<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>

class Solution {

    bool isValid(int y,int x,int r,int c){
        return y>=0&&y<r&&x>=0&&x<c;
    }

public:
    int minCost(vector<vector<int>>& grid) {
        
        int r = grid.size(), c = grid[0].size();

        int i,j,k;

        int dp[r][c][5];

        for(i=0;i<r;i++) {
            for(j=0;j<c;j++) {
                for(k=0;k<5;k++) {
                    dp[i][j][k] = INT_MAX;
                }
            }
        }
        priority_queue<vector<int>,vector<vector<int>>,greater<vector<int>>> pq;

        for(i=1;i<=4;i++) {
            if(i == grid[0][0]) {
                dp[0][0][i] = 0;
                pq.push({0,0,0,i});
            }
        }

        int dirs[5][2] = {{0,0},{0,1},{0,-1},{1,0},{-1,0}};
   
        while(!pq.empty()) {
            auto top = pq.top();pq.pop();
            int cost = top[0];
            int y = top[1];
            int x = top[2];
            int d = top[3];
            // cout<<y<<" "<<x<<" "<<d<<" "<<cost<<endl;;

            if(cost != dp[y][x][d]) continue;
            
            for(i=1;i<=4;i++) {
                if(d == i) {
                    int yy = y + dirs[i][0], xx = x + dirs[i][1];

                    if(isValid(yy,xx,r,c) && dp[yy][xx][grid[yy][xx]] > cost) {
                        dp[yy][xx][grid[yy][xx]] = cost;
                        pq.push({cost,yy,xx,grid[yy][xx]});
                    }
                }
            }

            for(i=1;i<=4;i++) {
                if(dp[y][x][i] > cost + 1) {
                    dp[y][x][i] = cost + 1;
                    pq.push({cost+1,y,x,i});
                }
            }

        }

        int res = INT_MAX;

        for(i=1;i<=4;i++) {
            res = min(res,dp[r-1][c-1][i]);
        }

        return res;
    }
};
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