<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>

//Approach-2 (BFS)
class Solution {
public:
    int n;
    bool bfs(map<int, vector<int>>& mp, int start, int end) {
        vector<bool> visited(n+1, false);
        queue<int> que;
        que.push(start);
        
        while(!que.empty()) {
            int curr = que.front();
            que.pop();
            visited[curr] = true;
            if(curr == end)
                return true;
            
            for(int &x : mp[curr]) {
                if(!visited[x]) {
                    que.push(x);
                }
            }
        }
        return false;
    }
    vector<int> findRedundantConnection(vector<vector<int>>& edges) {
        map<int, vector<int>> mp;
        n = edges.size();
        
        for(int i = 0; i<n; i++) {
            int u = edges[i][0];
            int v = edges[i][1];
            
            //if u and v are already in graph, then we check
            //if this current edge connects them again ?
            //If yes, this edge is unwanted (redundant)
            if(mp.find(u) != mp.end() && mp.find(v) != mp.end() && bfs(mp, u, v))
                return edges[i];
            
            //Else add them to graph
            mp[u].push_back(v);
            mp[v].push_back(u);
        }
        
        return {};
    }
};
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