<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>

# Approach a DFS solution to check if you can go from (0,0) to (m-1,n-1)

# Condition to go from a cell1 to cell2:
# cell1 leads to cell 2
# cell2 "Accepts" cell1


class Solution:
    def hasValidPath(self, grid: List[List[int]]) -> bool:
        vis = set()
        dest = (len(grid)-1, len(grid[0])- 1) 
        def getNeighbor(curr) -> list:
            left = (0, -1)
            right = (0, 1)
            up = (-1, 0)
            down = (1, 0)
            dirs = {
                1: [left, right],
                2: [up, down],
                3: [left, down],
                4: [right, down],
                5: [left, up],
                6: [right, up]
            }

            accepts = {
                left: [1,4,6],
                right:[1,3,5],
                up: [2,3,4],
                down:[2,5,6],
            }

    
            currStreet = grid[curr[0]][curr[1]]
            neighbors = []
            for direction in dirs[currStreet]:
                di, dj = direction
                newi = curr[0] + di
                newj = curr[1] + dj

                if (newi >= 0 and newi < len(grid) and newj >= 0 and newj < len(grid[0]) 
                    and grid[newi][newj] in accepts[direction]):
                    neighbors.append((newi, newj))

            return neighbors


        def dfs(curr) -> bool:
            vis.add(curr)
            if curr == dest:
                return True
            
            for neighbor in getNeighbor(curr):
                if neighbor not in vis:
                    if dfs(neighbor):
                        return True

            vis.remove(curr)
            return False


        return dfs((0,0))
            
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