<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:
        # dirs
        mid = [1,1]
        up = [0,1]
        down = [2,1]
        left = [1,0]
        right = [1,2]
        street_map = {1:[left, mid, right], 2: [up, mid, down],
                    3:[left, mid, down], 4:[down,mid,right], 5: [left, mid, up],
                    6:[up, mid, right]}
        # print(street_map)
        row = len(grid)
        col = len(grid[0])
        def dfs(i, j, matrix):
            if i<0 or j<0 or i>=row*3 or j>=col*3 or matrix[i][j]==0 or matrix[i][j]==2:
                return False
            if i == row*3-2 and j == col*3-2:
                return True
            flag = False # default can't reach, but if there's 1 way reach true, return T
            # mark checked road
            matrix[i][j] = 2
            flag |= dfs(i+1, j, matrix)
            flag |= dfs(i-1, j, matrix)
            flag |= dfs(i, j+1, matrix)
            flag |= dfs(i, j-1, matrix)
            matrix[i][j] = 1
            return flag
        check = [[0 for i in range(col*3)] for j in range(row*3)]
        for i in range(row):
            for j in range(col):
                cur_dir = grid[i][j]
                # all box goes to 1 if it's road
                mark_dir = street_map.get(cur_dir)
                # print(f"mark all {mark_dir}")
                for each in mark_dir:
                    new_i = i*3+each[0]
                    new_j = j*3+each[1]
                    check[new_i][new_j] = 1
        # print(check)
        return dfs(1,1, check)
        
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