Minimum Cost to Make at Least One Valid Path in a Grid
<p>Given an <code>m x n</code> grid. Each cell of the grid has a sign pointing to the next cell you should visit if you are currently in this cell. The sign of <code>grid[i][j]</code> can be:</p>

<ul>
	<li><code>1</code> which means go to the cell to the right. (i.e go from <code>grid[i][j]</code> to <code>grid[i][j + 1]</code>)</li>
	<li><code>2</code> which means go to the cell to the left. (i.e go from <code>grid[i][j]</code> to <code>grid[i][j - 1]</code>)</li>
	<li><code>3</code> which means go to the lower cell. (i.e go from <code>grid[i][j]</code> to <code>grid[i + 1][j]</code>)</li>
	<li><code>4</code> which means go to the upper cell. (i.e go from <code>grid[i][j]</code> to <code>grid[i - 1][j]</code>)</li>
</ul>

<p>Notice that there could be some signs on the cells of the grid that point outside the grid.</p>

<p>You will initially start at 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> following the signs on the grid. The valid path does not have to be the shortest.</p>

<p>You can modify the sign on a cell with <code>cost = 1</code>. You can modify the sign on a cell <strong>one time only</strong>.</p>

<p>Return <em>the minimum cost to make the grid have at least one valid path</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/1300-1399/1368.Minimum%20Cost%20to%20Make%20at%20Least%20One%20Valid%20Path%20in%20a%20Grid/images/grid1.png" style="width: 400px; height: 390px;" />
<pre>
<strong>Input:</strong> grid = [[1,1,1,1],[2,2,2,2],[1,1,1,1],[2,2,2,2]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> You will start at point (0, 0).
The path to (3, 3) is as follows. (0, 0) --&gt; (0, 1) --&gt; (0, 2) --&gt; (0, 3) change the arrow to down with cost = 1 --&gt; (1, 3) --&gt; (1, 2) --&gt; (1, 1) --&gt; (1, 0) change the arrow to down with cost = 1 --&gt; (2, 0) --&gt; (2, 1) --&gt; (2, 2) --&gt; (2, 3) change the arrow to down with cost = 1 --&gt; (3, 3)
The total cost = 3.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1300-1399/1368.Minimum%20Cost%20to%20Make%20at%20Least%20One%20Valid%20Path%20in%20a%20Grid/images/grid2.png" style="width: 350px; height: 341px;" />
<pre>
<strong>Input:</strong> grid = [[1,1,3],[3,2,2],[1,1,4]]
<strong>Output:</strong> 0
<strong>Explanation:</strong> You can follow the path from (0, 0) to (2, 2).
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1300-1399/1368.Minimum%20Cost%20to%20Make%20at%20Least%20One%20Valid%20Path%20in%20a%20Grid/images/grid3.png" style="width: 200px; height: 192px;" />
<pre>
<strong>Input:</strong> grid = [[1,2],[4,3]]
<strong>Output:</strong> 1
</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;= 100</code></li>
	<li><code>1 &lt;= grid[i][j] &lt;= 4</code></li>
</ul>
Breadth-First Search; Graph; Array; Matrix; Shortest Path; Heap (Priority Queue)
func minCost(grid [][]int) int {
	m, n := len(grid), len(grid[0])
	q := doublylinkedlist.New()
	q.Add([]int{0, 0, 0})
	dirs := [][]int{{0, 0}, {0, 1}, {0, -1}, {1, 0}, {-1, 0}}
	vis := make([][]bool, m)
	for i := range vis {
		vis[i] = make([]bool, n)
	}
	for !q.Empty() {
		v, _ := q.Get(0)
		p := v.([]int)
		q.Remove(0)
		i, j, d := p[0], p[1], p[2]
		if i == m-1 && j == n-1 {
			return d
		}
		if vis[i][j] {
			continue
		}
		vis[i][j] = true
		for k := 1; k <= 4; k++ {
			x, y := i+dirs[k][0], j+dirs[k][1]
			if x >= 0 && x < m && y >= 0 && y < n {
				if grid[i][j] == k {
					q.Insert(0, []int{x, y, d})
				} else {
					q.Add([]int{x, y, d + 1})
				}
			}
		}
	}
	return -1
}