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Divide Nodes Into the Maximum Number of Groups

Updated on 23 May, 2025
Divide Nodes Into the Maximum Number of Groups header image

Problem Statement

In this task, you're presented with an undirected graph defined by n nodes and multiple edges forming connections between these nodes. Each node is understandably linked with others through bidirectional edges described by the edges array where each element provides a pair of interconnected nodes.

Your challenge is to segment these n nodes into distinct groups, satisfying specific grouping rules:

  • Each node should be present in one and only one group.
  • Every connected node pair via an edge should reside in consecutive groups. For instance, if node ai is in group x, then node bi (connected to ai) must be in group x+1 or x-1.

The goal is to determine the maximum number of possible groups (m) these nodes can be distributed into, given the aforementioned constraints. If no suitable grouping exists that obeys the edge-to-group adjacency condition, the output should be -1.

Examples

Example 1

Input:

n = 6, edges = [[1,2],[1,4],[1,5],[2,6],[2,3],[4,6]]

Output:

4

Explanation:

As shown in the image we:
- Add node 5 to the first group.
- Add node 1 to the second group.
- Add nodes 2 and 4 to the third group.
- Add nodes 3 and 6 to the fourth group.
We can see that every edge is satisfied.
It can be shown that that if we create a fifth group and move any node from the third or fourth group to it, at least on of the edges will not be satisfied.

Example 2

Input:

n = 3, edges = [[1,2],[2,3],[3,1]]

Output:

-1

Explanation:

If we add node 1 to the first group, node 2 to the second group, and node 3 to the third group to satisfy the first two edges, we can see that the third edge will not be satisfied.
It can be shown that no grouping is possible.

Constraints

  • 1 <= n <= 500
  • 1 <= edges.length <= 104
  • edges[i].length == 2
  • 1 <= ai, bi <= n
  • ai != bi
  • There is at most one edge between any pair of vertices.

Approach and Intuition

The problem essentially centres around coloring the nodes of a graph so that adjacent nodes have related (consecutive numeric) colors. This closely parallels the bipartite nature test for a graph but extends into grouping potentially more than two sets.

  1. BFS or DFS for Assignment: Use a Breadth-First Search (BFS) or Depth-First Search (DFS) starting from any unvisited node. Each node you start with should be a cue to explore a new potential group. Label it with a group index (starting group), and attempt to label each connected node with either the previous or next group index.

    • If during your traversal you encounter a node that should change to an already established group index but conflicts with its current group, the arrangement is not possible (return -1).
    • Keep expanding and adjusting group indices as you traverse node connections.

  2. Handling Disconnections: Given the graph might be disconnected, each disconnected subset of the graph needs its own BFS/DFS initiation. This separates the graph into isolated components, each of which can be grouped independently of the others.

  3. Check for the Cycle Condition: If there exists any cycle of odd length in the graph, it's not possible to assign groups such that each connected node pair can strictly have group numbers differing by 1. This stems from the bipartite testing where an odd cycle precludes a two-color (or in this expanded sense, any strict +/-1 cycle grouping) solution.

  4. Maximizing m (number of groups): The maximum number of groups m can be visualized as the maximal distance you would need to appropriately label nodes while satisfying the requirement for all edges. In a certain view, this could also be related to finding the longest path in terms of groups, constrained by the given rules.

This approach essentially breaks down to examining the structure and properties of the graph with a focus on grouping and adjacency, using classical graph traversal methods, modified to check and adjust for the specific conditions stipulated by the problem.

