<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://assets.leetcode.com/uploads/2020/02/13/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://assets.leetcode.com/uploads/2020/02/13/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://assets.leetcode.com/uploads/2020/02/13/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>


struct Node{
    int x, y;
    // Define operator< for using Node as a key in map
    bool operator<(const Node& other) const {
        return tie(x, y) < tie(other.x, other.y);
    }
};

struct Node2{
    int dist, x, y;
    // Define operator< for using Node2 in priority_queue with greater<>
    bool operator<(const Node2& other) const {
        return dist > other.dist; // Inverted for min-heap
    }
};

class Solution {
public:
    int n, m;
    int minCost(vector<vector<int>>& grid) {
        n = grid.size();
        m = grid[0].size();

        map<Node, vector<pair<Node, int>>> mp;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (i)     mp[{i, j}].push_back({{i-1, j}, grid[i][j] == 4 ? 0 : 1});
                if (j)     mp[{i, j}].push_back({{i, j-1}, grid[i][j] == 2 ? 0 : 1});
                if (i+1 < n) mp[{i, j}].push_back({{i+1, j}, grid[i][j] == 3 ? 0 : 1});
                if (j+1 < m) mp[{i, j}].push_back({{i, j+1}, grid[i][j] == 1 ? 0 : 1});
            }
        }

        priority_queue<Node2> pq;

        vector<vector<int>> dist(n, vector<int>(m, 1e9));
        pq.push({0, 0, 0});
        dist[0][0] = 0;

        while (!pq.empty()) {
            auto node = pq.top();
            pq.pop();
            int dis = node.dist, x = node.x, y = node.y;

            for (auto p : mp[{x, y}]) {
                Node child = p.first;
                int w = p.second;

                if (dis + w < dist[child.x][child.y]) {
                    dist[child.x][child.y] = dis + w;
                    pq.push({dis + w, child.x, child.y});
                }
            }
        }

        return (dist[n-1][m-1] == 1e9 ? -1 : dist[n-1][m-1]);
    }
};

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