2493. Divide Nodes Into the Maximum Number of Groups

Problem Description

This problem presents a scenario in which we have to divide n nodes of an undirected graph into groups based on specific rules. The nodes are numbered from 1 to n, and the edges are represented as pairs where each pair [a, b] signifies a bidirectional connection between node a and node b. The goal is to create the maximum number of groups, indexed from 1 to m, such that each node is in one group, and for every pair of connected nodes, the absolute difference in their group indexes is 1. In case it's not possible to arrange the nodes while satisfying these conditions, we should return -1.

Flowchart Walkthrough

Certainly, let's analyze the problem on leetcode 2493 using the algorithm flowchart. First, let's understand the problem context to guide our traversal through the flowchart.

In leetcode 2493, we're tasked with dividing nodes in a graph into the maximum number of groups based on certain connection conditions. Essentially, the problem is about grouping nodes such that no two nodes within the same group are directly connected, maximizing the number of such groups.

Now, let's utilize the Flowchart to determine the appropriate algorithm:

Is it a graph?

Yes: The nodes and their connections naturally form a graph structure.

Is it a tree?

No: Since the graph may contain cycles (as there is no specification of acyclicity and no inherent hierarchical structuring indicated), it isn't a tree.

Is the problem related to directed acyclic graphs (DAGs)?

No: The problem does not specifically involve directed acyclic graphs nor does it involve processes or tasks that need ordering like in a DAG.

Is the problem related to shortest paths?

No: The purpose is not finding shortest paths but to maximize the grouping of the nodes.

Does the problem involve connectivity?

Yes: The problem involves determining connectivity to ensure no connected nodes are in the same group.

Is the graph weighted?

No: The graph seems unweighted as the problem statement does not mention any weights associated with the edges.

Conclusion: Following the flowchart suggests using BFS for this unweighted connectivity problem. Using BFS, we can traverse each component of the graph, marking nodes as visited and assigning them to groups such that no two adjacent nodes share the same group. This will help in solving the problem of dividing the nodes into the maximum number of groups.

Intuition

The process of solving this problem can be understood as finding connected components and then verifying if we can arrange them linearly in such a way that neighboring nodes in each component differ in their group indexes by exactly 1. The steps in the solution can be described as follows:

We first create a representation of the graph using adjacency lists, which enables us to explore the graph easily.
We use depth-first search (DFS) to find all connected components within the graph. This is accomplished by iterating through each node that hasn't been visited and performing DFS to collect all the nodes in the connected component related to the current node.
After obtaining a connected component, we use breadth-first search (BFS) to assign distances to each of the nodes from a certain starting node. In this context, the distances represent group indexes. While doing so, we validate whether the assignment is possible following the rules given in the problem. If we enc ounter a situation where the absolute difference between the group indexes of connected nodes isn't equal to 1, we return -1 as the grouping isn't possible.
We keep track of the maximum distance which correlates to the largest group index obtained for the particular component. We do this because the answer to the problem is the sum of the maximum group indexes in each component since nodes of different components do not influence each other's grouping.
We add the maximum group indexes obtained from each component to generate the final answer which is the maximum number of groups we can divide the nodes into.
In case any of the components cannot be arranged as per the given rules, the function will return -1.

By the end of the process, we will either have a maximum number of groups m that can be formed under the given constraints or -1 if the task is impossible.

Learn more about Breadth-First Search, Union Find and Graph patterns.

Solution Approach

The solution involves several algorithms and data structures, primarily Depth-First Search (DFS), Breadth-First Search (BFS), and the use of a graph representation. Here's a step-by-step explanation:

