<p>There is a <strong>bi-directional</strong> graph with <code>n</code> vertices, where each vertex is labeled from <code>0</code> to <code>n - 1</code> (<strong>inclusive</strong>). The edges in the graph are represented as a 2D integer array <code>edges</code>, where each <code>edges[i] = [u<sub>i</sub>, v<sub>i</sub>]</code> denotes a bi-directional edge between vertex <code>u<sub>i</sub></code> and vertex <code>v<sub>i</sub></code>. Every vertex pair is connected by <strong>at most one</strong> edge, and no vertex has an edge to itself.</p>

<p>You want to determine if there is a <strong>valid path</strong> that exists from vertex <code>source</code> to vertex <code>destination</code>.</p>

<p>Given <code>edges</code> and the integers <code>n</code>, <code>source</code>, and <code>destination</code>, return <code>true</code><em> if there is a <strong>valid path</strong> from </em><code>source</code><em> to </em><code>destination</code><em>, or </em><code>false</code><em> otherwise</em><em>.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/08/14/validpath-ex1.png" style="width: 141px; height: 121px;" />
<pre>
<strong>Input:</strong> n = 3, edges = [[0,1],[1,2],[2,0]], source = 0, destination = 2
<strong>Output:</strong> true
<strong>Explanation:</strong> There are two paths from vertex 0 to vertex 2:
- 0 &rarr; 1 &rarr; 2
- 0 &rarr; 2
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/08/14/validpath-ex2.png" style="width: 281px; height: 141px;" />
<pre>
<strong>Input:</strong> n = 6, edges = [[0,1],[0,2],[3,5],[5,4],[4,3]], source = 0, destination = 5
<strong>Output:</strong> false
<strong>Explanation:</strong> There is no path from vertex 0 to vertex 5.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 2 * 10<sup>5</sup></code></li>
	<li><code>0 &lt;= edges.length &lt;= 2 * 10<sup>5</sup></code></li>
	<li><code>edges[i].length == 2</code></li>
	<li><code>0 &lt;= u<sub>i</sub>, v<sub>i</sub> &lt;= n - 1</code></li>
	<li><code>u<sub>i</sub> != v<sub>i</sub></code></li>
	<li><code>0 &lt;= source, destination &lt;= n - 1</code></li>
	<li>There are no duplicate edges.</li>
	<li>There are no self edges.</li>
</ul>

class DSU():

    def __init__(self, x):
        self.vertex = x
        self.parent = self
        self.rank = 0

    def findRoot(self):
        if self.parent == self:
            return self
        else:
            self.parent = self.parent.findRoot()
            return self.parent

    def merge(self, y):
        rootX = self.findRoot()
        rootY = y.findRoot()

        if rootX.rank > rootY.rank:
            rootY.parent = rootX
        elif rootX.rank < rootY.rank:
            rootX.parent = rootY
        else:
            rootX.parent = rootY
            rootY.rank += 1


class Solution:
    def validPath(self, n: int, edges: List[List[int]], source: int, destination: int) -> bool:
        
        # Через DSU()

        # Назначаем Vertices
        vr = [DSU(x) for x in range(n)]

        # Прописываем рёбра
        for x, y in edges:
            if vr[x].findRoot() != vr[y].findRoot():
                vr[x].merge(vr[y])
        
        return vr[source].findRoot() == vr[destination].findRoot()


        # Стандартный обход пожаром. Довольно долго, 2499 ms
        # Составляем вершины
        vertices = [list() for _ in range(n)]
        for u, v in edges:
            vertices[u].append(v)
            vertices[v].append(u)

        # Делаем путь
        tovisitS = set([source])
        visitedS = set()

        while tovisitS:
            vertex = tovisitS.pop()
            visitedS.add(vertex)
            if vertex == destination:
                return True
            for direction in vertices[vertex]:
                if direction not in visitedS and direction not in tovisitS:
                    tovisitS.add(direction)

        return False

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