Divide Nodes Into the Maximum Number of Groups

Divide Nodes Into the Maximum Number of Groups

What is this problem about?

The Divide Nodes Into the Maximum Number of Groups interview question is a complex graph problem. You are given an undirected graph and you need to partition its nodes into as many groups as possible such that:

1. Every node belongs to exactly one group.
2. For every edge (u, v), the group number of u and v must differ by exactly 1. If such a partition is impossible, return -1.

Why is this asked in interviews?

This is a "Hard" Graph interview pattern asked by companies like Uber and Amazon. It tests several layers of graph theory: Bipartiteness check (the condition |group(u) - group(v)| = 1 implies the graph must be bipartite) and BFS for diameter (to maximize groups, you need to find the longest path within each connected component). It evaluates your ability to break down a difficult problem into sub-problems like BFS, connectivity, and cycle detection.

Algorithmic pattern used

1. Connectivity: Find all connected components of the graph.
2. Bipartiteness: For each component, check if it is bipartite. If not, return -1 immediately.
3. Maximize Groups per Component: The maximum number of groups for a component is its diameter (longest shortest-path) plus 1.
4. To find this, run a BFS from every node in the component and track the maximum depth reached.
5. Result: The total number of groups is the sum of the maximum groups found for each component.

Example explanation

Imagine a line graph: 1 - 2 - 3.

• Start BFS at 1: levels are {1: group 1, 2: group 2, 3: group 3}. Max groups = 3. Imagine a triangle graph: 1 - 2 - 3 - 1.
• If 1 is in group 1, then 2 must be in group 2.
• Since 3 is connected to 2, it must be group 1 or 3.
• Since 3 is connected to 1, it must be group 2.
• Contradiction! (Odd cycle means not bipartite). Result: -1.

Common mistakes candidates make

• Ignoring Bipartiteness: Failing to recognize that the "difference of 1" constraint means the graph cannot have any odd-length cycles.
• Only one BFS: Running BFS from a single arbitrary node per component. This finds a valid grouping, but not necessarily the maximum number of groups. You must check every node as a potential "level 1" node.
• Complexity: Not realizing that $O(N \cdot (V+E))$ is acceptable given typical constraints ($N=500$).

Interview preparation tip

This problem is all about the properties of bipartite graphs. Remember that a graph is bipartite if and only if it contains no odd cycles. If you find an edge between two nodes at the same BFS level, the graph is not bipartite.