Representation: The graph is represented using a dictionary g, which maps each node to a list of its adjacent nodes.
- This is a typical adjacency list representation which allows for efficient traversal of the graph.
Depth-First Search (DFS): A recursion-based dfs function is used to traverse the graph. It helps identify all nodes belonging to the same connected component.
- vis is a list of boolean values indicating whether a node has been visited during DFS.
- arr collects all nodes of the current connected component.
- As DFS proceeds, it marks nodes as visited and adds them to arr.
Breadth-First Search (BFS): Once a connected component is identified by DFS, BFS is used to try to assign each node a group index.
- bfs operates on node i, starting by assigning distance (group index) 1 to it.
- Then it uses a queue to process nodes level by level, maintaining the invariant that the difference between group indexes of adjacent nodes is 1.
- dist is a list where dist[i] is the group index of the i-th node or inf if the node hasn't been assigned a group yet.
- BFS continues until all nodes in arr have been assigned a valid group index or a situation is encountered that violates the conditions for such assignment (returning -1).
- ans within the bfs function is the largest group index assigned to any node in the current connected component.
Main Logic:
- Iterate over all nodes, if a node hasn't been visited, it means we're encountering a new connected component. Perform dfs from this node.
- After identifying all nodes in a connected component via dfs, bfs is executed for each node v in arr. This is to ensure that if it's possible to start the assignment of groups from different nodes to maximize the number of groups.
- We keep a variable t to store the highest number of groups achievable from the connected component based on the various starting points, and this value is added to the overall ans.
Final Step:
- After all the components have been processed, the final answer ans contains the maximum number of groups that the nodes can be divided into according to the problem's conditions.
- If at any point the grouping conditions cannot be met, the solution function returns -1.

In essence, the algorithm is a combination of DFS to identify components and BFS to validate and find the maximal grouping of the nodes within those components. The adjacency list allows for efficient graph traversal, and the queue in BFS ensures that nodes are processed in the correct order for group assignment.

Ready to land your dream job?

Unlock your dream job with a 2-minute evaluator for a personalized learning plan!

Start Evaluator

Example Walkthrough

Suppose we have an undirected graph with 5 nodes, and edges as follows: [1, 2], [1, 3], [2, 4], [3, 5]. The graph can be visualized as:

1 - 2 - 4
|   
3 - 5

We need to divide these nodes into groups while ensuring connected nodes have group indexes differing by exactly one. Here’s a step-by-step example using the given approach:

Representation: We create an adjacency list for the graph.

g = {1: [2, 3], 2: [1, 4], 3: [1, 5], 4: [2], 5: [3]}

Depth-First Search (DFS): Initialize a vis list to keep track of visited nodes, starting with all nodes set to False.

vis = [False, False, False, False, False] // Indexed from 1, so index 0 is ignored.

We start DFS from node 1, leading to the discovery of all nodes in the connected component.

arr = [1, 2, 3, 4, 5]

Breadth-First Search (BFS): We perform BFS starting from node 1.
- Assign group index 1 to node 1 and set dist to keep track of group assignments, initializing other distances to inf.
```
dist = [inf, 1, inf, inf, inf]
```
- Use a queue and begin with node 1. Then, process adjacent nodes. For each node, determine a valid group index based on its neighbors.
- After BFS, we might end up with these group assignments:
```
dist = [inf, 1, 2, 1, 2]
```
- The largest group index ans within this component is 2.
Main Logic: Since the graph is connected and we start with the first node, all nodes belong to the same connected component and are visited in the DFS step. Hence, there's no need to iterate over other nodes for the DFS step. The BFS step assigns group indexes such that adjacent nodes have indexes differing by 1.
Final Step:
- Having traversed all components and found a valid assignment, we add the largest group index from each component (which is 2 in our single-component example) to our overall answer ans.
- In this case, the answer is the maximum group index, which is 2.

Therefore, the nodes can be divided into a maximum of 2 groups under the given constraints, and all nodes in this graph follow the rule that every connected pair of nodes has group indexes differing by 1.

