2493. Divide Nodes Into the Maximum Number of Groups - Detailed Explanation

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Problem Statement

Given an undirected graph with n nodes labeled from 1 to n, and a list of undirected edges, we want to divide all nodes into m groups (numbered 1 through m) such that two conditions hold:

Every node belongs to exactly one group.
If there is an edge between node a and node b, and a is in group x and b is in group y, then the absolute difference between the group numbers must be 1 (i.e., |y - x| = 1). In other words, connected nodes must lie in consecutive groups.

Our task is to determine the maximum number of groups m that we can create under these conditions. If it is impossible to group the nodes to satisfy the edge conditions, we should return -1.

Example 1

Input: n = 6, edges = [[1,2],[1,4],[1,5],[2,6],[2,3],[4,6]]
Output: 4
Explanation: One optimal grouping is:
- Group 1: {5}
- Group 2: {1}
- Group 3: {2, 4}
- Group 4: {3, 6}
  All given edges connect nodes in consecutive groups. For instance, edge [1,5] connects node 1 in group 2 to node 5 in group 1 (difference 1), and edge [2,4] connects nodes in group 3 (difference 0, but note that 2 and 4 are in the same group because they are not directly connected by an edge). It’s not possible to have 5 groups; any attempt to add a fifth group would force some edge to connect nodes with a group difference greater than 1.

Example 2

Input: n = 3, edges = [[1,2],[2,3],[3,1]]
Output: -1
Explanation: The graph forms a triangle. If we try to place nodes 1, 2, 3 in groups 1, 2, 3 respectively (to satisfy edges [1,2] and [2,3]), the edge [3,1] will connect group 3 to group 1, which has a difference of 2 (not allowed). No matter how we assign groups, at least one edge will violate the consecutive-group rule, so no valid grouping exists.

Constraints

1 <= n <= 500 – Number of nodes in the graph.
1 <= edges.length <= 10^4 – Number of edges in the graph.
edges[i] = [a_i, b_i] defines an undirected edge between nodes a_i and b_i.
There are no self-loops (a_i != b_i), and at most one edge between any pair of nodes.
The graph may be disconnected (multiple components).

Hints

Think in terms of graph levels: The condition |y - x| = 1 for any edge [a, b] suggests that if you arrange nodes in sequential groups (like levels or layers), every edge must connect nodes on adjacent levels. Try modeling the problem as a graph and consider performing a breadth-first search (BFS) from some nodes. How do BFS levels relate to group assignments?
Check for impossible conditions: What kind of graph structure would make it impossible to satisfy the group difference rule? (Hint: If the graph contains an odd-length cycle, like a triangle, you will never be able to assign groups because at some point an edge will connect nodes whose groups differ by more than 1. This is related to the concept of a bipartite graph.)
Maximize the number of groups: If grouping is possible, how can we make the number of groups as large as possible? Consider the longest path (in terms of edges) in the graph. Intuitively, the longest shortest path between any two nodes (the graph's diameter) might indicate how many layers you can stretch the grouping. Each additional edge in a path can give another group level. Think about using BFS to determine the maximum distance between nodes in each component of the graph.

Using these hints, you might realize:

First, ensure the graph is bipartite (contains no odd cycle). If not, answer is -1 (no valid grouping possible).
If it is bipartite, then use BFS to layer each connected component. The maximum number of groups for that component will correspond to the number of BFS levels (the longest distance from one node to another within that component, plus one). We can then combine results from each disconnected component.

Approach 1: Brute Force (All Possible Group Assignments)

Idea: A naive brute-force approach would attempt to assign each node to a group (from 1 up to n at most, since we cannot have more groups than nodes) and then check if all edges meet the consecutive group requirement. This essentially means trying all possible labelings of n nodes with up to n group numbers.

However, this approach is impractical because the number of ways to assign n nodes into groups is enormous (even if we limit group count, it's combinatorial). For each node, we could choose a group label in [1, n], leading to up to n^n combinations in the worst case – far too many to try exhaustively when n can be up to 500.

