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

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

1. 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?

2. 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.)

3. 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:

1. 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.
2. 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:

1. Build the graph: Use an adjacency list for n nodes.

2. 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.

3. 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).

4. 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.

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.