<p>In this problem, a tree is an <strong>undirected graph</strong> that is connected and has no cycles.</p>

<p>You are given a graph that started as a tree with <code>n</code> nodes labeled from <code>1</code> to <code>n</code>, with one additional edge added. The added edge has two <strong>different</strong> vertices chosen from <code>1</code> to <code>n</code>, and was not an edge that already existed. The graph is represented as an array <code>edges</code> of length <code>n</code> where <code>edges[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> indicates that there is an edge between nodes <code>a<sub>i</sub></code> and <code>b<sub>i</sub></code> in the graph.</p>

<p>Return <em>an edge that can be removed so that the resulting graph is a tree of </em><code>n</code><em> nodes</em>. If there are multiple answers, return the answer that occurs last in the input.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/05/02/reduntant1-1-graph.jpg" style="width: 222px; height: 222px;" />
<pre>
<strong>Input:</strong> edges = [[1,2],[1,3],[2,3]]
<strong>Output:</strong> [2,3]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/05/02/reduntant1-2-graph.jpg" style="width: 382px; height: 222px;" />
<pre>
<strong>Input:</strong> edges = [[1,2],[2,3],[3,4],[1,4],[1,5]]
<strong>Output:</strong> [1,4]
</pre>

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

<ul>
	<li><code>n == edges.length</code></li>
	<li><code>3 &lt;= n &lt;= 1000</code></li>
	<li><code>edges[i].length == 2</code></li>
	<li><code>1 &lt;= a<sub>i</sub> &lt; b<sub>i</sub> &lt;= edges.length</code></li>
	<li><code>a<sub>i</sub> != b<sub>i</sub></code></li>
	<li>There are no repeated edges.</li>
	<li>The given graph is connected.</li>
</ul>

class Solution {
    int find(int x,vector<int>& parent){
        if(x==parent[x])
            return x;
        parent[x]=find(parent[x],parent);
        return parent[x];
    }
    void unionFunc(vector<int>& parent,vector<int>& rank,int x, int y){
        int x_ref=find(x,parent);
        int y_ref=find(y,parent);
        if(x_ref==y_ref)
            return;
        if(x_ref!=y_ref){
            if(rank[x_ref]>rank[y_ref])
                parent[y_ref]=x_ref;
            else if(rank[x_ref]<rank[y_ref])
                parent[x_ref]=y_ref;
            else{
                parent[x_ref]=y_ref;
                rank[y_ref]++;
            }
        }
    }
    vector<int> soln(vector<vector<int>>& edges){
        int n=edges.size();
        vector<int> ans;
        for(int i=n-1;i>=0;i--){
            vector<vector<int>> adjList(n+1);
            for(int j=0;j<n;j++){
                if(j==i)
                    continue;
                adjList[edges[j][0]].push_back(edges[j][1]);
            }
            vector<int> parent(n+1);
            for(int k=1;k<n;k++){
                parent[k]=k;
            }
            vector<int> rank(n+1,0);
            for(int l=1;l<adjList.size();l++){
                for(int k=0;k<adjList[l].size();k++){
                    unionFunc(parent,rank,l,adjList[l][k]);
                }
            }
            int temp=find(1,parent);
            if(find(edges[i][0],parent)!=find(edges[i][1],parent))
                continue;
            for(int l=2;l<=n;l++){
                if(temp==find(l,parent)){
                    ans.push_back(edges[i][0]);
                    ans.push_back(edges[i][1]);
                    return ans;
                }
            }
        }
        return ans;
    }
public:
    vector<int> findRedundantConnection(vector<vector<int>>& edges) {
        return soln(edges);
    }
};
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