<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 DSU{
    public:
    int * parent;
    int size;

    DSU(int n)
    {
        size=n;
        parent=new int[n+1];
        for(int i=0;i<=n;i++) parent[i]=-1;
    }
    int find(int x)
    {
        while(parent[x]>0) x=parent[x];
        return x;
    }

    void union2(int u,int v)
    {
        int parentU=find(u);
        int parentV=find(v);
        if(abs(parent[parentU])>abs(parent[parentV]))
        {
            parent[parentU]+=parent[parentV];
            parent[parentV]=parentU;
        }else{
            parent[parentV]+=parent[parentU];
            parent[parentU]=parentV;
        }
    }
};
class Solution {
private:
   int findComponent(int skip,vector<vector<int>>& edges)
   {
        int n=edges.size();
        DSU obj(n);
        int c=n;
        for(int i=0;i<n;i++)
        {
            if(i!=skip)
            {
                if(obj.find(edges[i][0])==obj.find(edges[i][1])) continue;
                c--;
                obj.union2(edges[i][0],edges[i][1]);      
            }
        }
        return c;
   }
public:
    vector<int> findRedundantConnection(vector<vector<int>>& edges) {
       
       int n=edges.size();
       vector<int>ed(2);
       for(int i=0;i<n;i++)
       {
           int c=findComponent(i,edges);
           if(c==1)
           {
              ed[0]=edges[i][0];
              ed[1]=edges[i][1];
           }
       }

       return ed;
    }
};
30103