<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 Solution:
    def hasValidPath(self, grid: List[List[int]]) -> bool:

        path = [[],
                [False, True, False, True],
                [True, False, True, False],
                [False, False, True, True],
                [False, True, True, False],
                [True, False, False, True],
                [True, True, False, False]]
            
        
        dir = [[-1,0], [0,1],[1,0],[0,-1]]


        m = len(grid)
        n = len(grid[0])
        djs = disjointSet(m*n)

        for i in range(m):
            for j in range(n):

                p = grid[i][j]

                for k in range(len(dir)):
                    dt = dir[k]
                    xx = dt[0] + i
                    yy = dt[1] + j
                    
                    if self.is_valid(xx,yy,m,n):
                        q = grid[xx][yy]
                        
                        if path[p][k] and path[q][(k+2)%4]:
                            
                            djs.union(i*n+j, xx*n+yy)
        
    
        return djs.connected(0, m*n-1)

        

    def is_valid(self,x,y,m,n):
        return x >= 0 and x <m and y >= 0 and y<n


class disjointSet:

    def __init__(self, n):
        self.root = [i for i in range(n)]
        self.rank = [0 for i in range(n)]
    

    def find(self, x):

        if self.root[x] == x:
            return x
        

        self.root[x] = self.find(self.root[x])

        return self.root[x]

    

    def union(self, a, b):

        A  = self.find(a)
        B = self.find(b)

        if not A == B:

            if self.rank[A] > self.rank[B]:
                self.root[B] = A
            
            elif self.rank[B] > self.rank[A]:
                self.root[A] = B
            else:
                self.root[A] = B
                self.rank[B] += 1
            
    def connected(self, a, b):
        return self.find(a) == self.find(b)
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