Is Graph Bipartite?

Is Graph Bipartite?

1. What is this problem about?

The Is Graph Bipartite? interview question asks you to determine if an undirected graph can be split into two independent sets such that no two nodes in the same set are connected by an edge. In other words, every edge in the graph must connect a node from set A to a node from set B.

2. Why is this asked in interviews?

Companies like Meta, Amazon, and LinkedIn use the Is Graph Bipartite coding problem to evaluate knowledge of Graph Coloring. It tests whether you can use traversal algorithms to detect cycles of odd length, which are the only structures that prevent a graph from being bipartite. It's a foundational Graph interview pattern.

3. Algorithmic pattern used

This problem is solved using Graph Coloring (BFS or DFS).

1. Initialize: Use a colors array where 0 is uncolored, 1 is Set A, and -1 is Set B.
2. Traversal: For each unvisited node, start a BFS/DFS.
3. Coloring Rule: Assign the current node a color. For all its neighbors:
   • If the neighbor is uncolored, give it the opposite color and recurse.
   • If the neighbor is already colored and has the same color as the current node, the graph is not bipartite.
4. Disjoint Components: Ensure you check every node in the graph, as it may contain multiple disconnected pieces.

4. Example explanation

Graph: 0-1, 1-2, 2-3, 3-0 (A square)

1. Color 0: Blue.
2. Neighbors of 0: 1 and 3. Color them Red.
3. Neighbor of 1: 2. Color it Blue.
4. Check 2's neighbor 3: 3 is Red, 2 is Blue. OK. Result: True. If it were a triangle 0-1, 1-2, 2-0:
5. 0: Blue, 1: Red, 2: Blue? No, 2 is connected to 0 (Blue). Result: False.

5. Common mistakes candidates make

• Ignoring Disconnected Graphs: Only running the traversal from node 0 and missing other components.
• Wrong Logic: Forgetting that if a neighbor is already colored, you must check if its color conflicts.
• Path Length: Thinking bipartiteness is about path length instead of cycle parity.

6. Interview preparation tip

Remember: Bipartite = 2-Colorable = No Odd Cycles. If you can't color a graph using only two colors without a conflict, it is not bipartite. This is a vital concept for many scheduling and matching problems.