Divide Nodes Into the Maximum Number of Groups
<p>You are given a positive integer <code>n</code> representing the number of nodes in an <strong>undirected</strong> graph. The nodes are labeled from <code>1</code> to <code>n</code>.</p>

<p>You are also given a 2D integer array <code>edges</code>, where <code>edges[i] = [a<sub>i, </sub>b<sub>i</sub>]</code> indicates that there is a <strong>bidirectional</strong> edge between nodes <code>a<sub>i</sub></code> and <code>b<sub>i</sub></code>. <strong>Notice</strong> that the given graph may be disconnected.</p>

<p>Divide the nodes of the graph into <code>m</code> groups (<strong>1-indexed</strong>) such that:</p>

<ul>
	<li>Each node in the graph belongs to exactly one group.</li>
	<li>For every pair of nodes in the graph that are connected by an edge <code>[a<sub>i, </sub>b<sub>i</sub>]</code>, if <code>a<sub>i</sub></code> belongs to the group with index <code>x</code>, and <code>b<sub>i</sub></code> belongs to the group with index <code>y</code>, then <code>|y - x| = 1</code>.</li>
</ul>

<p>Return <em>the maximum number of groups (i.e., maximum </em><code>m</code><em>) into which you can divide the nodes</em>. Return <code>-1</code> <em>if it is impossible to group the nodes with the given conditions</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2400-2499/2493.Divide%20Nodes%20Into%20the%20Maximum%20Number%20of%20Groups/images/example1.png" style="width: 352px; height: 201px;" />
<pre>
<strong>Input:</strong> n = 6, edges = [[1,2],[1,4],[1,5],[2,6],[2,3],[4,6]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 3, edges = [[1,2],[2,3],[3,1]]
<strong>Output:</strong> -1
<strong>Explanation:</strong> 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.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 500</code></li>
	<li><code>1 &lt;= edges.length &lt;= 10<sup>4</sup></code></li>
	<li><code>edges[i].length == 2</code></li>
	<li><code>1 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt;= n</code></li>
	<li><code>a<sub>i</sub> != b<sub>i</sub></code></li>
	<li>There is at most one edge between any pair of vertices.</li>
</ul>
Depth-First Search; Breadth-First Search; Union Find; Graph
/**
 * @param {number} n
 * @param {number[][]} edges
 * @return {number}
 */
var magnificentSets = function (n, edges) {
    const g = Array.from({ length: n }, () => []);
    for (const [a, b] of edges) {
        g[a - 1].push(b - 1);
        g[b - 1].push(a - 1);
    }
    const d = Array(n).fill(0);
    for (let i = 0; i < n; ++i) {
        const q = [i];
        const dist = Array(n).fill(0);
        dist[i] = 1;
        let mx = 1;
        let root = i;
        while (q.length) {
            const a = q.shift();
            root = Math.min(root, a);
            for (const b of g[a]) {
                if (dist[b] === 0) {
                    dist[b] = dist[a] + 1;
                    mx = Math.max(mx, dist[b]);
                    q.push(b);
                } else if (Math.abs(dist[b] - dist[a]) !== 1) {
                    return -1;
                }
            }
        }
        d[root] = Math.max(d[root], mx);
    }
    return d.reduce((a, b) => a + b);
};