Even if we try to be smarter by limiting groups, we'd have to try 1 group, 2 groups, 3 groups, ..., etc., and check each configuration, which is still exponential in complexity. Clearly, we need to use the graph's structure to reduce the search space.

Why brute force fails: The edge constraints impose a strong structure (essentially, a linear layering of the graph). We can exploit that structure rather than blindly assigning groups. A direct brute force would not finish in any reasonable time for non-trivial n.

Since a direct brute-force generation of assignments is infeasible, we should move to a strategy that uses graph traversal to implicitly assign groups. The next approach leverages BFS to systematically assign group levels.

Complexity: This theoretical brute force is exponential (O(n^n) in the worst formulation) and not tractable for n = 500. Even more clever brute force (like trying all BFS starting positions) would degrade quickly for larger graphs. We need a better approach.

Approach 2: Optimized BFS Layering (Graph + Bipartite + Diameter)

Idea: We can solve this by breaking the problem into two parts:

Feasibility (Bipartite Check): First, check if the graph can be grouped at all. For the condition to hold, the graph must be bipartite. A graph is bipartite if it can be colored using 2 colors such that no edge connects nodes of the same color. If the graph is not bipartite, it contains an odd cycle (like the triangle in Example 2), and grouping is impossible, so return -1. We can check bipartiteness using BFS or DFS by trying to 2-color the graph.
Maximize Groups via BFS Levels: If the graph is bipartite (no odd cycles), we then want to maximize the number of groups. Essentially, we want to "stretch" the grouping as much as possible. This can be done by leveraging the longest shortest path in each connected component (the diameter of each component). We can use BFS to determine the maximum distance between any two nodes in a component:
- For each connected component, pick an arbitrary node and run BFS to find the farthest node from it.
- Then run BFS from that farthest node to find the maximum distance to any other node in the component. This distance (let's call it d) is the diameter (in edges) of that component.
- The maximum number of groups for that component is d + 1 because if the farthest distance is d edges, you can assign each step along that longest path a new group number, resulting in d+1 layers.
- Do this for every component and sum up the number of groups. We sum them (rather than take a max) because disconnected components can be assigned disjoint sets of group numbers. Since there are no edges between components, they don't constrain each other’s groupings. We can start numbering each component's groups where the last one left off to use unique group indices for the entire graph.

Using BFS in this way naturally assigns nodes to levels (group indices) such that every edge connects adjacent levels (because BFS traverses one edge at a time, increasing the distance level by 1). This satisfies the condition |y - x| = 1 for edges by construction. By choosing the longest path, we maximize the number of levels.

Steps:

Build the graph: Use an adjacency list for n nodes.
Check bipartiteness: Create a color array for nodes, initially uncolored (0). For each node not yet colored, perform BFS (or DFS) and color it, say color 1. Color its neighbors with the opposite color (e.g., -1 or a different marker), and so on. If we ever find an edge that connects two nodes of the same color, the graph isn't bipartite – return -1 immediately.
Why bipartite? In a bipartite graph, all cycles are even-length. The consecutive-group constraint is essentially a stricter version of bipartite: it's like a multi-layer bipartite graph. If an odd cycle exists, as shown above, it's impossible to assign even two alternating groups without eventually breaking the consecutive rule.
For each connected component, find the longest distance (layers): We can do this either by:
- BFS from every node in the component to find the longest shortest path (this is simpler to implement but a bit less efficient), or
- BFS twice: once from an arbitrary node to find the farthest node A, then from A to find its farthest B and distance d. This two-pass BFS yields the diameter d of that component.
- During this, also keep track of which nodes belong to the current component (so we don't recount them later).
Sum the results for all components: For each component, the maximum groups = diameter + 1 (number of layers from one end of the graph to the other). Add these up for the final answer.

