<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 minReorder(self, n: int, connections: List[List[int]]) -> int:
        
        # we have 0th node to n-1st node

        # connection represents
        # [cityA to cityB]
        # [source to destination]

        # there is an event at capital city 0

        # we have to re orient the roads such that 
        # all roads can lead to city 0

        # first we must be able to traverse the nodes
        # such that we can expand outwards and 
        # sequentially correct the order incase it is 
        # in the wrong order

        # connection graph
        conn_graph = {}

        graph = {}

        # lets generated the graph first
        for connection in connections:

            source = connection[0]
            destination = connection[1]

            if source not in graph:
                graph[source] = set()

            if source not in conn_graph:
                conn_graph[source] = set()

            if destination not in graph:
                graph[destination] = set()
            
            graph[source].add(destination)
            conn_graph[source].add(destination)

            graph[destination].add(source)

        queue = deque()

        visited = set()

        # start from node 0
        queue.appendleft(0)

        toChange = 0

        while len(queue) > 0:

            node = queue.pop()

            if node in visited:
                continue
            
            visited.add(node)

            # get neighbors
            neighbors = graph[node]

            for neighbor in neighbors:
                if neighbor not in visited:
                    # if this is not present in the connection graph
                    if neighbor not in conn_graph or node not in conn_graph[neighbor]:
                        toChange += 1
                    queue.appendleft(neighbor)

        return toChange
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