<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 {
public:
    vector<int> findRedundantConnection(vector<vector<int>>& edges) {
        // create adj matrix
        // number of nodes in a tree = edge + 1
        vector<int> adj_list[edges.size() + 1];
        int degree[edges.size() + 1];
        int i;
        
        memset(degree, 0 , sizeof(int) * ( edges.size() + 1 ) );

        for(int i = 0 ; i <edges.size() ; i++)
        {
            int x,y;
            x = edges[i][0];
            y = edges[i][1];
            adj_list[x].push_back(y);
            adj_list[y].push_back(x);


        }
        

        //now we find the degree of each node
        for(int i = 0 ; i <edges.size() ; i++)
        {
            int x,y;
            x = edges[i][0];
            y = edges[i][1];

            degree[x]++;
            degree[y]++;

        }

        //put all the nodes with degree 1 to queue
        queue<int> q;
        for(int i = 1 ; i <= edges.size() ; i++ )
        {
           // cout<<" i -> "<<i<<" degree = "<<degree[i]<<endl;
            if(degree[i] == 1)
            {
                q.push(i);
            }
        }


        while(!q.empty())
        {
            int node;
            node = q.front();
            q.pop();
            for(int i = 0 ; i < adj_list[node].size(); i++)
            {
                degree[adj_list[node][i]]--;
                //cout<<"node ->"<<node<<" neighbour -> "<<adj_list[node][i]<< " size() "<< adj_list[node].size() <<endl;
                if(degree[adj_list[node][i]] == 1)
                {
                    q.push(adj_list[node][i]);
                }
            }
        }


        for( i = edges.size()-1; i >= 0 ; i--)
        {
            int degree_1,degree_2;
        
            degree_1 = degree[edges[i][0]];
            degree_2 = degree[edges[i][1]];

            if(degree_1 == 2 && degree_2 == 2 )
            {
                break;
            }
        }

        return edges[i];
    }
};
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