<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 {
public:

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


int minCost(vector<vector<int>>& grid) {
        int  n=grid.size();
        int m=grid[0].size();
        if(m==1 && n==1){
            return 0;
        }
        map<pair<int,int>,vector<pair<int,pair<int,int>>>>mpp;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(grid[i][j]==3){
                    mpp[{i, j}].push_back({1, {1, 0}});
                    mpp[{i, j}].push_back({0, {-1, 0}});
                    mpp[{i, j}].push_back({0, {0, 1}});
                    mpp[{i, j}].push_back({0, {0, -1}});         
                }
                if(grid[i][j]==4){
                    mpp[{i, j}].push_back({0, {1, 0}});
                    mpp[{i, j}].push_back({1, {-1, 0}});
                    mpp[{i, j}].push_back({0, {0, 1}});
                    mpp[{i, j}].push_back({0, {0, -1}});                
                }
                if(grid[i][j]==1){
                    mpp[{i, j}].push_back({0, {1, 0}});
                    mpp[{i, j}].push_back({0, {-1, 0}});
                    mpp[{i, j}].push_back({1, {0, 1}});
                    mpp[{i, j}].push_back({0, {0, -1}});                
                }
                if(grid[i][j]==2){
                    mpp[{i, j}].push_back({0, {1, 0}});
                    mpp[{i, j}].push_back({0, {-1, 0}});
                    mpp[{i, j}].push_back({0, {0, 1}});
                    mpp[{i, j}].push_back({1, {0, -1}});                  
                }
            }
        }
        priority_queue<pair<int,pair<int,int>>,vector<pair<int,pair<int,int>>>,greater<pair<int,pair<int,int>>>>pq;
        int dis[n][m];
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                dis[i][j]=1e9;
            }
        }
        pq.push({0,{0,0}});
        while(!pq.empty()){
            auto it=pq.top();
            pq.pop();
            int curr_dis=it.first;
            int a=it.second.first;
            int b=it.second.second;
            for(auto it:mpp[{a,b}]){
                int x=it.second.first;
                int y=it.second.second;
                int change=it.first;
                if( isValid(a+x,b+y,n,m) &&curr_dis+!change<dis[a+x][b+y]){
                    dis[a+x][b+y]=curr_dis+!change;
                    pq.push({dis[a+x][b+y],{a+x,b+y}});
                }
            }

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