<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:
    void print_v(const vector<int>& v) {
        cout << "{ ";
        for (int n : v) {
            cout << n << ", ";
        }
        cout << "}" << endl;
    }
    // returns idx in steak that causes cycle, or -1
    int dfs(unordered_map<int, vector<int>>& adj, vector<int>& steak, unordered_map<int, int> steak2, int prev) {
        int cur = steak.back();
        for (int neigh : adj[cur]) {
            if (neigh == prev) continue;
            if (steak2.find(neigh) != steak2.end()) {
                return steak2[neigh];
            }
            steak2[neigh] = steak.size();
            steak.push_back(neigh);
            int res = dfs(adj, steak, steak2, cur);
            if (res != -1) return res;
            steak2.erase(neigh);
            steak.pop_back();
        }
        return -1;
    }
    vector<int> findRedundantConnection(vector<vector<int>>& edges) {
        unordered_map<int, vector<int>> adj;
        unordered_map<int, int> edgeIdx;
        for (int i = 0; i < edges.size(); i++) {
            int a = edges[i][0];
            int b = edges[i][1];
            edgeIdx[a*1000+b] = i;
            adj[a].push_back(b);
            adj[b].push_back(a);
        }
        vector<int> steak;
        unordered_map<int, int> steak2;
        steak.push_back(edges[0][0]);
        steak2[edges[0][0]] = 0;
        int res = dfs(adj, steak, steak2, -2);
        //print_v(steak);
        int ansIdx = 0;
        for (int i = res; i < steak.size(); i++) {
            int j = i - 1;
            if (i == res) j = int(steak.size()) - 1;
            int a = steak[j];
            int b = steak[i];
            if (a > b) swap(a, b);
            int idx = edgeIdx[a*1000+b];
            if (idx > ansIdx) {
                ansIdx = idx;
            }
        }
        return edges[ansIdx];
    }
};
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