<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>

from collections import deque

class Solution:
    # key idea: we can check if we can reach 
    def hasValidPath(self, grid: List[List[int]]) -> bool:
        start = (0, 0)
        rows, columns = len(grid), len(grid[0])

        paths = {
            1: ['l', 'r'],
            2: ['u', 'd'],
            3: ['d', 'l'],
            4: ['d', 'r'],
            5: ['u', 'l'],
            6: ['u', 'r'],
        }
        # coordinates to next
        direction = {
            'l': (0, -1),
            'r': (0, 1),
            'u': (-1, 0),
            'd': (1, 0),
        }
        opposites = {
            'l': 'r',
            'r': 'l',
            'u': 'd',
            'd': 'u',
        }
        visited = set()

        def dfs(cur: tuple) -> bool:
            r, c = cur
            # check we reached the end:
            if r == rows-1 and c == columns-1:
                return True
            # check in bounds
            if r < 0 or c < 0:
                return False
            if r == rows or c == columns:
                return False
            # mark cur as visited
            visited.add(cur)
            street = grid[r][c]

            for cur_path in paths[street]:
                dr, dc = direction[cur_path]
                nextr, nextc = r+dr, c+dc
                if nextr < 0 or nextr == rows:
                    continue
                if nextc < 0 or nextc == columns:
                    continue
                
                next_node = (nextr, nextc)
                # check if we visited "next" already
                if next_node in visited:
                    continue

                # check that we can actually reach next
                next_street = grid[nextr][nextc]
                # need to use opposites of current "path" 
                # ie if next street 4 goes down and left
                # can reach next next if path is up or right
                if opposites[cur_path] not in paths[next_street]:
                    continue
                # going from
                # if we can reach the end by going on this path, no need to evaluate the other path
                if dfs(next_node):
                    return True
                
            return False

        return dfs(start)
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