Let's break down why summing components is valid. Suppose the graph has two disconnected components. We can assign group numbers 1...a for component 1 and then a+1...a+b for component 2, and so on. Since there are no edges between components, there's no restriction tying the group numbers of one component to the other. Therefore, we maximize groups by treating each component independently and then combining the results (ensuring we use distinct group labels for different components). For example, if component1 can have 3 groups (labeled 1,2,3) and component2 can have 2 groups, we could label component2's groups as 4 and 5, yielding a total of 5 groups used.

Complexity Analysis: Checking if the graph is bipartite takes O(n + e) time (essentially a BFS/DFS over all nodes and edges). Finding the longest path via BFS also takes O(n + e) for each component (or O(n + e) overall if done efficiently with two BFS per component, since each edge/node is processed a constant number of times). In the worst case, doing BFS from every node (the simpler method) would be O(n * (n + e)), but given n <= 500, that is at most 500 * (500 + 10000) ≈ 5 million operations, which is acceptable. The optimized two-BFS method per component is closer to O(n + e) overall. Space complexity is O(n + e) for storing the graph and O(n) for BFS queues and color arrays. Given the constraints, these operations are efficient.

In summary:

Time Complexity: O(n + e) on average (treating each edge and node with BFS/DFS a few times). In worst-case analysis, O(n * (n + e)) if one did redundant BFS for each node, but with optimizations it's linear.
Space Complexity: O(n + e) for graph storage and auxiliary arrays.

Step-by-Step Walkthrough of the Optimal Approach

Let's walk through the optimal approach using Example 1 to see how we arrive at 4 groups. The graph from Example 1 has 6 nodes and edges [[1,2],[1,4],[1,5],[2,6],[2,3],[4,6]]. First, we'll check that the graph is bipartite, then find the maximum layering.

Bipartite Check: Start BFS (for coloring) at node 1. Color node 1 as Red (we can call this Group A for the bipartite check). Neighbors of 1 are nodes 2, 4, 5, so color them Blue (Group B). Continue BFS: take node 2 (Blue), its neighbors are 1 (Red, already colored), 6, and 3, so color 6 and 3 Red. Next, node 4 (Blue) has neighbors 1 (Red) and 6 (Red, already colored) – no conflict so far. Next, node 5 (Blue) has neighbor 1 (Red) – fine. Next, node 6 (Red) has neighbors 2 (Blue) and 4 (Blue) – fine. Next, node 3 (Red) has neighbor 2 (Blue) – fine. We successfully colored with two colors and found no edge connecting same colors, so the graph is bipartite (no odd cycle).
Find BFS Layers: Now, to maximize groups, we find the longest shortest path in this component. We can take any node to start; let's pick node 3 (one of the "leaf" nodes in this component). Perform a BFS from node 3 to determine distances (levels):

BFS Level (Distance) Nodes at this level Group Assignment
Level 0 (distance = 0) 3 Group 1
Level 1 (distance = 1) 2 Group 2
Level 2 (distance = 2) 1, 6 Group 3
Level 3 (distance = 3) 4, 5 Group 4
- Start with node 3 in Group 1 (level 0).
- Neighbors of 3: node 2 goes to Group 2.
- Neighbors of 2 (that are not visited yet): nodes 1 and 6 go to Group 3.
- Neighbors of 1 and 6 (not visited yet): node 4 (neighbor of 1 and 6) and node 5 (neighbor of 1) go to Group 4.
- All nodes are now grouped. We have 4 levels (Group 1 through Group 4). Every edge connects nodes from consecutive rows in this table:
  - Edge 1–2 connects Group 3 to Group 2 (|3−2|=1).
  - Edge 1–4 connects Group 3 to Group 4 (|4−3|=1).
  - Edge 1–5 connects Group 3 to Group 4 (|4−3|=1).
  - Edge 2–6 connects Group 2 to Group 3 (|3−2|=1).
  - Edge 2–3 connects Group 2 to Group 1 (|2−1|=1).
  - Edge 4–6 connects Group 4 to Group 3 (|4−3|=1).
- This satisfies all edge constraints and uses 4 groups, which matches the output.
Could we use 5 groups? To have 5 groups, one of the above levels would have to be split further (creating an empty level or moving a node to a new level). As the problem explanation notes, if we try to create a fifth group and move any node into it, some edge will end up connecting groups with a gap . For example, if we tried to put node 5 into Group 1 (and shift others accordingly to stretch to Group 5), the edge 1–5 would connect Group 2 to Group 1, but we already had that; the tricky part is some other edge like 2–3 or 4–6 would break. Thus, 4 is maximal for this component.
Disconnected Components: In this example, the graph was one connected component. If there were multiple components, we would repeat the above for each component to get its group count, then add them (offsetting group numbers for different components). For instance, if another independent component could form 3 groups, we would label those groups 5, 6, 7 (continuing from 4) and add 3 to the total count.

