<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 {

    // Directions:
    // -1: start
    // 0: left
    // 1: top
    // 2: right
    // 3: bottom
    bool dfs(
        int row,
        int col,
        int from_direction, // Informs us which direction we came from
        vector<vector<int>> &grid,
        vector<vector<bool>> &visited) 
    {
        // If we go out of bounds, then immediately end the current path
        if (row >= grid.size() || col >= grid[0].size())
            return false;
        
        // Don't revisit the same tile
        if (visited[row][col] && (row != grid.size() - 1 || col != grid[0].size() - 1))
            return false;
        visited[row][col] = true;

        int type = grid[row][col];

        // We can only accept incoming roads from certain directions
        switch (from_direction) {
            case 0:
                if (type == 2 || type == 4 || type == 6)
                    return false;
                break;
            case 1:
                if (type == 1 || type == 3 || type == 4)
                    return false;
                break;
            case 2:
                if (type == 2 || type == 3 || type == 5)
                    return false;
                break;
            case 3:
                if (type == 1 || type == 5 || type == 6)
                    return false;
                break;
        }

        // Return true to indicate that our path brought us to the bottom right tile
        if (row == grid.size() - 1 && col == grid[0].size() - 1)
            return true;

        // Depending on the type of tile, we can only search in two directions
        switch (type) {
            case 1:
                return dfs(row, col+1, 0, grid, visited)
                    || dfs(row, col-1, 2, grid, visited);
            case 2:
                return dfs(row+1, col, 1, grid, visited)
                    || dfs(row-1, col, 3, grid, visited);
            case 3:
                return dfs(row+1, col, 1, grid, visited)
                    || dfs(row, col-1, 2, grid, visited);
            case 4:
                return dfs(row+1, col, 1, grid, visited)
                    || dfs(row, col+1, 0, grid, visited);
            case 5:
                return dfs(row-1, col, 3, grid, visited)
                    || dfs(row, col-1, 2, grid, visited);
            default:
                return dfs(row-1, col, 3, grid, visited)
                    || dfs(row, col+1, 0, grid, visited);
        }
    }

public:
    bool hasValidPath(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();
        if (m == 1 && n == 1) return true;

        vector<vector<bool>> visited(m, vector<bool>(n));
            
        return dfs(0, 0, -1, grid, visited);
    }
};
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