Check if There is a Valid Path in a Grid
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1300-1399/1391.Check%20if%20There%20is%20a%20Valid%20Path%20in%20a%20Grid/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1300-1399/1391.Check%20if%20There%20is%20a%20Valid%20Path%20in%20a%20Grid/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1300-1399/1391.Check%20if%20There%20is%20a%20Valid%20Path%20in%20a%20Grid/images/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>
Depth-First Search; Breadth-First Search; Union Find; Array; Matrix
func hasValidPath(grid [][]int) bool {
	m, n := len(grid), len(grid[0])
	p := make([]int, m*n)
	for i := range p {
		p[i] = i
	}
	var find func(x int) int
	find = func(x int) int {
		if p[x] != x {
			p[x] = find(p[x])
		}
		return p[x]
	}
	left := func(i, j int) {
		if j > 0 && (grid[i][j-1] == 1 || grid[i][j-1] == 4 || grid[i][j-1] == 6) {
			p[find(i*n+j)] = find(i*n + j - 1)
		}
	}
	right := func(i, j int) {
		if j < n-1 && (grid[i][j+1] == 1 || grid[i][j+1] == 3 || grid[i][j+1] == 5) {
			p[find(i*n+j)] = find(i*n + j + 1)
		}
	}
	up := func(i, j int) {
		if i > 0 && (grid[i-1][j] == 2 || grid[i-1][j] == 3 || grid[i-1][j] == 4) {
			p[find(i*n+j)] = find((i-1)*n + j)
		}
	}
	down := func(i, j int) {
		if i < m-1 && (grid[i+1][j] == 2 || grid[i+1][j] == 5 || grid[i+1][j] == 6) {
			p[find(i*n+j)] = find((i+1)*n + j)
		}
	}
	for i, row := range grid {
		for j, e := range row {
			if e == 1 {
				left(i, j)
				right(i, j)
			} else if e == 2 {
				up(i, j)
				down(i, j)
			} else if e == 3 {
				left(i, j)
				down(i, j)
			} else if e == 4 {
				right(i, j)
				down(i, j)
			} else if e == 5 {
				left(i, j)
				up(i, j)
			} else {
				right(i, j)
				up(i, j)
			}
		}
	}
	return find(0) == find(m*n-1)
}