- Difficulty: Hard
- Tags: LeetCode, Hard, Breadth-First Search, Union Find, Graph, leetcode-2493, O(n^2), O(n), Medium, BFS, DFS
Problem
You are given a positive integer n
representing the number of nodes in an undirected graph. The nodes are labeled from 1
to n
.
You are also given a 2D integer array edges
, where edges[i] = [ai, bi]
indicates that there is a bidirectional edge between nodes ai
and bi
. Notice that the given graph may be disconnected.
Divide the nodes of the graph into m
groups (1-indexed) such that:
- Each node in the graph belongs to exactly one group.
- For every pair of nodes in the graph that are connected by an edge
[ai, bi]
, ifai
belongs to the group with indexx
, andbi
belongs to the group with indexy
, then|y - x| = 1
.
Return the maximum number of groups (i.e., maximum m
) into which you can divide the nodes. Return -1
if it is impossible to group the nodes with the given conditions.
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.