<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://assets.leetcode.com/uploads/2020/06/04/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://assets.leetcode.com/uploads/2020/06/04/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://assets.leetcode.com/uploads/2020/06/04/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>

#define pii pair<int,int>
#define vsi vector<set<pii>>
#define idx first
#define val second

class Solution {

    struct cmp {
        bool operator() (pii a, pii b) {
            return a.val > b.val;
        }
    };

    int findMST (const vsi& adjList, int n) {
        vector<int> dists(n, INT_MAX);
        priority_queue<pii, vector<pii>, cmp> heapify;
        heapify.push({0,0});
        int ans = 0;
        while (!heapify.empty()) {
            pii curr = heapify.top();
            heapify.pop();
            if (dists[curr.idx] != INT_MAX) continue;
            dists[curr.idx] = curr.val;
            ans += curr.val;
            for (auto& edge : adjList[curr.idx]) if (dists[edge.idx] == INT_MAX) heapify.push(edge);
        }
        for (int i=0; i<n; i++) if (dists[i] == INT_MAX) return INT_MAX;
        return ans;
    }

    bool checkIfEdgePresentInSomeMST(const vsi& adjList, int n, vector<int>& testEdge, int threshold) {
        vector<int> dists(n, INT_MAX);
        priority_queue<pii, vector<pii>, cmp> heapify;
        dists[testEdge[0]] = dists[testEdge[1]] = 0;
        for (auto& edge : adjList[testEdge[0]]) if (dists[edge.idx] == INT_MAX) heapify.push(edge);
        for (auto& edge : adjList[testEdge[1]]) if (dists[edge.idx] == INT_MAX) heapify.push(edge);
        int ans = 0;
        while (!heapify.empty()) {
            pii curr = heapify.top();
            heapify.pop();
            if (dists[curr.idx] != INT_MAX) continue;
            dists[curr.idx] = curr.val;
            ans += curr.val;
            for (auto& edge : adjList[curr.idx]) if (dists[edge.idx] == INT_MAX) heapify.push(edge);
        }
        return (threshold == ans + testEdge[2]);
    }

public:
    vector<vector<int>> findCriticalAndPseudoCriticalEdges(int n, vector<vector<int>>& edges) {
        //Max number of edges = 200
        //If removing an edge causes Minimum-ST value to increase, then it is a critical edge
        vsi adjList(n);
        for (auto& edge : edges) {
            adjList[edge[0]].insert({edge[1], edge[2]});
            adjList[edge[1]].insert({edge[0], edge[2]});
        } 
        int originalMST = findMST (adjList, n);
        vector<vector<int>> ans(2);
        int cnt = 0;
        for (auto &edge : edges) {
            adjList[edge[0]].erase({edge[1], edge[2]});
            adjList[edge[1]].erase({edge[0], edge[2]});
            int newMST = findMST (adjList, n);
            adjList[edge[0]].insert({edge[1], edge[2]});
            adjList[edge[1]].insert({edge[0], edge[2]});
            if (originalMST < newMST) ans[0].push_back(cnt);
            else {
                bool isOptional = checkIfEdgePresentInSomeMST(adjList, n, edge, originalMST);
                if (isOptional) ans[1].push_back(cnt);
            }
            cnt++;
        }
        return ans;
    }
};
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