Solution Implementation

1from collections import defaultdict, deque
2
3class Solution:
4    def magnificentSets(self, n: int, edges: List[List[int]]) -> int:
5      
6        # Depth-first search function to explore all the nodes within a connected component
7        def dfs(node):
8            component_nodes.append(node)
9            visited[node] = True
10            for neighbor in graph[node]:
11                if not visited[neighbor]:
12                    dfs(neighbor)
13
14        # Breadth-first search function to calculate the maximum distance from the start node
15        def bfs(start_node):
16            maximum_distance = 1
17            distances = [float('inf')] * (n + 1)
18            distances[start_node] = 1
19            queue = deque([start_node])
20          
21            while queue:
22                current_node = queue.popleft()
23                for neighbor in graph[current_node]:
24                    if distances[neighbor] == float('inf'):
25                        maximum_distance = distances[neighbor] = distances[current_node] + 1
26                        queue.append(neighbor)
27          
28            # Assign an incremental distance value for disconnected nodes in the component
29            for node in component_nodes:
30                if distances[node] == float('inf'):
31                    maximum_distance += 1
32                    distances[node] = maximum_distance
33          
34            # Verify the resultant distances conform to the problem constraints
35            for node in component_nodes:
36                for neighbor in graph[node]:
37                    if abs(distances[node] - distances[neighbor]) != 1:
38                        return -1
39            return maximum_distance
40
41        # Initialize an adjacency list graph and visited list
42        graph = defaultdict(list)
43        for edge in edges:
44            a, b = edge
45            graph[a].append(b)
46            graph[b].append(a)
47      
48        visited = [False] * (n + 1)
49        total_sets = 0
50      
51        # Traverse all nodes to find disconnected components
52        for i in range(1, n + 1):
53            if not visited[i]:
54                component_nodes = []
55                dfs(i)  # DFS to populate component_nodes
56              
57                # Perform BFS on all nodes in the component and find the maximum distance
58                max_dist = max(bfs(node) for node in component_nodes)
59              
60                if max_dist == -1:
61                    return -1  # Early return if any component doesn't fulfill the requirements
62              
63                total_sets += max_dist
64      
65        # Return the total potential magnificent sets across all components
66        return total_sets
67

1class Solution {
2    private List<Integer>[] graph; // The graph represented as an adjacency list
3    private List<Integer> componentNodes = new ArrayList<>(); // List to store nodes of the currently visited component
4    private boolean[] visited; // Array to track visited nodes
5    private int totalNodes; // Total number of nodes in the graph
6
7    // Method to compute the magnificent sets value for the given graph
8    public int magnificentSets(int n, int[][] edges) {
9        totalNodes = n;
10        graph = new List[n + 1];
11        Arrays.setAll(graph, k -> new ArrayList<>()); // Initialize adjacency list for each node
12        for (int[] edge : edges) { // Build the graph
13            int nodeA = edge[0], nodeB = edge[1];
14            graph[nodeA].add(nodeB);
15            graph[nodeB].add(nodeA);
16        }
17
18        visited = new boolean[n + 1];
19        int totalMagnificentSets = 0;
20        for (int i = 1; i <= n; ++i) {
21            if (!visited[i]) {
22                dfs(i); // Perform Depth-First Search to find all nodes in the component
23                int largestDepth = -1;
24                // For each node in the component, use BFS to find the largest depth
25                for (int node : componentNodes) {
26                    largestDepth = Math.max(largestDepth, bfs(node));
27                }
28                if (largestDepth == -1) {
29                    return -1; // If it's not a magnificent set, return -1
30                }
31                totalMagnificentSets += largestDepth; // Add the largest depth to the total
32                componentNodes.clear(); // Clear nodes list for next component
33            }
34        }
35        return totalMagnificentSets; // Return the total magnificent sets of all components
36    }
37
38    // Helper method for BFS to calculate the largest depth of the BFS tree from the starting node
39    private int bfs(int startNode) {
40        int[] depth = new int[totalNodes + 1];
41        Arrays.fill(depth, Integer.MAX_VALUE); // Set initial depth to a high value
42        depth[startNode] = 1; // Depth of start node is 1
43        Deque<Integer> queue = new ArrayDeque<>();
44        queue.offer(startNode); // Initialize the queue with the starting node
45
46        int maxDepth = 1; // Track the maximum depth
47        while (!queue.isEmpty()) {
48            int currentNode = queue.poll();
49            for (int neighbor : graph[currentNode]) {
50                if (depth[neighbor] == Integer.MAX_VALUE) {
51                    depth[neighbor] = depth[currentNode] + 1; // Update the depth of the neighbor
52                    maxDepth = depth[neighbor]; // Update the max depth
53                    queue.offer(neighbor); // Add the neighbor node to the queue
54                }
55            }
56        }
57
58        // Update the depth for any node that hasn't been reached by BFS
59        for (int node : componentNodes) {
60            if (depth[node] == Integer.MAX_VALUE) {
61                depth[node] = ++maxDepth;
62            }
63        }
64
65        // Verify that all edges in the component have depths differing by 1
66        for (int node : componentNodes) {
67            for (int neighbor : graph[node]) {
68                if (Math.abs(depth[node] - depth[neighbor]) != 1) {
69                    return -1; // This component isn't magnificent
70                }
71            }
72        }
73
74        return maxDepth; // Return the largest depth in the BFS tree
75    }
76
77    // Helper method for DFS to traverse all nodes in a connected component
78    private void dfs(int currentNode) {
79        componentNodes.add(currentNode); // Add current node to component list
80        visited[currentNode] = true; // Mark current node as visited
81        for (int neighbor : graph[currentNode]) { // Visit all unvisited neighbors
82            if (!visited[neighbor]) {
83                dfs(neighbor); // Recursively visit neighbors
84            }
85        }
86    }
87}
88

