<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 rows ,cols  ; 
    int directions[4][2] = {{0,1} , {0,-1}, {1,0}, {-1, 0}} ; 
    bool withinBounds( int x , int y) { 
        return x >=0 && y>= 0 && x < rows && y < cols ; 
    }
    vector<vector<int>> visited  ; 
public:
    int minCost(vector<vector<int>>& grid) {
        this->rows = grid.size(); 
        this->cols = grid[0].size() ; 
        visited.resize(rows, vector<int>(cols , 0)) ;

        deque<vector<int>> bfs  ; 

        bfs.push_back({0 , 0, 0}) ; 
        int cost  = 0 ; 
        while(!bfs.empty()) { 
            auto top = bfs.front() ; 
            bfs.pop_front() ; 

            if( top[0] == rows-1 && top[1] == cols-1) return top[2]  ; 

            visited[top[0]][top[1]] = 1 ; 


            for( int index = 0 ; index < 4 ; index++) { 
                int nx = top[0] + directions[index][0] , ny  = top[1] + directions[index][1] ; 
                int weight = 1 ; 
                if( index == grid[top[0]][top[1]]-1 ) {
                    weight  = 0 ; 
                }
                if( withinBounds(nx , ny) && !visited[nx][ny]) {
                    if( !weight) {
                        bfs.push_front({nx,ny,0 + top[2]}) ; 
                    }else{
                        bfs.push_back({nx,ny,1 + top[2]});
                    }
                    //visited[nx][ny] = visited[top[0]][top[1]] + weight ; 
                }
            }

        }

        return 0 ; 
    }
};
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