<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 minCost(vector<vector<int>>& grid) {
            int n = grid.size();
            int m = grid[0].size();
            // store the cells by converting them to single number 
            deque<pair<int, int>> q{{0, 0}};
            unordered_set<int> vis;
            int mincost = 0;
            vector<pair<int, int>> directions = {{0, 1}, {0, -1}, {1, 0}, {-1, 0
            }};
            while(!q.empty()) {
                auto t = q.front();
                q.pop_front();

                int current_row = (t.first)/m;
                int current_col = (t.first)%m;

                // if we havent visited this cell, then this distance willl be the shortest distance to that cell
                if(vis.insert(t.first).second) {
                    mincost = t.second;
                } 

                // if the last cell is reached then this will be the minimum cost so return the cell 
                if(current_row == n-1 && current_col == m-1) {
                    return mincost;
                } 

                // traverse the four possiblities 
                for(auto it:directions) {
                    int nrow = current_row + it.first;
                    int ncol = current_col + it.second;
                    int posvalue = nrow * m + ncol;
                    // now see if continuing with the same direction will take us to the destination 
                    if(nrow < 0 ||  nrow >= n || ncol < 0 || ncol >= m || vis.count(posvalue)) continue;
                    int cost;
                    if(grid[current_row][current_col] == 1 && it.first == 0 && it.second == 1) cost = 0;
                    else if(grid[current_row][current_col] == 2 && it.first == 0 && it.second == -1) cost = 0;
                    else if(grid[current_row][current_col] == 3 && it.first == 1 && it.second == 0) cost = 0;
                    else if(grid[current_row][current_col] == 4 && it.first == -1 && it.second == 0) cost = 0;
                    else cost = 1;

                    // if cost is 1 then it implies we need to change the direction so insert it at the back 
                    if(cost == 1) {
                        q.push_back({posvalue, t.second + cost});
                    }
                    else {
                        // insert at the front to move to the next cell 
                        q.push_front({posvalue, t.second + cost});
                    }

                }

            }
            return mincost;
        }
    };
25998