1class Solution {
2public:
3    // Function to find the number of magnificent sets in a graph
4    int magnificentSets(int n, vector<vector<int>>& edges) {
5        // Graph represented as an adjacency list
6        vector<vector<int>> graph(n + 1);
7        // Building the graph
8        for (auto& edge : edges) {
9            int from = edge[0], to = edge[1];
10            graph[from].emplace_back(to);
11            graph[to].emplace_back(from);
12        }
13      
14        vector<int> nodes;
15        bool visited[n + 1] = {false}; // Array to keep track of visited nodes
16        int totalCount = 0; // To store the sum of magnificent sets found
17      
18        // Depth First Search (DFS) to traverse components of the graph
19        function<void(int)> depthFirstSearch = [&](int node) {
20            nodes.emplace_back(node);
21            visited[node] = true;
22            for (int neighbor : graph[node]) {
23                if (!visited[neighbor]) {
24                    depthFirstSearch(neighbor);
25                }
26            }
27        };
28      
29        // Breadth First Search (BFS) to find the largest distance in a single component
30        auto breadthFirstSearch = [&](int startNode) {
31            int localCount = 1;
32            int distances[n + 1];
33            memset(distances, 0x3f, sizeof distances);
34            distances[startNode] = 1;
35            queue<int> queue;
36            queue.push(startNode);
37          
38            while (!queue.empty()) {
39                int currentNode = queue.front();
40                queue.pop();
41                for (int neighbor : graph[currentNode]) {
42                    if (distances[neighbor] == 0x3f3f3f3f) {
43                        localCount = distances[neighbor] = distances[currentNode] + 1;
44                        queue.push(neighbor);
45                    }
46                }
47            }
48          
49            // Update distances for all nodes to ensure they are reachable
50            for (int& node : nodes) {
51                if (distances[node] == 0x3f3f3f3f) {
52                    distances[node] = ++localCount;
53                }
54            }
55          
56            // Check if there is a violation in magnificent set property
57            for (int& node : nodes) {
58                for (int& neighbor : graph[node]) {
59                    if (abs(distances[node] - distances[neighbor]) != 1) {
60                        return -1;
61                    }
62                }
63            }
64            return localCount;
65        };
66      
67        // Loop over all nodes to explore every component of the graph
68        for (int i = 1; i <= n; ++i) {
69            if (!visited[i]) {
70                depthFirstSearch(i);
71                int maxDistance = -1;
72                for (int& node : nodes) {
73                    maxDistance = max(maxDistance, breadthFirstSearch(node));
74                }
75                if (maxDistance == -1) return -1; // If impossible to find a magnificent set
76                totalCount += maxDistance; // Add the maximum distance found in this component
77                nodes.clear();
78            }
79        }
80        return totalCount;
81    }
82};
83

