#2493

Hard

auto_awesomeGraph traversal with depth-first search

LeetCode Problem Workspace

Divide Nodes Into the Maximum Number of Groups

Determine the maximum number of groups nodes can form in a graph using depth-first traversal without violating edge connections.

depth first search breadth first search union find graph

Problem Statement

Given an undirected graph with n nodes labeled from 1 to n and a list of edges, divide the nodes into the maximum number of 1-indexed groups. Each edge must connect nodes in different groups, and nodes within the same group must not violate edge constraints.

Return the largest possible number of groups that satisfy the edge constraints. If it is impossible to assign groups without violating any edge, return -1. The graph may contain multiple disconnected components, requiring careful traversal and component analysis.

Examples

Example 1

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

Output: 4

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

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.

Solution Approach

Check Bipartite Components

Traverse each connected component using DFS or BFS. Assign alternating levels to nodes and verify that no edge connects nodes at the same level. If any component fails the bipartite check, return -1 immediately.

Track Maximum Depth per Component

While performing DFS or BFS, record the maximum distance from the starting node in each component. The depth determines the number of groups possible in that component. Accumulate depths from all components to get the total maximum groups.

Use Union Find for Disconnected Graphs

Apply Union Find to quickly identify disconnected components. This helps to manage multiple DFS/BFS traversals efficiently and ensures all nodes are considered when computing the total number of groups.

Complexity Analysis

Metric	Value
Time	O(n \times (n + m))
Space	O(n + m)

Time complexity is O(n * (n + m)) due to DFS/BFS traversals over each component and checking edges. Space complexity is O(n + m) for adjacency lists and level tracking arrays.

What Interviewers Usually Probe

Checks if you detect non-bipartite graphs and return -1 correctly.
Looks for clear DFS/BFS level assignment to calculate group depth.
Wants handling of disconnected components using Union Find or multiple traversals.

Common Pitfalls or Variants

Common pitfalls

Assuming the graph is always connected and missing disconnected components.
Failing to check that edges do not connect nodes in the same group.
Incorrectly calculating maximum depth, leading to underestimated group counts.

Follow-up variants

Maximize groups in a weighted undirected graph with additional constraints on edge weights.
Compute minimum groups needed when edges define forbidden connections between nodes.
Handle dynamic graphs where edges can be added or removed and recalculate groups efficiently.

FAQ

Why does checking for bipartite structure matter in this problem?

A non-bipartite component cannot be grouped without violating edge constraints, so detecting it ensures correct -1 returns.

Can BFS be used instead of DFS for grouping?

Yes, BFS works similarly by assigning levels and tracking maximum depth, producing the same result for group counts.

How do disconnected components affect the maximum groups?

Each component can be grouped independently, so the total maximum groups is the sum of depths from all components.

What is the role of Union Find in this problem?

Union Find efficiently identifies disconnected components, ensuring every node is considered without repeated traversals.

What common mistakes occur with 'Divide Nodes Into the Maximum Number of Groups'?

Mistakes include ignoring non-bipartite detection, miscalculating depth for group counts, or skipping disconnected components.

terminal

Solution

Solution 1: BFS + Enumeration

Given that the graph provided by the problem may be disconnected, we need to process each connected component, find the maximum number of groups in each connected component, and accumulate them to get the final result.

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class Solution:
    def magnificentSets(self, n: int, edges: List[List[int]]) -> int:
        g = [[] for _ in range(n)]
        for a, b in edges:
            g[a - 1].append(b - 1)
            g[b - 1].append(a - 1)
        d = defaultdict(int)
        for i in range(n):
            q = deque([i])
            dist = [0] * n
            dist[i] = mx = 1
            root = i
            while q:
                a = q.popleft()
                root = min(root, a)
                for b in g[a]:
                    if dist[b] == 0:
                        dist[b] = dist[a] + 1
                        mx = max(mx, dist[b])
                        q.append(b)
                    elif abs(dist[b] - dist[a]) != 1:
                        return -1
            d[root] = max(d[root], mx)
        return sum(d.values())

Continue Practicing

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