<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:
    # def dfs(self, adj, visited, minChange, city):
    #     connections = adj[city]
    #     for neighbor, connection in connections:
    #         print(neighbor, connection)
    #         if neighbor not in visited:
    #             visited.add(neighbor)
    #             if connection == 1:
    #                 minChange += 1
    #             print(minChange)
    #             minChange = self.dfs(adj, visited, minChange, neighbor)
    #     return minChange

    # def minReorder(self, n: int, connections: List[List[int]]) -> int:
    #     adj = [[] for _ in range(n)]
    #     for fromCity, toCity in connections:
    #         adj[fromCity].append([toCity, 1])
    #         adj[toCity].append([fromCity, -1])
    #     print(adj)
    #     minChange = 0
    #     visited = set()
    #     minChange = self.dfs(adj, visited, minChange, 0)
    #     return minChange

    # Exp
    def dfs(self, adj: List[List[Tuple[int, int]]], visited: List[bool], minChange: List[int], currCity: int) -> None:
        visited[currCity] = True
        for neighbourCity in adj[currCity]:
            if not visited[neighbourCity[0]]:
                if neighbourCity[1] == 1:
                    minChange[0] += 1
                self.dfs(adj, visited, minChange, neighbourCity[0])
    
    def minReorder(self, n: int, connections: List[List[int]]) -> int:
        adj = [[] for _ in range(n)]
        for connection in connections:
            adj[connection[0]].append((connection[1], 1))
            adj[connection[1]].append((connection[0], -1))
        visited = [False] * n
        minChange = [0]
        self.dfs(adj, visited, minChange, 0)
        return minChange[0]
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