<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>

#include <vector>
#include <unordered_map>
#include <list>
#include <queue>
#include <cstring>

using namespace std;

class Solution {
private:
    // Function to check for a cycle using BFS
    bool isCycle(int src, unordered_map<int, list<int>>& adj, vector<int>& visited) {
        queue<pair<int, int>> q;
        q.push({src, -1});  // Push the source node with no parent (-1)
        visited[src] = 1;

        while (!q.empty()) {
            int node = q.front().first;
            int parent = q.front().second;
            q.pop();

            for (auto adjNode : adj[node]) {
                if (!visited[adjNode]) {
                    visited[adjNode] = 1;
                    q.push({adjNode, node});
                } else if (adjNode != parent) {
                    // If an adjacent node is visited and it's not the parent, cycle detected
                    return true;
                }
            }
        }

        return false;
    }

public:
    vector<int> findRedundantConnection(vector<vector<int>>& edges) {
        unordered_map<int, list<int>> adj;
        int n = edges.size();
        vector<int> visited(n + 1, 0);  // Use n+1 since nodes are 1-indexed
        vector<int> ans;

        for (auto& edge : edges) {
            int u = edge[0];
            int v = edge[1];

            // Add the edge temporarily
            adj[u].push_back(v);
            adj[v].push_back(u);

            fill(visited.begin(), visited.end(), 0);  // Reset visited array before checking for a cycle

            // Check if this edge creates a cycle
            if (isCycle(u, adj, visited)) {
                ans = edge;  // If a cycle is detected, this is the redundant edge
                break;
            }
        }

        return ans;
    }
};


20121