<p>You are given an <code>m x n</code> <code>grid</code>. Each cell of <code>grid</code> represents a street. The street of <code>grid[i][j]</code> can be:</p>

<ul>
	<li><code>1</code> which means a street connecting the left cell and the right cell.</li>
	<li><code>2</code> which means a street connecting the upper cell and the lower cell.</li>
	<li><code>3</code> which means a street connecting the left cell and the lower cell.</li>
	<li><code>4</code> which means a street connecting the right cell and the lower cell.</li>
	<li><code>5</code> which means a street connecting the left cell and the upper cell.</li>
	<li><code>6</code> which means a street connecting the right cell and the upper cell.</li>
</ul>
<img alt="" src="https://assets.leetcode.com/uploads/2020/03/05/main.png" style="width: 450px; height: 708px;" />
<p>You will initially start at the street of the upper-left cell <code>(0, 0)</code>. A valid path in the grid is a path that starts from the upper left cell <code>(0, 0)</code> and ends at the bottom-right cell <code>(m - 1, n - 1)</code>. <strong>The path should only follow the streets</strong>.</p>

<p><strong>Notice</strong> that you are <strong>not allowed</strong> to change any street.</p>

<p>Return <code>true</code><em> if there is a valid path in the grid or </em><code>false</code><em> otherwise</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/03/05/e1.png" style="width: 455px; height: 311px;" />
<pre>
<strong>Input:</strong> grid = [[2,4,3],[6,5,2]]
<strong>Output:</strong> true
<strong>Explanation:</strong> As shown you can start at cell (0, 0) and visit all the cells of the grid to reach (m - 1, n - 1).
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/03/05/e2.png" style="width: 455px; height: 293px;" />
<pre>
<strong>Input:</strong> grid = [[1,2,1],[1,2,1]]
<strong>Output:</strong> false
<strong>Explanation:</strong> As shown you the street at cell (0, 0) is not connected with any street of any other cell and you will get stuck at cell (0, 0)
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> grid = [[1,1,2]]
<strong>Output:</strong> false
<strong>Explanation:</strong> You will get stuck at cell (0, 1) and you cannot reach cell (0, 2).
</pre>

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

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 300</code></li>
	<li><code>1 &lt;= grid[i][j] &lt;= 6</code></li>
</ul>

class UnionFind:
    def __init__(self):
        self.nodes = {}
        self.sizes = {}
    
    def parent(self, node):
        if node not in self.nodes:
            self.nodes[node] = node
            self.sizes[node] = 1
        
        while self.nodes[node] != node:
            parent = self.nodes[node]
            self.nodes[node] = self.nodes[parent]
            node = parent
        
        return node
    
    def union(self, node1, node2):
        p1 = self.parent(node1)
        p2 = self.parent(node2)

        if p1 == p2:
            return
        elif self.sizes[p1] < self.sizes[p2]:
            p1, p2 = p2, p1
        
        self.nodes[p2] = p1
        self.sizes[p1] += self.sizes[p2]
    
    def connected(self, node1, node2):
        return self.parent(node1) == self.parent(node2)

class Solution:
    def hasValidPath(self, grid: List[List[int]]) -> bool:
        n = len(grid)
        m = len(grid[0])

        adj = {
            1: [(0, 1), (0, -1)],
            2: [(1, 0), (-1, 0)],
            3: [(0, -1), (1, 0)],
            4: [(0, 1), (1, 0)],
            5: [(-1, 0), (0, -1)],
            6: [(-1, 0), (0, 1)],
        }

        def is_connected(c1, c2):
            x1, y1 = c1
            x2, y2 = c2

            neigh1 = set((x1 + dx, y1 + dy) for dx, dy in adj[grid[x1][y1]])
            neigh2 = set((x2 + dx, y2 + dy) for dx, dy in adj[grid[x2][y2]])

            return c2 in neigh1 and c1 in neigh2
        
        def is_valid_cell(c):
            x, y = c
            return 0 <= x < n and 0 <= y < m

        def adjacent_cells(c):
            x, y = c
            return filter(is_valid_cell, ((x + dx, y + dy) for dx, dy in [(1, 0), (-1, 0), (0, 1), (0, -1)]))
        
        uf = UnionFind()
        for x in range(n):
            for y in range(m):
                for neigh in adjacent_cells((x, y)):
                    if is_connected((x, y), neigh):
                        uf.union((x, y), neigh)

        return uf.connected((0, 0), (n - 1, m - 1))
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