<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 {
private:
    std::unordered_map<int, std::vector<std::pair<int, int>>> graph;
    int dfs(int node, int prevNode) {
        int flips = 0;
        for (int i = 0; i < graph[node].size(); i++) {
            int nextNode = graph[node][i].first;
            if (nextNode != prevNode) {
                flips += graph[node][i].second + dfs(nextNode, node);
            }
        }

        return flips;
    }
public:
    int minReorder(int n, vector<vector<int>>& connections) {
        /**
        We are given n nodes and n - 1 edges representing a graph of cities
        We need to find the find the minimum number of edges flipped such that
        each city has a path to the capital
        It may be helpful to first identify all the cities that already have a path to the capital
        We may be able to solve this problem with a DFS
        We can start at the capital as this node should be our root
        We then perform a DFS, if we can't traverse to the next node because it is headed towards us,
        we flip the edge and increment our counter by 1
        */
        for (auto connection : connections) {
            int a = connection[0], b = connection[1];
            if (!graph.contains(a)) {
                graph[a] = std::vector<std::pair<int, int>>();
            }
            if (!graph.contains(b)) {
                graph[b] = std::vector<std::pair<int, int>>();
            }

            graph[a].push_back(std::pair<int, int> {b, 1});
            graph[b].push_back(std::pair<int, int> {a, 0});
        }

        return dfs(0, -1);
    }
};
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