<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:
    struct pair_hash {
        template <class T1, class T2>
        std::size_t operator()(const std::pair<T1, T2>& p) const {
            auto h1 = std::hash<T1>{}(p.first);
            auto h2 = std::hash<T2>{}(p.second);

            // Mainly for demonstration purposes, i.e. works but is overly
            // simple In the real world, use sth. like boost.hash_combine
            return h1 ^ h2;
        }
    };
    bool findConnection(
        std::unordered_map<int, std::vector<int>>& adjList,
        std::unordered_set<std::pair<int, int>, pair_hash>& visitedConnections,
        int current, int target) {
        std::cout << "current:" << current << "target:" << target << std::endl;
        if (current == target) {
            std::cout << "found" << std::endl;
            return true;
        }

        auto neighbours = adjList[current];
        for (const auto& neighbour : neighbours) {
            auto connection = std::pair<int, int>(current, neighbour);
            if (visitedConnections.find(connection) !=
                visitedConnections.end()) {
                continue;
            }
            visitedConnections.insert(connection);
            if (findConnection(adjList, visitedConnections, neighbour, target)) {
                return true;
            }
        }
        return false;
    }
    vector<int> findRedundantConnection(vector<vector<int>>& edges) {
        std::unordered_map<int, std::vector<int>> adjList;
        for (const auto& edge : edges) {
            std::unordered_set<std::pair<int, int>, pair_hash> visitedConnections;
            if (findConnection(adjList, visitedConnections, edge[0], edge[1])) {
                return edge;
            }
            // add to the adjacency list
            if (adjList.find(edge[0]) != adjList.end()) {
                adjList[edge[0]].emplace_back(edge[1]);
            } else {
                adjList[edge[0]] = std::vector<int>({edge[1]});
            }

            if (adjList.find(edge[1]) != adjList.end()) {
                adjList[edge[1]].emplace_back(edge[0]);
            } else {
                adjList[edge[1]] = std::vector<int>({edge[0]});
            }
        }
        return {};
    }
};
26171