<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 Disjoint {
    int *rank, *parent, n;
public:
    Disjoint(int n)
    {
        rank = new int[n];
        parent = new int[n];
        this->n = n;
        makeSet();
    }
 
    void makeSet()
    {
        for (int i = 0; i < n; i++) {
            parent[i] = i;
        }
    }
    int findset(int x)
    {
        if (parent[x] != x) {
            parent[x] = findset(parent[x]);
        }
        return parent[x];
    }
    void merge(int x, int y)
    {
        int xset = findset(x);
        int yset = findset(y);
        if (xset == yset)
            return;
        if (rank[xset] < rank[yset]) {
            parent[xset] = yset;
        }
        else if (rank[xset] > rank[yset]) {
            parent[yset] = xset;
        }
        else {
            parent[yset] = xset;
            rank[xset] = rank[xset] + 1;
        }
    }
};

class Solution {
public:
    vector<int> findRedundantConnection(vector<vector<int>>& edges) {
        int n;
        for(vector<int> e: edges)
        {
            n = max(n, e[0]);
            n = max(n, e[1]);
        }
        Disjoint *d = new Disjoint(n);
        for(vector<int> e: edges)
        {
            if(d->findset(e[0]-1)!=d->findset(e[1]-1))
                d->merge(e[1]-1, e[0]-1);
            else
                return e;
        }
        vector<int> x;
        return x;
    }
};
24658