Find Critical and Pseudo-Critical Edges in Minimum Spanning Tree
<p>Given a weighted undirected connected graph with <code>n</code>&nbsp;vertices numbered from <code>0</code> to <code>n - 1</code>,&nbsp;and an array <code>edges</code>&nbsp;where <code>edges[i] = [a<sub>i</sub>, b<sub>i</sub>, weight<sub>i</sub>]</code> represents a bidirectional and weighted edge between nodes&nbsp;<code>a<sub>i</sub></code>&nbsp;and <code>b<sub>i</sub></code>. A minimum spanning tree (MST) is a subset of the graph&#39;s edges that connects all vertices without cycles&nbsp;and with the minimum possible total edge weight.</p>

<p>Find <em>all the critical and pseudo-critical edges in the given graph&#39;s minimum spanning tree (MST)</em>. An MST edge whose deletion from the graph would cause the MST weight to increase is called a&nbsp;<em>critical edge</em>. On&nbsp;the other hand, a pseudo-critical edge is that which can appear in some MSTs but not all.</p>

<p>Note that you can return the indices of the edges in any order.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1400-1499/1489.Find%20Critical%20and%20Pseudo-Critical%20Edges%20in%20Minimum%20Spanning%20Tree/images/ex1.png" style="width: 259px; height: 262px;" /></p>

<pre>
<strong>Input:</strong> n = 5, edges = [[0,1,1],[1,2,1],[2,3,2],[0,3,2],[0,4,3],[3,4,3],[1,4,6]]
<strong>Output:</strong> [[0,1],[2,3,4,5]]
<strong>Explanation:</strong> The figure above describes the graph.
The following figure shows all the possible MSTs:
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1400-1499/1489.Find%20Critical%20and%20Pseudo-Critical%20Edges%20in%20Minimum%20Spanning%20Tree/images/msts.png" style="width: 540px; height: 553px;" />
Notice that the two edges 0 and 1 appear in all MSTs, therefore they are critical edges, so we return them in the first list of the output.
The edges 2, 3, 4, and 5 are only part of some MSTs, therefore they are considered pseudo-critical edges. We add them to the second list of the output.
</pre>

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

<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1400-1499/1489.Find%20Critical%20and%20Pseudo-Critical%20Edges%20in%20Minimum%20Spanning%20Tree/images/ex2.png" style="width: 247px; height: 253px;" /></p>

<pre>
<strong>Input:</strong> n = 4, edges = [[0,1,1],[1,2,1],[2,3,1],[0,3,1]]
<strong>Output:</strong> [[],[0,1,2,3]]
<strong>Explanation:</strong> We can observe that since all 4 edges have equal weight, choosing any 3 edges from the given 4 will yield an MST. Therefore all 4 edges are pseudo-critical.
</pre>

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

<ul>
	<li><code>2 &lt;= n &lt;= 100</code></li>
	<li><code>1 &lt;= edges.length &lt;= min(200, n * (n - 1) / 2)</code></li>
	<li><code>edges[i].length == 3</code></li>
	<li><code>0 &lt;= a<sub>i</sub> &lt; b<sub>i</sub> &lt; n</code></li>
	<li><code>1 &lt;= weight<sub>i</sub>&nbsp;&lt;= 1000</code></li>
	<li>All pairs <code>(a<sub>i</sub>, b<sub>i</sub>)</code> are <strong>distinct</strong>.</li>
</ul>
Union Find; Graph; Minimum Spanning Tree; Sorting; Strongly Connected Component
class Solution {
    public List<List<Integer>> findCriticalAndPseudoCriticalEdges(int n, int[][] edges) {
        for (int i = 0; i < edges.length; ++i) {
            int[] e = edges[i];
            int[] t = new int[4];
            System.arraycopy(e, 0, t, 0, 3);
            t[3] = i;
            edges[i] = t;
        }
        Arrays.sort(edges, Comparator.comparingInt(a -> a[2]));
        int v = 0;
        UnionFind uf = new UnionFind(n);
        for (int[] e : edges) {
            int f = e[0], t = e[1], w = e[2];
            if (uf.union(f, t)) {
                v += w;
            }
        }
        List<List<Integer>> ans = new ArrayList<>();
        for (int i = 0; i < 2; ++i) {
            ans.add(new ArrayList<>());
        }
        for (int[] e : edges) {
            int f = e[0], t = e[1], w = e[2], i = e[3];
            uf = new UnionFind(n);
            int k = 0;
            for (int[] ne : edges) {
                int x = ne[0], y = ne[1], z = ne[2], j = ne[3];
                if (j != i && uf.union(x, y)) {
                    k += z;
                }
            }
            if (uf.getN() > 1 || (uf.getN() == 1 && k > v)) {
                ans.get(0).add(i);
                continue;
            }
            uf = new UnionFind(n);
            uf.union(f, t);
            k = w;
            for (int[] ne : edges) {
                int x = ne[0], y = ne[1], z = ne[2], j = ne[3];
                if (j != i && uf.union(x, y)) {
                    k += z;
                }
            }
            if (k == v) {
                ans.get(1).add(i);
            }
        }
        return ans;
    }
}

class UnionFind {
    private int[] p;
    private int n;

    public UnionFind(int n) {
        p = new int[n];
        this.n = n;
        for (int i = 0; i < n; ++i) {
            p[i] = i;
        }
    }

    public int getN() {
        return n;
    }

    public boolean union(int a, int b) {
        if (find(a) == find(b)) {
            return false;
        }
        p[find(a)] = find(b);
        --n;
        return true;
    }

    public int find(int x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }
}