<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>

"""
Old Notes:

- If the city itself is a neighbour, we reverse the edge
or route from the source city to be the destination city.
- Keep reversing and accumulating this sum of reversals
until the entirety of the search space is exhausted.
- Search space is exhausted once the depth first search is
complete, i.e., this while loop is exited.
            

Alternate approach - treat this as an undirected graph. Examine
all of the edges. Edges in which the node is destination rather
than the source/origin are flipped.

Is the city itself a neighbour of others (aside from being as such 
as a result of our depth first search)?
"""

class Solution:
    def minReorder(self, n: int, connections: List[List[int]]) -> int:
        # Undirected graph, which would allow the agent
        # to reach every city through a DFS
        graph = {a:[] for a, _ in zip(*list(zip(*connections)))}
        graph.update({b:[] for _, b in zip(*list(zip(*connections)))})
        for i in range(len(connections)):
            a, b = connections[i]
            graph[a].append(b)
            graph[b].append(a)
            connections[i] = tuple(connections[i])

        """
        Graph indicating reverse edges.
        For a key or city n_1, any set of values indicate 
        reverse edges from these cities.
        """
        connections_set = {*connections}

        visited = set()
        visited.add(0)
        stack = [(0, 0)]
        edge_reversal_count = 0
        prev = 0

        while len(stack) != 0:
            current = stack.pop()
            print(current) # Process current city
            for neighbour in graph[current[1]]:
                print(graph[current[1]])
                if neighbour not in visited:
                    visited.add(current[1]) # Add to visited set
                    stack.append((current[1], neighbour))

            if current in connections_set:
                edge_reversal_count += 1
    
        return edge_reversal_count



                










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