<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 DisjointSet{

public:
    vector<int> rank,parent,size;
    DisjointSet(int n){
        rank.resize(n+1,0);
        size.resize(n+1,1);
        parent.resize(n+1);
        for(int i=0;i<=n;i++){
            parent[i] = i;
        }
    }
    
    int findUPar(int node){
        if(parent[node]==node)return node;
        
        return parent[node] = findUPar(parent[node]);
    }
    
    void unionByRank(int u, int v){
        int ulp_u = findUPar(u);
        int ulp_v = findUPar(v);
        if(ulp_u==ulp_v)return;
        if(rank[ulp_u] < rank[ulp_v]){
            parent[ulp_u] = ulp_v;
        }
        else if(rank[ulp_u] > rank[ulp_v]){
            parent[ulp_v] = ulp_u;
        }
        else{
            parent[ulp_u] = ulp_v;
            rank[ulp_v]++;
        }
    }
    
    void unionBySize(int u, int v){
        int ulp_u = findUPar(u);
        int ulp_v = findUPar(v);
        if(ulp_u==ulp_v)return;
        if(size[ulp_u] < size[ulp_v]){
            parent[ulp_u] = ulp_v;
            size[ulp_v] += size[ulp_u];
        }
        else{
            parent[ulp_v] = ulp_u;
            size[ulp_u] += size[ulp_v];
        }
    }
};

class Solution {
public:
    vector<int> findRedundantConnection(vector<vector<int>>& edges) {
        int n = edges.size();
        DisjointSet ds(n);
        vector<int> ans;
        for(int i=0;i<n;i++){
            int u = edges[i][0];
            int v = edges[i][1];
            if(ds.findUPar(u)==ds.findUPar(v)){
                ans = edges[i];
            }
            else{
                ds.unionBySize(u,v);
            }
        }
        return ans;
    }
};
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