<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 {
private:
    bool dfs(int node, int parent, unordered_map<int, vector<int>>& adj, unordered_map<int, bool>& visited) {
        visited[node] = true;
        for (int neighbor : adj[node]) {
            if (!visited[neighbor]) {
                if (dfs(neighbor, node, adj, visited)) {
                    return true;  // Cycle found
                }
            } else if (neighbor != parent) {
                return true;  // Cycle found
            }
        }
        return false;  // No cycle found
    }

public:
    vector<int> findRedundantConnection(vector<vector<int>>& edges) {
        for (int i = edges.size() - 1; i >= 0; i--) {
            unordered_map<int, vector<int>> adj;
            unordered_map<int, bool> visited;
            for (int k = 0; k < edges.size(); k++) {
                if (k != i) {
                    int u = edges[k][0];
                    int v = edges[k][1];
                    adj[u].push_back(v);
                    adj[v].push_back(u);
                }
            }
            bool hasCycle = false;
            for (int j = 1; j <= edges.size(); j++) {
                if (!visited[j] && adj.find(j) != adj.end()) {
                    if (dfs(j, -1, adj, visited)) {
                        hasCycle = true;
                        break;
                    }
                }
            }
            
            // If no cycle is found after removing this edge, return this edge
            if (!hasCycle) {
                return edges[i];
            }
        }
        
        return {};  // In case no redundant connection is found
    }
};

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