<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:

        # method: try to use hashmap 
        graph = {}
        def add_edge(i, j):
            if i in graph: 
                graph[i].add(j)
            else:
                graph[i] = set([j])
        for i, j in connections:
            add_edge(i, j)
            add_edge(j, i)

        path = set()
        visited = set()
        seen = set([0])
        while seen:
            node = seen.pop()
            neighs = graph[node]
            new = neighs - visited - seen
            if new:
                x = new.pop()
                seen.add(node)
                seen.add(x)
                path.add((node, x))
            else:
                visited.add(node)

        res = 0
        for i, j in connections:
            if (i, j) in path:
                res += 1
        
        return res

                








        # # method: only use queue -> too slow
        # # potentially every loop only fix 1 elements -> O(n^2)
        # res = 0
        # comp = set([0])
        # from collections import deque
        # comp_connections = deque(connections)
        # while len(comp) != n:
        #     i, j = comp_connections.popleft()
        #     if i in comp and j in comp:
        #         continue
        #     if i in comp:
        #         res += 1
        #         comp.add(j)
        #     elif j in comp:
        #         comp.add(i)
        #     else:
        #         comp_connections.append([i,j])
        # return res




            




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