BFS Level (Distance)	Nodes at this level	Group Assignment
Level 0 (distance = 0)	3	Group 1
Level 1 (distance = 1)	2	Group 2
Level 2 (distance = 2)	1, 6	Group 3
Level 3 (distance = 3)	4, 5	Group 4

This step-by-step layering via BFS ensures we both satisfy the conditions and achieve the maximum layers. Now, let's implement this in code.

Solution Implementation in Python

Python3

. . . .

Explanation: In the code above, we first build the graph. Then we iterate over each node to ensure we cover all components. When we find an unvisited node, we perform a BFS (queue) to color its component in two colors (1 and -1). We track all nodes in this component in component_nodes. If a conflict is found (an edge connecting two nodes of the same color), we immediately return -1.

If the component is bipartite, we then perform two BFS traversals to compute the diameter:

The first BFS finds an arbitrary farthest node from the starting node.
The second BFS finds the farthest distance from that farthest node. This distance max_dist is the number of edges in the longest path of the component. We add max_dist + 1 to our answer (because if the longest path has, say, 3 edges, that path covers 4 nodes which can be placed in 4 consecutive groups).

Finally, after checking all components, max_groups_total holds the maximum number of groups we can achieve, which we return.

Solution Implementation in Java

Java

. . . .

Explanation: The Java implementation mirrors the Python logic. We use an adjacency list (List<Integer>[]) to represent the graph. We maintain a color array for bipartite checking. We iterate through nodes, and for each unvisited node, do a BFS to color the component. If a conflict is found, we return -1. Otherwise, we perform two BFS traversals to find the longest path in that component and accumulate the group count. The main method demonstrates the function with the given examples, which should output 4 and -1 respectively.

Complexity Analysis of Approaches

Brute Force Approach: As discussed, attempting all possible group assignments is infeasible. Even a semi-bruteforce that tries to incrementally assign groups without using graph logic would have exponential complexity. For instance, if we tried to assign groups using backtracking, the worst-case time complexity would be O( $m^n$ ) where $m$ is the number of groups you try and $n$ is number of nodes (or something on the order of factorials/combinatorics if reducing symmetries). This is not tractable for $n=500$ . Thus, a pure brute force approach is not used in practice.
Optimized BFS Approach: The BFS-based solution runs in polynomial time. Checking bipartiteness is O( $n+e$ ) (linear in the size of the graph). BFS to find longest paths is also O( $n+e$ ) for each component. In the worst case of a single component, it's O( $n+e$ ) for two BFS passes, which is still O( $n+e$ ). If we did BFS from every node (less optimal approach), that would be O( $n*(n+e)$ ), but with $n=500$ and $e$ up to 10000, this is around 5 million operations, which is fine. So either way, it's efficient. The space complexity is O( $n+e$ ) for adjacency lists and O( $n$ ) for BFS queues and color/dist arrays. This is quite manageable for the given limits.

In Big-O terms:

Time Complexity: Approximately O(n + e) (linear) for the optimized solution. For $n=500$ and $e$ up to $10^4$ , this is very fast.
Space Complexity: O(n + e) for graph representation and additional arrays.

