<p>There are <code>n</code> cities numbered from <code>0</code> to <code>n - 1</code> and <code>n - 1</code> roads such that there is only one way to travel between two different cities (this network form a tree). Last year, The ministry of transport decided to orient the roads in one direction because they are too narrow.</p>

<p>Roads are represented by <code>connections</code> where <code>connections[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> represents a road from city <code>a<sub>i</sub></code> to city <code>b<sub>i</sub></code>.</p>

<p>This year, there will be a big event in the capital (city <code>0</code>), and many people want to travel to this city.</p>

<p>Your task consists of reorienting some roads such that each city can visit the city <code>0</code>. Return the <strong>minimum</strong> number of edges changed.</p>

<p>It&#39;s <strong>guaranteed</strong> that each city can reach city <code>0</code> after reorder.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/05/13/sample_1_1819.png" style="width: 311px; height: 189px;" />
<pre>
<strong>Input:</strong> n = 6, connections = [[0,1],[1,3],[2,3],[4,0],[4,5]]
<strong>Output:</strong> 3
<strong>Explanation: </strong>Change the direction of edges show in red such that each node can reach the node 0 (capital).
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/05/13/sample_2_1819.png" style="width: 509px; height: 79px;" />
<pre>
<strong>Input:</strong> n = 5, connections = [[1,0],[1,2],[3,2],[3,4]]
<strong>Output:</strong> 2
<strong>Explanation: </strong>Change the direction of edges show in red such that each node can reach the node 0 (capital).
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 3, connections = [[1,0],[2,0]]
<strong>Output:</strong> 0
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= n &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>connections.length == n - 1</code></li>
	<li><code>connections[i].length == 2</code></li>
	<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt;= n - 1</code></li>
	<li><code>a<sub>i</sub> != b<sub>i</sub></code></li>
</ul>

class Solution {
public:
    int minReorder(int n, vector<vector<int>>& connections) {
        vector<int> g[n];
        vector<int> temp[n];

        for(int i=0;i<connections.size();i++){
            g[connections[i][0]].push_back(connections[i][1]);
            temp[connections[i][0]].push_back(connections[i][1]);
            temp[connections[i][1]].push_back(connections[i][0]);
        }

        queue<pair<int, int>> q;
        q.push({0, -1});
        vector<int> vis(n, 0);
        vector<int> par[n];
        while(!q.empty()) {
            pair<int, int> p = q.front();
            if(p.second != -1) {
                par[p.second].push_back(p.first);
            }
            vis[p.first] = 1;
            q.pop();
            for(auto &it : temp[p.first]) {
                if(!vis[it]) {
                    q.push({it, p.first});

                }
            }
        }

        queue<int> q1;
        vector<int> vis1(n, 0);
        q1.push(0);
        int ans=0;
        while(!q1.empty()) {
            int val = q1.front();
            vis1[val] = 1;
            q1.pop();

            for(auto &it : par[val]) {
                int fl=0;
                for(auto &it1 : g[it]) {
                    if(it1 == val) {
                        fl=1;
                        break;
                    } 
                }

                if(!fl) ans++;

                if(!vis1[it]) {
                    q1.push(it);
                }
            }
        }

        return ans;
    }
};
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