if (!this.dfsCheck(neighbor, color, 1 - c)) return false; // assign opposite color
      } else if (color.get(neighbor) === c) {
        return false; // neighbor has same color, not bipartite
      }
    }
    return true;
  }
}

// Driver code
const edges: [number, number][] = [[1, 2], [2, 3], [3, 4], [4, 1]];
const graph = new Graph(edges);
console.log(graph.isBipartite() ? "Graph is bipartite" : "Graph is not bipartite");

color.set(node, c); // assign color to current node

assign color to current node to mark its group

for (const neighbor of this.adjacencyList.get(node)!) {

check all neighbors of current node

if (!color.has(neighbor)) {

if neighbor not colored, assign opposite color recursively

if (!this.dfsCheck(neighbor, color, 1 - c)) return false; // assign opposite color

recursive call to color neighbor with opposite color

else if (color.get(neighbor) === c) {

if neighbor has same color, graph is not bipartite

Output

Graph is bipartite

Complexity Analysis

Time: O(V + E) because we visit every node and edge once during DFS

Space: O(V) for storing colors and recursion stack in worst case

vs Alternative: Compared to BFS approach, DFS uses recursion but both have same time and space complexity

Edge Cases

Empty graph

Returns true since no edges mean no conflicts

DSA Typescript

for (const node of this.adjacencyList.keys()) { if (!color.has(node)) { ... } }

Graph with one node and no edges

Returns true as single node can be colored any color

DSA Typescript

color.set(node, c); // assign color to current node

Graph with odd cycle (e.g., triangle 1-2-3-1)

Returns false because odd cycle cannot be bipartite

DSA Typescript

else if (color.get(neighbor) === c) { return false; }

When to Use This Pattern

When a problem asks if nodes can be split into two groups with no internal edges, use bipartite check with graph coloring to detect conflicts.

Common Mistakes

Mistake: Not checking all connected components, missing disconnected nodes
Fix: Loop over all nodes and run DFS/BFS if not colored yet

Mistake: Assigning same color to neighbors instead of opposite
Fix: Assign opposite color using 1 - current color

Summary

Checks if a graph's nodes can be divided into two groups with no edges inside the same group.

Use when you need to separate nodes into two sets without internal connections.

The key is to color nodes alternately and detect any color conflicts.