Common Mistakes and Pitfalls

Not recognizing the bipartite condition: A common mistake is to not realize that an odd cycle makes the problem impossible. Trying to assign groups in a graph with an odd cycle without explicitly checking for it can lead to incorrect results or infinite loops. Always check for bipartiteness first. If the graph isn't bipartite, return -1 immediately.
Confusing "difference exactly 1" with "difference at most 1": The problem specifically requires the difference to be exactly 1, not just ≤1. This means nodes connected by an edge cannot be in the same group. If you misinterpret it as "at most 1", you might think putting all nodes in one group is fine (or grouping connected nodes together), which is wrong. They must be in adjacent distinct groups, essentially forming a chain-like layering.
Assuming only 2 groups (just bipartite coloring): While a bipartite graph can always be colored in 2 groups, the problem asks for the maximum number of groups. For example, a simple path of 3 edges (4 nodes) is bipartite but can actually be divided into 4 groups (each node in its own group, forming a chain). A mistake is to stop at 2 just because the graph is bipartite. You need to extend the coloring into multiple layers to maximize the count.
Not handling disconnected components: If you compute the answer for only one connected component (like just take the diameter of the whole graph assuming it's connected), you might miss that you can reuse group numbers for different components to increase the total count. Each component can contribute its own layers to the total as long as you assign their group numbers disjointly. A common bug is to take the maximum diameter among components instead of summing them, which undercounts the answer.
Off-by-one errors in counting groups: Remember that if the longest path has d edges, it spans d+1 nodes, which corresponds to d+1 groups in an optimal assignment. Be careful to add 1 when converting a distance (edge count) into number of groups (node count along that path).
Mixing 0-indexing and 1-indexing: The nodes are labeled 1 to n. If you use 0-based indexing in arrays (common in programming languages), be careful to allocate size n+1 or adjust indices appropriately. Many errors come from off-by-one issues with indexing the nodes and arrays.
Using DFS for diameter without care: An alternative to BFS is DFS to find longest path, but since this is an arbitrary graph (not necessarily a tree), DFS needs to handle cycles and visited tracking. BFS is simpler for shortest paths in an unweighted graph. If one uses DFS to find the longest path in a tree, that's fine, but in a general graph BFS for diameter (longest shortest path) is more appropriate. Some might confuse longest simple path (NP-hard in general graphs) with longest shortest path (diameter). We want the latter here because we require edges to be satisfied (which effectively means we're dealing with shortest paths given the constraint structure).

Edge Cases to Consider

No edges (empty graph): If there are no edges, you can put every node in its own group. The difference condition is trivially satisfied (there are no edges to check), and the maximum number of groups is just n (each node isolated in a distinct group). For example, n=4 and edges=[] would yield output 4 (groups 1,2,3,4 each containing one of the nodes). Our algorithm handles this: the graph is bipartite (trivially, since no edges), and each component is just a single node (diameter 0, so 1 group per node, summing to 4).
Single node: If n=1, obviously the answer is 1 (one node by itself in group 1). No edges to worry about.
Graph that is already a path or a tree: For a linear chain of nodes or any tree, the graph is bipartite and the maximum groups equals the number of nodes (you can alternate groups along the path). For example, a path 1–2–3–4 (3 edges) can be grouped as 1–2–3–4 (each in its own group) which is 4 groups. A star graph (one central node connected to many leaves) is also bipartite; the diameter is 2 (leaf -> center -> another leaf), so the maximum groups is 3. Our BFS approach will correctly find the diameter in these cases.
Even cycle vs. odd cycle: An even cycle (like a square 4-cycle) is bipartite and can be grouped. For instance, 4 nodes in a cycle can actually be arranged in 3 groups (one possible grouping: 1->Group1, 2->Group2, 3->Group1, 4->Group2 – that only uses 2 groups though, but if you want more, you can do 1->G1, 2->G2, 3->G3, 4->G2 to use 3 groups). Our algorithm would find the diameter of the cycle (which is 2 edges) and yield 3 groups. An odd cycle will trigger the not bipartite case and return -1.
Multiple components: Already discussed – handle each separately and sum. For example, if you have two components each of which is just a single edge (two nodes connected), each component can have 2 groups (an edge is just 2 nodes in groups 1 and 2). Two such components would result in 2 + 2 = 4 groups total (one component might occupy Group1-2, the other Group3-4, for instance).
Dense graph that is bipartite: e.g., a complete bipartite graph (like two sets of nodes each connected to all nodes in the other set). Such a graph is bipartite. The diameter in a complete bipartite graph K_{a,b} is 2 (any node to any other on the same side goes via one node on the other side), so maximum groups would be 3. Even if the graph has 100 nodes on each side, you still only get 3 groups max (one side might be group1, the other group2, but to maximize you could offset one node to group3 perhaps, but then an edge from that node back to group1 would break because that's distance 2 difference, so indeed 3 is max). Our algorithm would handle this since BFS from one side to another yields distance 2.

