<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:
        row_length = len(grid)
        col_length = len(grid[0])
        visited = set()

        def viable_path(si, sj, sc, ei, ej, ec):
            if si is None and sj is None:
                return True
            if ei < 0 or ei >= row_length or ej < 0 or ej >= col_length:
                return False
            
            if sc == 1:
                if ej > sj and (ec == 1 or ec == 3 or ec == 5):
                    return True
                if ej < sj and (ec == 1 or ec == 4 or ec == 6):
                    return True
            if sc == 2:
                if ei > si and (ec == 2 or ec == 5 or ec == 6):
                    return True
                if ei < si and (ec == 2 or ec == 3 or ec == 4):
                    return True
            if sc == 3:
                if ei > si and (ec == 2 or ec == 5 or ec == 6):
                    return True
                if ej < sj and (ec == 1 or ec == 4 or ec == 6):
                    return True
            if sc == 4:
                if ei > si and (ec == 2 or ec == 5 or ec == 6):
                    return True
                if ej > sj and (ec == 1 or ec == 3 or ec == 5):
                    return True
            if sc == 5:
                if ej < sj and (ec == 6 or ec == 4 or ec == 1):
                    return True
                if ei < si and (ec == 3 or ec == 4 or ec == 2):
                    return True
            if sc == 6:
                if ej > sj and (ec == 5 or ec == 3 or ec == 1):
                    return True
                if ei < si and (ec == 4 or ec == 3 or ec == 2):
                    return True
            return False
                

        
        def dfs(pi, pj, pc, i, j):
            if i < 0 or i >= row_length or j < 0 or j >= col_length:
                return False

            if not viable_path(pi, pj, pc, i, j, grid[i][j]):
                return False

            if i == row_length - 1 and j == col_length - 1:
                return True

            if (i,j) in visited:
                return False

            cell = grid[i][j]

            visited.add((i,j))
            if cell == 1:
                return dfs(i, j, 1, i, j-1) or dfs(i, j, 1, i, j+1)
            if cell == 2:
                return dfs(i, j, 2, i+1, j) or dfs(i, j, 2, i-1, j)
            if cell == 3:
                return dfs(i, j, 3, i, j-1) or dfs(i, j, 3, i+1, j)
            if cell == 4:
                return dfs(i, j, 4, i+1, j) or dfs(i, j, 4, i, j+1)
            if cell == 5:
                return dfs(i, j, 5, i, j-1) or dfs(i, j, 5, i-1, j)
            if cell == 6:
                return dfs(i, j, 6, i-1, j) or dfs(i, j, 6, i, j+1)

        return dfs(None, None, None, 0,0)

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