<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:
    int helper1(int &i,int &j,int &n,int &m){
        return i*m+j;
    }

    bool isValid(int &i,int &j,int &n,int &m){
        if((i<0 || j<0) || (i>=n || j>=m))return false;
        return true;
    }

    void helper2(vector<vector<pair<int,int>>>&v,vector<int>&dr,vector<int>&dc,int &a,int x,int &i,int &j,
    int &n,int &m){
        for(int k=0;k<4;k++){
            int z=i+dr[k];
            int y=j+dc[k];
            int b=helper1(z,y,n,m);
            if(isValid(z,y,n,m)==false)continue;
            if(x==1){
                if(z==i && y==j+1){
                    v[a].push_back({0,b});
                }else{
                    v[a].push_back({1,b});
                }
            }
            else if(x==2){
                if(z==i && y==j-1){
                    v[a].push_back({0,b});
                }else{
                    v[a].push_back({1,b});
                }
            }
            else if(x==3){
                if(z==i+1 && y==j){
                    v[a].push_back({0,b});
                }else{
                    v[a].push_back({1,b});
                }
            }
            else if(x==4){
                if(z==i-1 && y==j){
                    v[a].push_back({0,b});
                }else{
                    v[a].push_back({1,b});
                }
            }
            
        }
    }

    int minCost(vector<vector<int>>& grid) {
        int n=grid.size();
        int m=grid[0].size();
        vector<vector<pair<int,int>>>v(n*m+1);
        vector<int>dist(n*m+1,INT_MAX);
        vector<int>dr={1,0,-1,0};
        vector<int>dc={0,1,0,-1};


        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                int a=helper1(i,j,n,m);
                if(grid[i][j]==1){
                    helper2(v,dr,dc,a,1,i,j,n,m);
                }
                else if(grid[i][j]==2){
                    helper2(v,dr,dc,a,2,i,j,n,m);
                }
                else if(grid[i][j]==3){
                    helper2(v,dr,dc,a,3,i,j,n,m);

                }else{
                    helper2(v,dr,dc,a,4,i,j,n,m);

                }
            }
        }


        priority_queue<pair<int,int>,vector<pair<int,int>>,greater<pair<int,int>>>pq;

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

        while(pq.size()>0){
            int node=pq.top().second;
            int cost=pq.top().first;
            pq.pop();
            if(node==n*m-1)return cost;
            if(dist[node]<cost)continue;
            dist[node]=cost;
            vis[node]=1;
            for(int j=0;j<v[node].size();j++){
                if(vis[v[node][j].second]==1)continue;
                if(dist[v[node][j].second]>(cost+v[node][j].first)){
                    dist[v[node][j].second]=(cost+v[node][j].first);
                    pq.push({cost+v[node][j].first,v[node][j].second});
                }
            }
        }
        
        return dist[n*m-1];

    }
};
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