In all these cases, the BFS approach adapts nicely.

Alternative Variations

Minimum number of groups: A different but related problem could be to ask for the minimum number of groups needed to satisfy the condition (instead of maximum). For the given constraint |y-x|=1, the minimum number of groups is actually 2 for any connected bipartite graph (because bipartite graphs can always be 2-colored). For graphs that are not bipartite, the answer would be impossible for this exact condition (since you at least need to satisfy exactly 1 difference). If the condition were "≤1" instead of exactly 1, then the minimum groups would be 1 (you could put everything in one group, since edges with difference 0 would be allowed). So, a variation in the problem condition changes the nature of the solution significantly.
Different allowed differences: Imagine a variation where edges must connect nodes with group difference at most 1 (instead of exactly 1). Then any bipartite grouping would work and you could potentially have fewer groups because edges could even connect within the same group. The problem becomes trivial in that case (you could always just use 1 group). If it was "|y-x| <= k" for some small $k$ , the grouping becomes a graph coloring problem where you have $k+1$ colors (groups) and edges impose a constraint on color difference. That would be a more general graph coloring or multi-coloring problem.
Exact distance layering in weighted graphs: A more complex variant could involve weighted graphs and requiring edges to connect nodes whose group numbers differ exactly by the weight or something along those lines. Then grouping might relate to shortest path distances equaling certain values.
Longest path in a tree vs. graph: If we drop the requirement of consecutive layering, finding the longest path in a general graph is an NP-hard problem (the longest simple path). But here we are effectively finding the longest shortest path (diameter), which is tractable via BFS in an unweighted graph, because of the consecutive constraint forcing us into a shortest path paradigm.
Graph diameter problems: The second part of the solution essentially finds the diameter of each component. A related variation is just asking for the diameter of a graph or tree. In a tree, it's classic to find the longest path with two DFS/BFS. In a general graph, diameter calculation can be expensive in worst cases, but since $n=500$ here it's fine.

While these variations differ, many use similar concepts: BFS, graph coloring, and diameter calculations.

To strengthen your understanding, you can try these related problems:

Is Graph Bipartite? (LeetCode 785) – Determine if a graph can be colored with 2 colors such that no edge connects same-colored nodes . This is essentially the first step of our solution. Practice with this will help you master bipartite checking.
Possible Bipartition (LeetCode 886) – Another problem about dividing a graph into 2 groups given dislike edges. It is a direct application of bipartite check (very similar in nature).
Binary Tree Level Order Traversal (LeetCode 102) – Although about trees, this problem uses BFS to traverse level by level. It’s a good exercise in BFS layering. In our problem, we applied BFS layering to a general graph.
Tree Diameter (e.g., LeetCode 1245 "Tree Diameter") – Find the longest path in a tree. Our approach of two BFS/DFS is the typical solution for tree diameter. Practicing it will make the diameter-finding step second nature.
Longest Path in a Graph – (This might be a more advanced topic, as longest path in arbitrary graphs is hard, but for trees or DAGs it's doable.) If you assume the graph is a tree or acyclic, finding the longest path becomes easier. This can tie into the concept of dynamic programming on trees or DAGs.