<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 {

    int n,m;
    bool valid (int i,int j){
        if( i>=0 && i<n && j>=0 && j<m) return true;
        return false;
    }
public:
    int minCost(vector<vector<int>>& v1) {
        n = v1.size(); m  =v1[0].size();
        vector<vector<int>> dis(n + 7, vector<int>(m, 1e9 + 7));
         vector<vector<int>> vis(n + 7, vector<int>(m, 0));
        set<pair<int, pair<int, int>>> st;
        st.insert({0, {0, 0}});
        dis[0][0] = 0;
        while (!st.empty()) {
            auto it = *st.begin();
            int curr_dis = it.first;
           
            int a = it.second.first;
            int b = it.second.second;
             st.erase(st.begin());
            // vis[a][b]=1;

            
             if (valid(a, b + 1) && v1[a][b] != 1  && vis[a][b+1]  != 1 ) {
                if (dis[a][b + 1] > curr_dis + 1) {
                    dis[a][b + 1] = curr_dis + 1;
                    st.insert({dis[a][b + 1], {a, b + 1}});
                }
             } else if (valid(a, b + 1) && v1[a][b] == 1 && vis[a][b+1]  != 1) {
                 if (dis[a][b + 1] > curr_dis) {
                     dis[a][b + 1] = curr_dis;
                    st.insert({dis[a][b + 1], {a, b+1}});
                 }
             }

             if (valid(a, b - 1) && v1[a][b] != 2 && vis[a][b-1]  != 1) {
                if (dis[a][b - 1] > curr_dis + 1) {
                    dis[a][b - 1] = curr_dis + 1;
                    st.insert({dis[a][b - 1], {a, b - 1}});
                }
             } else if (valid(a, b - 1) && v1[a][b] == 2  && vis[a][b-1]  != 1) {
                 if (dis[a][b - 1] > curr_dis) {
                     dis[a][b - 1] = curr_dis;
                    st.insert({dis[a][b - 1], {a, b-1}});
                 }
             }

             if (valid(a+1, b ) && v1[a][b] != 3  && vis[a+1][b]  != 1) {
                if (dis[a+1][b] > curr_dis + 1) {
                    dis[a+1][b] = curr_dis + 1;
                    st.insert({dis[a+1][b], {a+1, b}});
                }
             } else if (valid(a+1, b ) && v1[a][b] == 3  && vis[a+1][b]  != 1) {
                 if (dis[a+1][b ] > curr_dis) {
                     dis[a+1][b ] = curr_dis;
                    st.insert({dis[a+1][b ], {a+1, b}});
                 }
             }

             if (valid(a-1, b) && v1[a][b] != 4 && vis[a-1][b]  != 1) {
                if (dis[a-1][b ] > curr_dis + 1) {
                    dis[a-1][b ] = curr_dis + 1;
                    st.insert({dis[a-1][b], {a-1, b }});
                }
             } else if (valid(a-1, b ) && v1[a][b] == 4 && vis[a-1][b]  != 1) {
                 if (dis[a-1][b ] > curr_dis) {
                     dis[a-1][b ] = curr_dis;
                    st.insert({dis[a-1][b ], {a-1, b }});
                 }
             }
           
        }

        return dis[n-1][m-1];
    }
};
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