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
type unionFind struct {
	p []int
	n int
}

func newUnionFind(n int) *unionFind {
	p := make([]int, n)
	for i := range p {
		p[i] = i
	}
	return &unionFind{p, n}
}

func (uf *unionFind) find(x int) int {
	if uf.p[x] != x {
		uf.p[x] = uf.find(uf.p[x])
	}
	return uf.p[x]
}

func (uf *unionFind) union(a, b int) bool {
	if uf.find(a) == uf.find(b) {
		return false
	}
	uf.p[uf.find(a)] = uf.find(b)
	uf.n--
	return true
}

func findCriticalAndPseudoCriticalEdges(n int, edges [][]int) [][]int {
	for i := range edges {
		edges[i] = append(edges[i], i)
	}
	sort.Slice(edges, func(i, j int) bool {
		return edges[i][2] < edges[j][2]
	})
	v := 0
	uf := newUnionFind(n)
	for _, e := range edges {
		f, t, w := e[0], e[1], e[2]
		if uf.union(f, t) {
			v += w
		}
	}
	ans := make([][]int, 2)
	for _, e := range edges {
		f, t, w, i := e[0], e[1], e[2], e[3]
		uf = newUnionFind(n)
		k := 0
		for _, ne := range edges {
			x, y, z, j := ne[0], ne[1], ne[2], ne[3]
			if j != i && uf.union(x, y) {
				k += z
			}
		}
		if uf.n > 1 || (uf.n == 1 && k > v) {
			ans[0] = append(ans[0], i)
			continue
		}
		uf = newUnionFind(n)
		uf.union(f, t)
		k = w
		for _, ne := range edges {
			x, y, z, j := ne[0], ne[1], ne[2], ne[3]
			if j != i && uf.union(x, y) {
				k += z
			}
		}
		if k == v {
			ans[1] = append(ans[1], i)
		}
	}
	return ans
}