<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 find(self,rep,  x):
        root = x
        while root != rep[root]:
            rep[root] = rep[rep[root]]
            root = rep[root]
        
        while x != root:
            rep[x], x = root, rep[x]           
            
        return root

    def union(self,rep, count, x, y):
        rep_x = self.find(rep, x)
        rep_y = self.find(rep, y)

        if rep_x != rep_y:
            if count[rep_x] > count[rep_y]:
                rep[rep_y] = rep_x
                count[rep_x] += count[rep_y]
            else:
                rep[rep_x] = rep_y
                count[rep_y] += count[rep_x]

    def connected(self,rep, x, y):
        return self.find(rep,x) == self.find(rep,y)

    def hasValidPath(self, grid: List[List[int]]) -> bool:
        directions =  {1:(0, 1, 0, 1), 2:(1, 0, 1, 0), 3:(0, 0, 1, 1), 4:(0, 1, 1, 0), 5:(1, 0, 0, 1), 6:(1, 1, 0, 0)}
        length_row = len(grid)
        length_col = len(grid[0])
        rep = {(i, j) : (i, j) for i in range(length_row) for j in range(length_col)}
        count = {(i, j) : 1 for i in range(length_row) for j in range(length_col)}
      
        def inbound(row, col):
            return (0 <= row < len(grid)) and (0 <= col < len(grid[0]))
        
        for row in range(length_row):
            for col in range(length_col):
                new_row = row + 1
                new_col = col + 1
                if inbound(new_row, col) and directions[grid[row][col]][2] and directions[grid[new_row][col]][0]:
                    self.union(rep,count, (row, col), (new_row, col))
                if inbound(row, new_col) and directions[grid[row][col]][1] and directions[grid[row][new_col]][3]:
                    self.union(rep,count, (row, col), (row, new_col))


        answer = self.connected(rep, (0, 0), (length_row-1, length_col-1))

        return answer
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