Solutions

  • C++
  • Java
  • Python
cpp 
class Solution {
public:
    int calculateMaximumGroups(int nodesCount, vector<vector<int>> &links) {
        vector<vector<int>> adjacencyList(nodesCount);
        vector<int> leader(nodesCount, -1);
        vector<int> rank(nodesCount, 0);

        // Construct the graph and perform union-find on nodes
        for (auto link : links) {
            adjacencyList[link[0] - 1].push_back(link[1] - 1);
            adjacencyList[link[1] - 1].push_back(link[0] - 1);
            performUnion(link[0] - 1, link[1] - 1, leader, rank);
        }

        unordered_map<int, int> componentMaxGroups;

        // Determine the max number of groups possible per component identified
        for (int vertex = 0; vertex < nodesCount; vertex++) {
            int groupCount = breadthFirstSearchGroupCount(adjacencyList, vertex, nodesCount);
            if (groupCount == -1) return -1;  // In case of an invalid group configuration
            int rootVertex = findRoot(vertex, leader);
            componentMaxGroups[rootVertex] = max(componentMaxGroups[rootVertex], groupCount);
        }

        // Sum up the maximum groups from each connected component
        int totalGroups = 0;
        for (auto [root, count] : componentMaxGroups) {
            totalGroups += count;
        }
        return totalGroups;
    }

private:
    int findRoot(int vertex, vector<int> &leader) {
        while (leader[vertex] != -1) vertex = leader[vertex];
        return vertex;
    }

    void performUnion(int vertexA, int vertexB, vector<int> &leader, vector<int> &rank) {
        vertexA = findRoot(vertexA, leader);
        vertexB = findRoot(vertexB, leader);

        if (vertexA == vertexB) return;

        if (rank[vertexA] < rank[vertexB]) swap(vertexA, vertexB);
        leader[vertexB] = vertexA;

        if (rank[vertexA] == rank[vertexB]) rank[vertexA]++;
    }

    int breadthFirstSearchGroupCount(vector<vector<int>> &adjacencyList, int startVertex, int totalVertices) {
        queue<int> nodeQueue;
        vector<int> visitedLayer(totalVertices, -1);
        nodeQueue.push(startVertex);
        visitedLayer[startVertex] = 0;
        int depth = 0;

        while (!nodeQueue.empty()) {
            int layerSize = nodeQueue.size();
            for (int i = 0; i < layerSize; i++) {
                int current = nodeQueue.front();
                nodeQueue.pop();
                for (int adjacent : adjacencyList[current]) {
                    if (visitedLayer[adjacent] == -1) {
                        visitedLayer[adjacent] = depth + 1;
                        nodeQueue.push(adjacent);
                    } else if (visitedLayer[adjacent] == depth) {
                        return -1;
                    }
                }
            }
            depth++;
        }
        return depth;
    }
};

The C++ solution provided involves dividing nodes into the maximum number of groups based on the given relationships or links between them. This problem is tackled using graph theory, specifically utilizing the union-find algorithm and breadth-first search (BFS).

  • The approach involves several key steps:

    1. Develop an adjacency list to represent the graph from the given nodes and links.
    2. Implement Union-Find to identify and merge connected components of the graph. This involves finding the root of each node and deciding the union based on the rank.
    3. Perform a BFS to determine the maximum number of groups that can be formed within each component of the graph.
    4. Use a map to track the maximum number of groups obtainable for each unique component determined by the Union-Find process.
    5. Sum all values from the map to get the total maximum number of groups across all components.

  • Key functions used in the solution:

    • findRoot: Helps in finding the representative or root of a node using path compression to optimize the union-find operations.
    • performUnion: Merges two nodes into one component based on the union-find logic, utilizing their ranks to keep the tree as flat as possible to optimize further find operations.
    • breadthFirstSearchGroupCount: Executes a BFS starting from each node to calculate the max depth or group number while avoiding cycles which would invalidate a group configuration.

  • The function returns the total number of groups formed or -1 if a valid group configuration cannot be achieved due to cycles within components.