1// Define the magnificentSets function with 'n' of type number and 
2// 'edges' as a tuple array of type number.
3const magnificentSets = (n: number, edges: [number, number][]): number => {
4  
5    // Initialize graph as an array of Sets to hold edges for each vertex.
6    const graph: Set<number>[] = Array.from({ length: n + 1 }, () => new Set<number>());
7  
8    // Populate the graph with the edges from the edges list.
9    for (const [u, v] of edges) {
10        graph[u].add(v);
11        graph[v].add(u);
12    }
13
14    // Create a hash map to keep track of the greatest distances.
15    const hash: Map<number, number> = new Map();
16
17    // Perform BFS to find the shortest paths.
18    for (let i = 1; i <= n; i++) {
19        let queue: number[] = [i];
20        const distances: number[] = Array(n + 1).fill(0);
21        distances[i] = 1;
22        let maxDistance: number = 1,
23            minVertex: number = n;
24
25        while (queue.length > 0) {
26            let next: number[] = [];
27            for (let u of queue) {
28                minVertex = Math.min(minVertex, u);
29                for (const v of graph[u]) {
30                    if (!distances[v]) {
31                        distances[v] = distances[u] + 1;
32                        maxDistance = Math.max(maxDistance, distances[v]);
33                        next.push(v);
34                    }
35                    if (Math.abs(distances[u] - distances[v]) !== 1) {
36                        return -1;
37                    }
38                }
39            }
40            queue = next;
41        }
42
43        // Update the hash for each node with the maximum distance.
44        hash.set(minVertex, Math.max(maxDistance, hash.get(minVertex) || 0));
45    }
46
47    // Calculate the sum of maximum distances.
48    let result: number = 0;
49    hash.forEach((value) => {
50        result += value;
51    });
52
53    // Return the total sum.
54    return result;
55};
56

Time and Space Complexity

Time Complexity

The given code consists of two Depth First Search (DFS) and Breadth First Search (BFS) traversals of a graph. Here, n is the number of nodes, and let's assume e is the number of edges.

DFS Part:
- The DFS is done once for each component of the graph. In the worst case, if the graph is fully connected, the DFS will visit all n nodes and it would traverse all edges twice (since it is an undirected graph and each edge will be considered from both ends), which gives us O(n + 2e).
- However, in a DFS, each node is visited exactly once, so the complexity boils down to O(n + e) regardless of how many times each edge is considered.
BFS Part:
- For BFS in the bfs function, in the worst case, it will visit all nodes and traverse all edges in the connected component it starts from. This again is O(n + e).
- The BFS is called once for each node in the connected component found by DFS. Therefore, in the worst case, its complexity is O(n*(n + e)).
Overall Time Complexity:
- Combining DFS and BFS, and since BFS is nested within the scope of DFS results, the overall time complexity is therefore O(n*(n + e)).

Note that max(bfs(v) for v in arr) is a potentially costly operation since it will execute BFS for each vertex v in arr. The above analysis assumes that the graph is sparse and that e is much smaller than n^2. If the graph were dense with e close to n^2, the time complexity would be O(n^3) because performing BFS from each vertex would amount to checking all edges for all vertices.

Space Complexity

The space complexity is determined by the storage required for the graph representation, visited array, arr, and the queue used in BFS.

Graph Representation (g): The adjacency list will have at most 2e entries, so it requires O(e) space.
Visited Array (vis): An array of length n + 1, requiring O(n) space.
Array (arr): In the worst case, it can hold all the nodes in the graph, so O(n).
Queue for BFS: At most, it can hold all nodes in a connected component. Again, this is O(n) in the worst case.

The auxiliary space for BFS and DFS is not considered since the queue and call stack would at most hold O(n) entries at any point.

Thus, Overall Space Complexity: O(n + e).

Learn more about how to find time and space complexity quickly using problem constraints.

Discover Your Strengths and Weaknesses: Take Our 2-Minute Quiz to Tailor Your Study Plan:

Question 1 out of 10

Which algorithm should you use to find a node that is close to the root of the tree?

DFS

BFS

Binary search

Backtracking

I don’t know