<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:
        
        prev_map_f = {1:{1,3,5}, 2:{2, 5, 6}, 3:{2,5,6}, 4:{1,3,5}, 5:{3, 2,4}, 6:{5,3,1}}
        prev_map_b = {1:{1,4,6}, 2:{2,3,4}, 3:{1,4,6}, 4:{2,5,6}, 5:{1,4,6}, 6:{4,2,3}}
        ROW = len(grid)
        COL = len(grid[0])
        visited = set()
        def dfs(r, c, prev, dir):
            if r == ROW or r < 0 or c == COL or c < 0 or (r,c) in visited:
                return False
            num = grid[r][c]
            if r == 0 and c == 0:
                pass
            else:
                if dir == 1:
                    if num not in prev_map_f[prev]:
                        print(f"{r} {c} {num} {prev} {num}")
                        return False
                else:
                    if num not in prev_map_b[prev]:
                        print(f"{r} {c} {grid[r][c]} {prev}")
                        return False

            if r == ROW - 1 and c == COL - 1:
                return True
            visited.add((r,c))
            if num == 1:
                if dfs(r, c+1, num, 1):
                    return True
                if dfs(r, c-1, num, 0):
                    return True
            elif num == 2:
                if dfs(r+1, c, num, 1):
                    return True
                if dfs(r-1, c, num, 0):
                    return True
            elif num == 3:
                if dfs(r+1, c, num, 1):
                    return True
                if dfs(r, c-1, num, 0):
                    return True
            elif num == 4:
                if dfs(r+1, c, num, 0):
                    return True
                if dfs(r, c+1, num, 1):
                    return True
            elif num == 5:
                if dfs(r-1, c, num, 1):
                    return True
                if dfs(r, c-1, num, 0):
                    return True
            else:
                if dfs(r-1, c, num, 0):
                    return True
                if dfs(r, c+1, num, 1):
                    return True
        return dfs(0,0,-1, 1)

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