This method efficiently uses graph theory and data structure management to solve the problem of node grouping based on given conditions in a clear and optimized manner.

java 
class GraphAnalyzer {

    // Method to determine the maximum possible number of distinct groups
    public int calculateMaxGroups(int nodesCount, int[][] connections) {
        List<List<Integer>> graph = new ArrayList<>();
        for (int i = 0; i < nodesCount; i++) {
            graph.add(new ArrayList<>());
        }
        int[] leader = new int[nodesCount];
        int[] rank = new int[nodesCount];
        Arrays.fill(leader, -1);

        // Constructing the graph and performing Union operations
        for (int[] connection : connections) {
            graph.get(connection[0] - 1).add(connection[1] - 1);
            graph.get(connection[1] - 1).add(connection[0] - 1);
            mergeSets(connection[0] - 1, connection[1] - 1, leader, rank);
        }

        Map<Integer, Integer> groupCountByRoot = new HashMap<>();

        // Evaluate groups for each graph component
        for (int i = 0; i < nodesCount; i++) {
            int groupCount = evaluateGroupCount(graph, i, nodesCount);
            if (groupCount == -1) return -1; // Infeasible grouping
            int componentRoot = findLeader(i, leader);
            groupCountByRoot.put(
                componentRoot,
                Math.max(
                    groupCountByRoot.getOrDefault(componentRoot, 0),
                    groupCount
                )
            );
        }

        // Sum up all maximum groups possible in each component
        int overallMaxGroups = 0;
        for (int groups : groupCountByRoot.values()) {
            overallMaxGroups += groups;
        }
        return overallMaxGroups;
    }

    // Helper function to assess feasible group counts in a component
    private int evaluateGroupCount(
        List<List<Integer>> graph,
        int startNode,
        int totalNodes
    ) {
        Queue<Integer> queue = new LinkedList<>();
        int[] visitedLevels = new int[totalNodes];
        Arrays.fill(visitedLevels, -1);
        queue.offer(startNode);
        visitedLevels[startNode] = 0;
        int levels = 0;

        // BFS to determine groups/levels
        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                int node = queue.poll();
                for (int neighbor : graph.get(node)) {
                    if (visitedLevels[neighbor] == -1) {
                        visitedLevels[neighbor] = levels + 1;
                        queue.offer(neighbor);
                    } else if (visitedLevels[neighbor] == levels) {
                        return -1;  // Not possible to split as required
                    }
                }
            }
            levels++;
        }
        return levels;
    }

    // Helper to find root leader of a node
    private int findLeader(int node, int[] leader) {
        while (leader[node] != -1) node = leader[node];
        return node;
    }

    // Helper to union two nodes
    private void mergeSets(int node1, int node2, int[] leader, int[] rank) {
        node1 = findLeader(node1, leader);
        node2 = findLeader(node2, leader);

        if (node1 == node2) return;  // Already connected

        // Union by rank
        if (rank[node1] < rank[node2]) {
            int temp = node1;
            node1 = node2;
            node2 = temp;
        }
        leader[node2] = node1;
        if (rank[node1] == rank[node2]) rank[node1]++;
    }
}

The Java program provided addresses the problem of dividing nodes into the maximum number of distinct groups represented by a graph. The solution involves constructing and analyzing a graph using disjoint set union-find and breadth-first search (BFS) algorithms. The main features and steps of the program include:

  • Graph Initialization: Nodes are initialized into an adjacency list format.

  • Union-Find Structure Setup: Each node is originally set as its own leader in the leader array and the rank array initializes to rank all nodes equally for later union operations.

  • Graph Construction and Merge Operation: Connections between nodes are added to the adjacency list, and the nodes in each connection are unioned, effectively grouping connected nodes together while tracking their root leader.

  • Group Evaluation per Component: For each node, a BFS is performed:

    • This checks the feasible numbers of levels (or groups) in which nodes can be arranged based on their mutual connections to avoid cycles that would make grouping at certain levels impossible.
    • The leader for each node’s component is found, and the maximum number of feasible groups is maintained within a hashing mechanism mapping each component's root to its potential grouping.

  • Error Handling: If any setup of nodes results in cycles that affect the required grouping, an immediate return value of -1 indicates that a feasible grouping is impossible.

  • Group Accumulation and Result Calculation: After iterating through all nodes, the maximum number of groups across all components of the graph is accumulated and represented as the final output.

