<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:
        '''
        from collections import deque
        result = 0
        reached = {0}
        dq = deque(connections)
        while dq:
            cur = dq.popleft()
            if cur[1] in reached:
                reached.add(cur[0])
            elif cur[0] in reached:
                reached.add(cur[1])
                result +=1
            else: dq.append(cur)
        return result
        '''
        result = 0
        reached = {0}
        while len(reached) < n:
            new_cons = []
            for current in reversed(connections):
                if current[1] in reached:
                    reached.add(current[0])
                elif current[0] in reached:
                    reached.add(current[1])
                    result +=1
                else:
                    new_cons.append(current)
            connections = new_cons
        
        return result
        
        '''
        
        connected = set([0])
        ct = 0
        graph = {i: [] for i in range(n)} # all bridges
        directed = set()
        for u, v in connections:
            graph[u].append(v)
            graph[v].append(u)
            directed.add((u, v))

        def DFS(node):
            nonlocal ct
            for neighbor in graph[node]:
                if neighbor not in connected:
                    connected.add(neighbor)
                    # Check if we need to reorder (i.e., if the edge is in the wrong direction)
                    if (node, neighbor) in directed:
                        ct += 1
                    DFS(neighbor)

        # Start DFS from node 0
        DFS(0)
        return ct
        '''

# class Solution:
#     def minReorder(self, n: int, connections: List[List[int]]) -> int:
#         connected = set([0])
#         ct = 0

#         def seek(k):
#             global ct
#             for i in range(n-1):
#                 if connections[i][0] == k and (connections[i][1] not in connected): 
#                     connected.add(connections[i][1])
#                     ct += 1
#                     return connections[i][1]
#                 elif connections[i][1] == k and (connections[i][0] not in connected): 
#                     connected.add(connections[i][0])
#                     return connections[i][0]
                
#         def DFS(i):
#             DFS(seek(i))
#         DFS(0)
#         return ct
            
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