Divide Nodes Into the Maximum Number of Groups
A hard-tier problem at 67% community acceptance, tagged with Depth-First Search, Breadth-First Search, Union Find. Reported in interviews at Accenture and 0 others.
You're given an undirected graph and need to partition its nodes into the maximum number of groups such that each edge connects nodes in adjacent groups. It's a hard problem that looks deceptively simple on first read. Accenture has asked it. The trap is confusing it with graph coloring or assuming a greedy layering works. The real constraint is bipartiteness: if the graph isn't bipartite, it has no valid grouping. If it is, you're doing BFS from every node to find the deepest level assignment, then merging connected components intelligently. Most candidates underestimate the graph-theory prerequisite and the complexity of coordinating BFS across components. If this hits your OA and you blank on the component-level coordination logic, StealthCoder surfaces a working solution in seconds.
Companies that ask "Divide Nodes Into the Maximum Number of Groups"
Divide Nodes Into the Maximum Number of Groups is the kind of problem that decides whether you pass. StealthCoder reads the problem on screen and surfaces a working solution in under 2 seconds. Invisible to screen share. The proctor sees nothing. Made by an Amazon engineer who watched the leaked-problem repo become an industry secret. He decided you should have it too.
Get StealthCoderThe algorithm sits on a hard geometric fact: a valid grouping exists if and only if the graph is bipartite. Check that first with DFS or BFS. If any cycle has odd length, return -1. If bipartite, each connected component can be split into two partitions (left and right). Now the trick: for each component, run BFS from every node and compute the maximum distance to any other node in that component. That distance plus one is the max groups that component can contribute. Sum across all components. The pitfall is skipping bipartiteness verification, or computing distances without respecting partition boundaries. Candidates often try greedy color-counting or ignore that components are independent. The Union Find topic appears in the tags but the actual solution favors BFS-based distance calculation with component iteration. If you're hitting the assessment unprepared on this exact pattern, StealthCoder runs invisibly and handles the bipartite check, BFS traversal, and component merging in real time.
Pattern tags
You know the problem.
Make sure you actually pass it.
Divide Nodes Into the Maximum Number of Groups recycles across companies for a reason. It's hard-tier, and most candidates blank under the timer. StealthCoder is the hedge: an AI overlay invisible during screen share. It reads the problem and surfaces a working solution in under 2 seconds. Made by an Amazon engineer who watched the leaked-problem repo become an industry secret. He decided you should have it too. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Divide Nodes Into the Maximum Number of Groups interview FAQ
Is this really bipartite checking plus BFS, or is there a harder trick?+
It's both. Bipartiteness is the gate: no valid grouping if the graph has odd cycles. Once confirmed bipartite, the core work is BFS distance calculation per component. The trick is realizing each component independently contributes its own max group count, then you sum them. Most candidates miss the component-independence insight.
How does this differ from standard graph coloring?+
Graph coloring assigns each node a single color such that adjacent nodes differ. This problem assigns each node a group number such that edges always span adjacent group numbers. That constraint is much stricter. It forces the graph to be bipartite and requires distance-based numbering, not just binary coloring.
Why does acceptance sit at 67% for a hard problem?+
The problem statement can sound ambiguous on first read, and many submissions fail the bipartite check or compute distances incorrectly across component boundaries. The acceptance rate reflects that implementation details matter significantly more than the core concept.
Is Union Find actually needed, or is it a red herring?+
Union Find can identify connected components, but DFS or BFS does the same job and is simpler to code here. The real work is bipartiteness verification and BFS distance calculation. Union Find appears in topic tags but isn't critical to the solution.
What happens if I just do BFS without checking bipartiteness first?+
Your BFS will either fail to assign valid group numbers or produce an invalid grouping. Checking bipartiteness up front catches odd cycles and lets you return -1 immediately, saving runtime and avoiding logic errors downstream.
Want the actual problem statement? View "Divide Nodes Into the Maximum Number of Groups" on LeetCode →