The utilization of union-find helps efficiently manage the identification of connected components and merging of sets during the BFS analysis. This approach ensures that the solution is optimal for large graphs by maintaining an efficient complexity for operations such as union, find, and graph traversal.

python 
class GraphOperations:
    # Function for calculating maximum groups across graph components
    def maxGroups(self, vertex_count, links):
        neighbors = [[] for _ in range(vertex_count)]
        root = [-1] * vertex_count
        rank = [0] * vertex_count

        # Create adjacency list and union-find operation for each link
        for link in links:
            neighbors[link[0] - 1].append(link[1] - 1)
            neighbors[link[1] - 1].append(link[0] - 1)
            self.union(link[0] - 1, link[1] - 1, root, rank)

        group_count = {}

        # Determining groups per component
        for vertex in range(vertex_count):
            group_total = self.findGroups(neighbors, vertex, vertex_count)
            if group_total == -1:
                return -1  # In case of impossible partition
            component_root = self.findRoot(vertex, root)
            group_count[component_root] = max(
                group_count.get(component_root, 0), group_total
            )

        # Sum of maximal groups across all components
        aggregated_groups = sum(group_count.values())
        return aggregated_groups

    # Function to determine number of partitionable groups from a vertex
    def findGroups(self, adjacency, source, total_vertices):
        from collections import deque
        queue = deque()
        visited = [-1] * total_vertices
        queue.append(source)
        visited[source] = 0
        max_depth = 0

        # Breadth-first search to figure out partition depth
        while queue:
            layer_size = len(queue)
            for _ in range(layer_size):
                current = queue.popleft()
                for neighbor in adjacency[current]:
                    if visited[neighbor] == -1:
                        visited[neighbor] = max_depth + 1
                        queue.append(neighbor)
                    elif visited[neighbor] == max_depth:
                        return -1  # Neighbor in same layer indicates invalid partition
            max_depth += 1
        return max_depth

    # Finding component root with path compression
    def findRoot(self, node, ancestor):
        while ancestor[node] != -1:
            node = ancestor[node]
        return node

    # Union operation with rank optimization
    def union(self, node1, node2, ancestor, depth):
        root1 = self.findRoot(node1, ancestor)
        root2 = self.findRoot(node2, ancestor)

        if root1 == root2:
            return

        # Union by rank
        if depth[root1] < depth[root2]:
            root1, root2 = root2, root1
        ancestor[root2] = root1
        if depth[root1] == depth[root2]:
            depth[root1] += 1

This code snippet defines a Python class GraphOperations aimed at solving the problem of dividing nodes in a graph into the maximum number of groups such that each group represents a distinct component. The main operations are encapsulated in several methods within this class:

  • maxGroups: Calculates the maximum groups possible across all graph components given a list of links between nodes and a total count of nodes.
  • findGroups: Determines the number of groups that can be formed starting from a specific vertex using a breadth-first search to ensure that no two nodes in the same group are immediate neighbors, providing a check for layer membership conflicts.
  • findRoot: Implements path compression as part of the union-find algorithm to find the root of a component efficiently.
  • union: Merges two components into a single component using the union-by-rank strategy.

Follow these key steps incorporated in the code:

  1. Initialize structures to represent the graph—specifically, lists for maintaining neighboring nodes (adjacency list), root information, and rank of nodes for the union-find algorithm.
  2. Populate the adjacency list and apply the union-find operation for each link to effectively determine connected components.
  3. Using a helper function findGroups, calculate the group size potential for each vertex and apply this to aggregate the count per component root.
  4. Sum the maximum groups from all components as the final result.

Important operations include ensuring components are merged efficiently without causing circles and components' roots are updated to reflect their latest connections. Additionally, depth-first searches are essential in finding if a proper partition of groups is feasible. If any two neighbors fall within the same partition layer, it becomes impossible to continue, and an error state is returned. The maximum depth achieved in the search outlines the potential groupings starting from each vertex.

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